arXiv:1001.0030v2 [math.CO] 28 Feb 2012CYCLIC SIEVING FOR GENERALISED NON-CROSSING PARTITIONS ASSOCIATED WITH COMPLEX REFLECTION GROUPS OF EXCEPTIONAL TYPE — THE DETAILS Christian Krattenthaler†andThomas W. M ¨uller‡ †Fakult¨ at f¨ ur Mathematik, Universit¨ at Wien, Nordbergstraße 15, A-1090 Vienna, Austria. WWW:http://www.mat.univie.ac.at/ ~kratt ‡School of Mathematical Sciences, Queen Mary & Westfield College, University of London, Mile End Road, London E1 4NS, United Kingdom. WWW:http://www.maths.qmw.ac.uk/ ~twm/ Dedicated to the memory of Herb Wilf Abstract. We prove that the generalised non-crossing partitions associated with well-generated complex reflection groups of exceptional type obe y two different cyclic sieving phenomena, as conjectured by Armstrong, and by Bessis a nd Reiner. This manuscript accompanies the paper “Cyclic sieving for generalised non-crossing parti- tions associated with complex reflection groups of exceptio nal type” [arχiv:1001.0028 ], for which it provides the computational details. 1.Introduction In his memoir [2], Armstrong introduced generalised non-crossing partitions asso- ciated with finite (real) reflection groups, thereby embedding Krew eras’ non-crossing partitions [22], Edelman’s m-divisible non-crossing partitions [12], thenon-crossing par- titions associated with reflection groups due to Bessis [6] and Brady and Watt [10] into one uniform framework. Bessis and Reiner [9] observed that Arms trong’s definition can be straightforwardly extended to well-generated complex reflection groups (see Section 2 for the precise definition). These generalised non-crossing partit ions possess a wealth of beautiful properties, and they display deep and surprising relat ions to other combi- natorial objects defined for reflection groups (such as the gene ralised cluster complex 2000Mathematics Subject Classification. Primary 05E15; Secondary 05A10 05A15 05A18 06A07 20F55. Key words and phrases. complex reflection groups, unitary reflection groups, m-divisible non- crossing partitions, generalised non-crossing partitions, Fuß–Ca talan numbers, cyclic sieving. †Research partially supported by the Austrian Science Foundation F WF, grants Z130-N13 and S9607-N13, the latter in the framework of the National Research Network “Analytic Combinatorics and Probabilistic Number Theory.” ‡Research supported by the Austrian Science Foundation FWF, Lise Meitner grant M1201-N13. 12 C. KRATTENTHALER AND T. W. M ¨ULLER of Fomin and Reading [13], or the extended Shi arrangement and the geometric multi- chains of filters of Athanasiadis [4, 5]); see Armstrong’s memoir [2] and the references given therein. Ontheotherhand, cyclic sieving isaphenomenonbroughttolightbyReiner, Stanton and White [30]. It extends the so-called “( −1)-phenomenon” of Stembridge [34, 35]. Cyclic sieving can be defined in three equivalent ways (cf. [30, Prop. 2.1]). The one which gives the name can be described as follows: given a set Sof combinatorial objects, an action on Sof a cyclic group G=/an}bracketle{tg/an}bracketri}htwith generator gof ordern, and a polynomial P(q) inqwith non-negative integer coefficients, we say that the triple (S,P,G)exhibits the cyclic sieving phenomenon , if the number of elements of Sfixed bygkequalsP(e2πik/n). In [30] it is shown that this phenomenon occurs in surprisingly many contexts, and several further instances have been discov ered since then. In [2, Conj. 5.4.7] (also appearing in [9, Conj. 6.4]) and [9, Conj. 6.5], Ar mstrong, respectively Bessis and Reiner, conjecture that generalised non- crossing partitions for irreducible well-generated complex reflection groups exhibit two diffe rent cyclic sieving phenomena (see Sections 3 and 7 for the precise statements). According to the classification of these groups due to Shephard an d Todd [32], there are two infinite families of irreducible well-generated complex reflectio n groups, namely the groups G(d,1,n) andG(e,e,n), wheren,d,eare positive integers, and there are 26 exceptional groups. For the infinite families of types G(d,1,n) andG(e,e,n), the two cyclic sieving conjectures follow from the results in [19]. The purpose of the present article is to prove the cyclic sieving conj ectures of Arm- strong, and of Bessis and Reiner, for the 26 exceptional types, t hus completing the proof of these conjectures. Since the generalised non-crossing partitions feature a pa- rameterm, from the outset this is nota finite problem. Consequently, we first need several auxiliary results to reduce the conjectures for each of t he 26 exceptional types to afiniteproblem. Subsequently, we use Stembridge’s Maplepackagecoxeter [36] and theGAPpackageCHEVIE[14, 28] to carry out the remaining finitecomputations. It is interesting to observe that, for the verification of the type E8case, it is essential to use the decomposition numbers in the sense of [17, 18, 20] becau se, otherwise, the necessary computations would not be feasible in reasonable time with the currently available computer facilities. We point out that, for the special case where the afore- mentioned parameter mis equal to 1, the first cyclic sieving conjecture has been proved in a uniform fashion by Bessis and Reiner in [9]. (See [3] for a — non-unifo rm — proof of cyclic sieving for non-crossing partitions associated with realreflection groups under the action of the so-called Kreweras map, a special case of the sec ond cyclic sieving phenomenon discussed in the present paper.) The crucial result on which the proof of Bessis andReiner is based is(5.5) below, andit plays animportant role in our reduction of the conjectures for the 26 exceptional groups to a finite prob lem. Our paper is organised as follows. In the next section, we recall the definition of generalised non-crossing partitions for well-generated complex re flection groups and of decomposition numbers in the sense of [17, 18, 20], and we review so me basic facts. The first cyclic sieving conjecture is subsequently stated in Section 3. In Section 4, we outline an elementary proof that the q-Fuß–Catalan number, which is the polynomial Pin the cyclic sieving phenomena concerning the generalised non-cros sing partitions for well-generated complex reflection groups, is always a polynomial with non-negative integer coefficients, asrequired bythedefinitionofcyclic sieving. (T he readerisreferredCYCLIC SIEVING FOR GENERALISED NON-CROSSING PARTITIONS 3 to the first paragraph of Section 4 for comments on other approa ches for establishing polynomiality with non-negative coefficients.) Section 5 contains the a nnounced auxil- iary results which, for the 26 exceptional types, allow a reduction o f the conjecture to a finite problem. The remaining case-by-case verification of the conj ecture is then carried out in Section 6. The second cyclic sieving conjecture is stated in Sec tion 7. Section 8 contains the auxiliary results which, for the 26 exceptional types, allow a reduction of the conjecture to a finite problem, while Section 9 contains the rema ining case-by-case verification of the conjecture. 2.Preliminaries Acomplex reflection group isa groupgeneratedby(complex) reflections in Cn. (Here, a reflection is a non-trivial element of GLn(C) which fixes a hyperplane pointwise and which hasfiniteorder.) Wereferto[24]foranin-depthexpositionof thetheorycomplex reflection groups. Shephard and Todd provided a complete classification of all finitecomplex reflection groups in [32] (see also [24, Ch. 8]). According to this classification, a n arbitrary complex reflection group Wdecomposes into a direct product of irreducible complex reflection groups, acting on mutually orthogonal subspaces of th e complex vector space onwhichWisacting. Moreover, thelistofirreduciblecomplexreflectiongroups consists of the infinite family of groups G(m,p,n), wherem,p,nare positive integers, and 34 exceptional groups, denoted G4,G5,...,G 37by Shephard and Todd. In this paper, we are only interested in finite complex reflection grou ps which are well-generated . A complex reflection group of rank nis called well-generated if it is generated by nreflections.1Well-generation can be equivalently characterised by a duality property due to Orlik and Solomon [29]. Namely, a complex reflec tion group of ranknhastwo sets ofdistinguished integers d1≤d2≤ ··· ≤dnandd∗ 1≥d∗ 2≥ ··· ≥d∗ n, called its degreesandcodegrees , respectively (see [24, p. 51 and Def. 10.27]). Orlik and Solomon observed, using case-by-case checking, that an irreduc ible complex reflection groupWof ranknis well-generated if and only if its degrees and codegrees satisfy di+d∗ i=dn for alli= 1,2,...,n. The reader is referred to [24, App. D.2] for a table of the degree s and codegrees of all irreducible complex reflection groups. Togeth er with the classi- fication of Shephard and Todd [32], this constitutes a classification o f well-generated complex reflection groups: the irreducible well-generated complex r eflection groups are — the two infinite families G(d,1,n) andG(e,e,n), whered,e,nare positive inte- gers, — the exceptional groups G4,G5,G6,G8,G9,G10,G14,G16,G17,G18,G20,G21of rank 2, — the exceptional groups G23=H3,G24,G25,G26,G27of rank 3, — the exceptional groups G28=F4,G29,G30=H4,G32of rank 4, — the exceptional group G33of rank 5, — the exceptional groups G34,G35=E6of rank 6, — the exceptional group G36=E7of rank 7, 1We refer to [24, Def. 1.29] for the precise definition of “rank.” Roug hly speaking, the rank of a complex reflection group Wis the minimal nsuch that Wcan be realized as reflection group on Cn.4 C. KRATTENTHALER AND T. W. M ¨ULLER — and the exceptional group G37=E8of rank 8. In this list, we have made visible the groups H3,F4,H4,E6,E7,E8which appear as exceptional groups in the classification of all irreducible realreflection groups (cf. [16]). LetWbe a well-generated complex reflection group of rank n, and letT⊆Wdenote theset of all(complex) reflections inthegroup. Let ℓT:W→Zdenotethewordlength in terms of the generators T. This word length is called absolute length orreflection length. Furthermore, we define a partial order ≤TonWby u≤Twif and only if ℓT(w) =ℓT(u)+ℓT(u−1w). (2.1) This partial order is called absolute order orreflection order . As is well-known and easy to see, the equation in (2.1) is equivalent to the statement tha t every shortest representation of uby reflections occurs as an initial segment in some shortest produc t representation of wby reflections. Now fix a (generalised) Coxeter element2c∈Wand a positive integer m. The m-divisible non-crossing partitions NCm(W) are defined as the set NCm(W) =/braceleftbig (w0;w1,...,w m) :w0w1···wm=cand ℓT(w0)+ℓT(w1)+···+ℓT(wm) =ℓT(c)/bracerightbig . A partial order is defined on this set by (w0;w1,...,w m)≤(u0;u1,...,u m) if and only if ui≤Twifor 1≤i≤m. We have suppressed the dependence on c, since we understand this definition up to isomorphism of posets. To be more precise, it can be shown that any two Coxeter elements are related to each other by conjugation and (possibly) a n automorphism on the field of complex numbers (see [33, Theorem 4.2] or [24, Cor. 11.2 5]), and hence the resulting posets NCm(W) are isomorphic to each other. If m= 1, thenNC1(W) can be identified with the set NC(W) of non-crossing partitions for the (complex) reflection groupWasdefined byBessis andCorran(cf.[8]and[7, Sec.13]; theirdefinit ionextends the earlier definition by Bessis [6] and Brady and Watt [10] for real r eflection groups). The following result has been proved by a collaborative effort of seve ral authors (see [7, Prop. 13.1]). Theorem 1. LetWbe an irreducible well-generated complex reflection group, and let d1≤d2≤ ··· ≤dnbe its degrees and h:=dnits Coxeter number. Then |NCm(W)|=n/productdisplay i=1mh+di di. (2.2) 2An element of an irreducible well-generated complex reflection group Wof ranknis called a Coxeter element if it isregularin the sense of Springer [33] (see also [24, Def. 11.21]) and of order dn. An element of Wis called regular if it has an eigenvector which lies in no reflecting hyperp lane of a reflection of W. It follows from an observation of Lehrer and Springer, proved un iformly by Lehrer and Michel [23] (see [24, Theorem 11.28]), that there is always a regu lar element of order dnin an irreducible well-generated complex reflection group Wof rankn. More generally, if a well-generated complex reflection group Wdecomposes as W∼=W1×W2×···×Wk, where the Wi’s are irreducible, then a Coxeter element of Wis an element of the form c=c1c2···ck, whereciis a Coxeter element of Wi,i= 1,2,...,k. IfWis arealreflection group, that is, if all generators in Thave order 2, then the notion of generalised Coxeter element given above reduces to that of a Coxeter element in the classical sense (cf. [16, Sec. 3.16]).CYCLIC SIEVING FOR GENERALISED NON-CROSSING PARTITIONS 5 Remark1.(1) The number in (2.2) is called the Fuß–Catalan number for the reflection groupW. (2) Ifcis a Coxeter element of a well-generated complex reflection group Wof rank n, thenℓT(c) =n. (This follows from [7, Sec. 7].) We conclude this section by recalling the definition of decomposition nu mbers from [17, 18, 20]. Although we need them here only for (very small) real re flection groups, and although, strictly speaking, they have been only defined for re al reflection groups in [17, 18, 20], this definition can be extended to well-generated comple x reflection groups without any extra effort, which we do now. Given a well-generated complex reflection group Wof rankn, typesT1,T2,...,T d(in the sense of the classification of well-generated complex reflection groups) such that the sumoftheranksofthe Ti’sequalsn, andaCoxeter element c, thedecompositionnumber NW(T1,T2,...,T d) is defined as the number of “minimal” factorisations c=c1c2···cd, “minimal” meaning that ℓT(c1) +ℓT(c2) +···+ℓT(cd) =ℓT(c) =n, such that, for i= 1,2,...,d, the type of cias a parabolic Coxeter element is Ti. (Here, the term “parabolic Coxeter element” means a Coxeter element in some parab olic subgroup. It follows from [31, Prop.6.3] that any element ciis indeed a Coxeter element in a unique parabolic subgroup of W.3By definition, the type of ciis the type of this parabolic subgroup.) Since any two Coxeter elements are related to each oth er by conjugation plus field automorphism, the decomposition numbers are independen t of the choice of the Coxeter element c. The decomposition numbers for real reflection groups have been c omputed in [17, 18, 20]. To compute the decomposition numbers for well-generated complex reflection groups is a task that remains to be done. 3.Cyclic sieving I In this section we present the first cyclic sieving conjecture due to Armstrong [2, Conj. 5.4.7], and to Bessis and Reiner [9, Conj. 6.4]. Letφ:NCm(W)→NCm(W) be the map defined by (w0;w1,...,w m)/mapsto→/parenleftbig (cwmc−1)w0(cwmc−1)−1;cwmc−1,w1,w2,...,w m−1/parenrightbig .(3.1) It is indeed not difficult to see that, if the ( m+ 1)-tuple on the left-hand side is an element ofNCm(W), then so is the ( m+1)-tuple on the right-hand side. For m= 1, this action reduces to conjugation by the Coxeter element c(applied to w1). Cyclic sieving arising from conjugation by chas been the subject of [9]. It is easy to see that φmhacts as the identity, where his the Coxeter number of W (see (5.1) and Lemma 29 below). By slight abuse of notation, let C1be the cyclic group of ordermhgenerated by φ. (The slight abuse consists in the fact that we insist on C1 to be a cyclic group of order mh, while it may happen that the order of the action of φgiven in (3.1) is actually a proper divisor of mh.) 3The uniqueness can be argued as follows: suppose that ciwere a Coxeter element in two parabolic subgroups of W, sayU1andU2. Then it must also be a Coxeter element in the intersection U1∩U2. On the other hand, the absolute length of a Coxeter element of a co mplex reflection group Uis always equal to rk( U), the rank of U. (This follows from the fact that, for each element uofU, we have ℓT(u) = codim/parenleftbig ker(u−id)/parenrightbig , with id denoting the identity element in U; see e.g. [31, Prop. 1.3]). We conclude that ℓT(ci) = rk(U1) = rk(U2) = rk(U1∩U2), This implies that U1=U2.6 C. KRATTENTHALER AND T. W. M ¨ULLER Given these definitions, we are now in the position to state the first c yclic sieving conjecture of Armstrong, respectively of Bessis and Reiner. By t he results of [19] and of this paper, it becomes the following theorem. Theorem 2. For an irreducible well-generated complex reflection group Wand any m≥1, the triple (NCm(W),Catm(W;q),C1), whereCatm(W;q)is theq-analogue of the Fuß–Catalan number defined by Catm(W;q) :=n/productdisplay i=1[mh+di]q [di]q, (3.2) exhibits the cyclic sieving phenomenon in the sense of Reine r, Stanton and White [30]. Here,nis the rank of W,d1,d2,...,d nare the degrees of W,his the Coxeter number ofW, and[α]q:= (1−qα)/(1−q). Remark2.We write Catm(W) for Catm(W;1). By definition of the cyclic sieving phenomenon, we have to prove that Catm(W;q) is a polynomial in qwith non-negative integer coefficients, and that |FixNCm(W)(φp)|= Catm(W;q)/vextendsingle/vextendsingle q=e2πip/mh, (3.3) for allpin the range 0 ≤p0. We begin with several auxiliary results.CYCLIC SIEVING FOR GENERALISED NON-CROSSING PARTITIONS 7 Proposition 3. For all non-negative integers nandk, theq-binomial coefficient [n k]q is a polynomial in qwith non-negative integer coefficients. Proof.This is a well-known fact, which can be derived either from the recurr ence rela- tion(s) satisfied by the q-binomial coefficients (generalising Pascal’s recurrence relation for binomial coefficients; cf. [1, eqs. (3.3.3) and (3.3.4)]), or from th e fact that the q- binomial coefficient [n k]qis the generating function for (integer) partitions with at most kparts all of which are at most n−k(cf. [1, Theorem 3.1]). /square Proposition 4. For all non-negative integers mandn, theq-Fuß–Catalan number of typeAn, 1 [(m+1)n+1]q/bracketleftbigg (m+1)n+1 n/bracketrightbigg q, is a polynomial in qwith non-negative integer coefficients. Proof.In [25, Sec. 3.3], Loehr proves that 1 [(m+1)n+1]q/bracketleftbigg (m+1)n+1 n/bracketrightbigg q =/summationdisplay v∈V(m) nqm(n 2)+/summationtext i≥0(m(vi 2)−ivi)/productdisplay i≥1qvi/summationtextm j=1(m−j)vi−j/bracketleftbigg vi+vi−1+···+vi−m−1 vi/bracketrightbigg q,(4.1) whereV(m) ndenotes the set of all sequences v= (v0,v1,...,v s) (for some s) of non- negative integers with v0>0,vs>0, andv0+v1+···+vs=n, and such that there is never a string of mor more consecutive zeroes in v. By convention, vi= 0 for all negativei. His proof works by showing that the expressions on both sides of ( 4.1) satisfy the same recurrence relation and initial conditions, using cla ssicalq-binomial identities. We refer the reader to [25] for details. By Proposition 3, the expression on the right-hand side of (4.1) is manifestly a polynomial in qwith non-negative integer coefficients. /square Lemma 5. Ifaandbare coprime positive integers, then [ab]q [a]q[b]q(4.2) is a polynomial in qof degree (a−1)(b−1), all of whose coefficients are in {0,1,−1}. Moreover, if one disregards the coefficients which are 0, then+1’s and(−1)’s alternate, and the constant coefficient as well as the leading coefficient o f the polynomial equal +1. Proof.LetΦn(q)denotethe n-thcyclotomicpolynomialin q. Usingtheclassicalformula 1−qn=/productdisplay d|nΦd(q), we see that (1−q)(1−qab) (1−qa)(1−qb)=/productdisplay d1|a,d1/ne}ationslash=1 d2|a,d2/ne}ationslash=1Φd1d2(q), so that, manifestly, the expression in (4.2) is a polynomial in q. The claim concerning the degree of this polynomial is obvious.8 C. KRATTENTHALER AND T. W. M ¨ULLER In order to establish the claim on the coefficients, we start with a sub -expression of (4.2), (1−qab) (1−qa)(1−qb)=/parenleftbiggb−1/summationdisplay i=0qia/parenrightbigg/parenleftbigg∞/summationdisplay j=0qjb/parenrightbigg =∞/summationdisplay k=0Ckqk, (4.3) say. The assumption that aandbare coprime implies that 0 ≤Ck≤1 fork≤ (a−1)(b−1). Multiplying both sides of (4.3) by 1 −q, we obtain the equation [ab]q [a]q[b]q= (1−q)(a−1)(b−1)/summationdisplay k=0Ckqk+(1−q)∞/summationdisplay k=(a−1)(b−1)+1Ckqk. (4.4) By our previous observation on the coefficients Ckwithk≤(a−1)(b−1), it is obvious that the coefficients of the first expression on the right-hand side of (4.4) are alternately +1 and−1, when 0’s are disregarded. Since we already know that the left-ha nd side is a polynomial in qof degree (a−1)(b−1), we may ignore the second expression. The proof is concluded by observing that the claims on the constant and leading coefficients are obvious. /square Corollary 6. Letaandbbe coprime positive integers, and let γbe an integer with γ≥(a−1)(b−1). Then the expression [γ]q[ab]q [a]q[b]q is a polynomial in qwith non-negative integer coefficients. Proof.Let [ab]q [a]q[b]q=(a−1)(b−1)/summationdisplay k=0Dkqk. We then have [γ]q[ab]q [a]q[b]q=(a−1)(b−1)+γ−1/summationdisplay N=0qNN/summationdisplay k=max{0,N−γ+1}Dk. (4.5) IfN≤γ−1, then, by Lemma 5, the sum over kon the right-hand side of (4.5) equals 1−1+1−1+···, which is manifestly non-negative. On the other hand, if N >γ−1, then we may rewrite the sum over kon the right-hand side of (4.5) as N/summationdisplay k=max{0,N−γ+1}Dk=(a−1)(b−1)/summationdisplay k=N−γ+1Dk=(a−1)(b−1)+γ−1−N/summationdisplay k=0D(a−1)(b−1)−k. Again, by Lemma 5, this sum equals 1 −1 + 1−1 +···, which is manifestly non- negative. /square Lemma 7. Letαandβbe positive integers with α≥6andβ≥8. Then the expression [α]q3[β]q4[72]q[3]q[4]q [8]q[9]q[12]q is a polynomial in qwith non-negative integer coefficients.CYCLIC SIEVING FOR GENERALISED NON-CROSSING PARTITIONS 9 Proof.We have [72]q[3]q[4]q [8]q[9]q[12]q = (1−q3+q9−q15+q18)(1−q4+q8−q12+q16−q20+q24−q28+q32). It should be observed that both factors on the right-hand side ha ve the property that coefficients are in {0,1,−1}and that (+1)’s and ( −1)’s alternate, if one disregards the coefficients which are 0. If we now apply the same idea as in the proof o f Corollary 6, then we see that [ α]q3times the first factor is a polynomial in qwith non-negative integer coefficients, as is [ β]q4times the second factor. Taken together, this establishes the claim. /square Lemma 8. Letαandβbe positive integers with α≥26andβ≥8. Then the expression [α]q[β]q4[15]q [3]q[5]q[72]q[3]q[4]q [8]q[9]q[12]q is a polynomial in qwith non-negative integer coefficients. Proof.We have [15]q[72]q[4]q [5]q[8]q[9]q[12]q = (1−q+q5−q6+q9−q11+q12−q13+q14−q15+q17−q20+q21−q25+q26) ×(1−q4+q8−q12+q16−q20+q24−q28+q32). Again, if we apply the same idea as in the proof of Corollary 6, then we s ee that [α]q times the first factor on the the right-hand side of the above equa tion is a polynomial inqwith non-negative integer coefficients, as is [ β]q4times the second factor. Taken together, this establishes the claim. /square Lemma 9. Letαandβbe positive integers with α≥18andβ≥3. Then the expression [α]q3[β]q4[90]q[3]q[4]q [5]q[6]q[9]q is a polynomial in qwith non-negative integer coefficients. Proof.We have [90]q[3]q[4]q [5]q[6]q[9]q = (1−q3+q9−q12+q18−q21+q27−q33+q36−q42+q45−q51+q54) ×(1−q4+q5+q6−q9+q11+q12−q14+q17+q18−q19+q23) = (1−q3+q9−q12+q18−q21+q27−q33+q36−q42+q45−q51+q54) ×/parenleftbig 1−q4+q12+q5(1−q4+q12)+q6(1−q8+q12)+q11(1−q8+q12)/parenrightbig . Again, if we apply the same idea as in the proof of Corollary 6, then we s ee that [α]q3 times the first factor on the the right-hand side of the above equa tion is a polynomial inqwith non-negative integer coefficients, as is [ β]q4times the second factor. Taken together, this establishes the claim. /square10 C. KRATTENTHALER AND T. W. M ¨ULLER Lemma 10. Letαandβbe positive integers with α≥20andβ≥18. Then the expression [α]q[β]q3[90]q[3]q [5]q[6]q[9]q is a polynomial in qwith non-negative integer coefficients. Proof.We have [90]q[3]q [5]q[6]q[9]q = (1−q3+q9−q12+q18−q21+q27−q33+q36−q42+q45−q51+q54) ×(1−q+q5−q7+q10−q13+q15−q19+q20). Again, if we apply the same idea as in the proof of Corollary 6, then we s ee that [α]q3times the second factor on the the right-hand side of the above eq uation is a polynomial in qwith non-negative integer coefficients, as is [ β]q3times the first factor. Taken together, this establishes the claim. /square Lemma 11. Letαbe a positive integer with α≥26. Then the expression [α]q[15]q [3]q[5]q[12]q3 [3]q3[4]q3 is a polynomial in qwith non-negative integer coefficients. Proof.We have [15]q [3]q[5]q[12]q3 [3]q3[4]q3 = 1−q+q5−q6+q9−q11+q12−q13+q14−q15+q17−q20+q21−q25+q26. Once again, the coefficients of the polynomial on the right-hand side are in{0,1,−1} and (+1)’s and ( −1)’s alternate, if one disregards the coefficients which are 0. The argument from the proof of Corollary 6 then implies that this is a polyn omial inqwith non-negative integer coefficients. /square Lemma 12. Letαbe a positive integer with α≥14. Then the expression [α]q[15]q [3]q[5]q[6]q3 [2]q3[3]q3 is a polynomial in qwith non-negative integer coefficients. Proof.We have [15]q [3]q[5]q[6]q3 [2]q3[3]q3= 1−q+q5−q7+q9−q13+q14. Repeating the arguments of the previous proof, this establishes t he claim. /squareCYCLIC SIEVING FOR GENERALISED NON-CROSSING PARTITIONS 11 Lemma 13. Letαandβbe positive integers with α≥30andβ≥20. Then the expression [α]q[β]q2[84]q[2]q [4]q[6]q[7]q is a polynomial in qwith non-negative integer coefficients. Proof.We have [84]q[2]q [4]q[6]q[7]q = (1−q+q6−q8+q12−q15+q18−q22+q24−q29+q30) ×(1−q2+q4−q6+q8−q10+q12−q14+q16−q18+q20−q22 +q24−q26+q28−q30+q32−q34+q36−q38+q40). Onceagain, ifweapplythesameideaasintheproofofCorollary6, the nweseethat[ α]q times the first factor on the the right-hand side of the above equa tion is a polynomial inqwith non-negative integer coefficients, as is [ β]q2times the second factor. Taken together, this establishes the claim. /square Lemma 14. Letαandβbe positive integers with α≥24andβ≥68. Then the expression [α]q[β]q[105]q [3]q[5]q[7]q is a polynomial in qwith non-negative integer coefficients. Proof.We have [105]q [3]q[5]q[7]q = (1−q+q5−q6+q7−q8+q10−q11+q12−q13+q14−q16 +q17−q18+q19−q23+q24) ×(1−q+q3−q4+q6−q7+q9−q10+q12−q13+q15−q16+q18−q19 +q21−q22+q24−q25+q27−q28+q30−q31+q33−q34+q35−q37 +q38−q40+q41−q43+q44−q46+q47−q49+q50−q52+q53−q55 +q56−q58+q59−q61+q62−q64+q65−q67+q68). Once again, if we apply the same idea as in the proof of Corollary 6, the n we see that [α]qtimes the first factor on the the right-hand side of the above equa tion is a polynomial in qwith non-negative integer coefficients, as is [ β]qtimes the second factor. Taken together, this establishes the claim. /square Lemma 15. Letαandβbe positive integers with α≥24andβ≥34. Then the expression [α]q[β]q[70]q [2]q[5]q[7]q is a polynomial in qwith non-negative integer coefficients.12 C. KRATTENTHALER AND T. W. M ¨ULLER Proof.We have [70]q [2]q[5]q[7]q = (1−q+q5−q6+q7−q8+q10−q11+q12−q13 +q14−q16+q17−q18+q19−q23+q24) ×(1−q+q2−q3+q4−q5+q6−q7+q8−q9+q10−q11+q12−q13 +q14−q15+q16−q17+q18−q19+q20−q21+q22−q23+q24−q25 +q26−q27+q28−q29+q30−q31+q32−q33+q34). Also here, if we apply the same idea as in the proof of Corollary 6, then we see that [ α]q times the first factor on the the right-hand side of the above equa tion is a polynomial inqwith non-negative integer coefficients, as is [ β]qtimes the second factor. Taken together, this establishes the claim. /square Lemma 16. Letαandβbe positive integers with α≥4andβ≥2. Then the expression [α]q2[β]q5[30]q[2]q[3]q[5]q [6]q[10]q[15]q is a polynomial in qwith non-negative integer coefficients. Proof.We have [30]q[2]q[3]q[5]q [6]q[10]q[15]q= 1+q−q3−q4−q5+q7+q8. If we multiply this expression by [ α]q2, then, forα= 4 we obtain 1+q+q2−q5−q9+q12+q13+q14, forα= 5 we obtain 1+q+q2−q5+q8−q11+q14+q15+q16, and, forα≥6, we obtain 1+q+q2−q5+q8+q10+p1(q)+q2α−4+q2α−2−q2α+1+q2α+4+q2α+5+q2α+6, wherep1(q) is a polynomial in qwith non-negative coefficients of order at least 11 and degree at most 2 α−5. In all cases it is obvious that the product of the result and [ β]q5, withβ≥2, is a polynomial in qwith non-negative coefficients. /square Lemma 17. Letαandβbe positive integers with α≥14andβ≥2. Then the expression [α]q[β]q5[14]q [2]q[7]q[30]q[2]q[3]q[5]q [6]q[10]q[15]q is a polynomial in qwith non-negative integer coefficients. Proof.We have [14]q [2]q[7]q[30]q[2]q[3]q[5]q [6]q[10]q[15]q= 1−q3−q5+q6+q7+q8−q9−q11+q14.CYCLIC SIEVING FOR GENERALISED NON-CROSSING PARTITIONS 13 If we multiply this expression by [ α]q, then, forα≥14, we obtain 1+q+q2−q5+q7+2q8+q9+q10+p2(q) +qα+3+qα+4+2qα+5+qα+6−qα+8+qα+11+qα+12+qα+13, wherep2(q) is a polynomial in qwith non-negative coefficients of order at least 11 and degree at most α+2. It is obvious that the product of the result and [ β]q5, withβ≥2, is a polynomial in qwith non-negative coefficients. /square Lemma 18. Letαandβbe positive integers with α≥32andβ≥12. Then the expression [α]q[β]q2[35]q [5]q[7]q[30]q[2]q[3]q[5]q [6]q[10]q[15]q is a polynomial in qwith non-negative integer coefficients. Proof.We have [35]q [5]q[7]q[30]q[2]q[3]q[5]q [6]q[10]q[15]q= 1−q2−q3+q5+q6+q7−q8−2q9+q11+q12−q16+q20+q21 −2q23−q24+q25+q26+q27−q29−q30+q32. If we multiply this expression by [ α]q, then, forα≥32, we obtain 1+q−q3−q4+q6+2q7+q8−q9−q10+q12+q13+q14+q15+q20+2q21+2q22 −q24+q26+2q27+2q28+q29+p3(q)+qα+2+2qα+4+2qα+5+qα+6−qα+8 +2qα+9+2qα+10+qα+11+qα+16+qα+17+qα+18+qα+19−qα+21−qα+22+qα+23 +2qα+24+qα+25−qα+27−qα+28+qα+30+qα+31, wherep3(q) is a polynomial in qwith non-negative coefficients of order at least 30 and degree at most α+1. It is obvious that the product of the result and [ β]q2, withβ≥12, is a polynomial in qwith non-negative coefficients. /square Lemma 19. Letαandβbe positive integers with α≥16andβ≥2. Then the expression [α]q2[β]q5[60]q[2]q[3]q[5]q [10]q[12]q[15]q is a polynomial in qwith non-negative integer coefficients. Proof.We have [60]q[2]q[3]q[5]q [10]q[12]q[15]q= 1+q−q3−q4−q5−q6+q8+q9+q10+q11+q12−q14−q15−q16 −q17−q18+q20+q21+q22+q23+q24−q26−q27−q28−q29+q31+q32. If we multiply this expression by [ α]q2, then, forα≥16, we obtain 1+q+q2−q5−q6−q7+q10+q11+2q12+q13+q14 −q17−q18−q19+q22+q23+2q24+p4(q)+s4(q),14 C. KRATTENTHALER AND T. W. M ¨ULLER wherep4(q) is a polynomial in qwith non-negative coefficients of order at least 25 and degree at most 2 α+ 5 ands4(q) completes the above expression to a symmetric polynomial. It is obvious that the product of the result and [ β]q5, withβ≥2, is a polynomial in qwith non-negative coefficients. /square Lemma 20. Letαandβbe positive integers with α≥56andβ≥4. Then the expression [α]q[β]q2[35]q [5]q[7]q[60]q[2]q[3]q[5]q [10]q[12]q[15]q is a polynomial in qwith non-negative integer coefficients. Proof.We have [35]q [5]q[7]q[60]q[2]q[3]q[5]q [10]q[12]q[15]q= 1−q2−q3+q5+q7−q9+q12−q13+q15−q16−q18 +q19+q20+q22−2q23+q27−q28+q29−2q33+q34+q36+q37−q38 −q40+q41−q43+q44−q47+q49+q51−q53−q54+q56. If we multiply this expression by [ α]q, then, forα≥56, we obtain 1+q−q3−q4+q7+q8+q12+q15−q18+q20+q21+2q22+q27 +q29+q30+q31+q32−q33+q36+2q37+q38+p5(q)+s5(q), wherep5(q) is a polynomial in qwith non-negative coefficients of order at least 39 and degree at most α+ 16 ands5(q) completes the above expression to a symmetric polynomial. It is obvious that the product of the result and [ β]q2, withβ≥4, is a polynomial in qwith non-negative coefficients. /square Lemma 21. Letαandβbe positive integers with α≥38andβ≥2. Then the expression [α]q[β]q5[14]q [2]q[7]q[60]q[2]q[3]q[5]q [10]q[12]q[15]q is a polynomial in qwith non-negative integer coefficients. Proof.We have [14]q [2]q[7]q[60]q[2]q[3]q[5]q [10]q[12]q[15]q= 1−q3−q5+q7+q8−q13+q19−q25+q30+q31−q33−q35+q38. If we multiply this expression by [ α]q, then, forα≥38, we obtain 1+q+q2−q5−q6+q8+q9+q10+q11+p6(q)+s6(q), wherep6(q) is a polynomial in qwith non-negative coefficients of order at least 12 and degree at most α+ 25 ands6(q) completes the above expression to a symmetric polynomial. It is obvious that the product of the result and [ β]q2, withβ≥2, is a polynomial in qwith non-negative coefficients. /square Lemma 22. Letαandβbe positive integers with α≥30andβ≥26. Then the expression [α]q[β]q3[126]q[3]q [6]q[7]q[9]qCYCLIC SIEVING FOR GENERALISED NON-CROSSING PARTITIONS 15 is a polynomial in qwith non-negative integer coefficients. Proof.We have [126]q[3]q [6]q[7]q[9]q= (1−q+q6−q8+q12−q15+q18−q22+q24−q29+q30) ×(1−q3+q9−q12+q18−q21+q27−q30+q36−q39 +q42−q48+q51−q57+q60−q66+q69−q75+q78). If we apply the same idea as in the proof of Corollary 6, then we see th at [α]qtimes the first factor on the the right-hand side of the above equation is a po lynomial in qwith non-negative integer coefficients, as is [ β]q3times the second factor. Taken together, this establishes the claim. /square Lemma 23. Letαandβbe positive integers with α≥66andβ≥54. Then the expression [α]q[β]q3[252]q[3]q [7]q[9]q[12]q is a polynomial in qwith non-negative integer coefficients. Proof.We have [252]q[3]q [7]q[9]q[12]q = (1−q+q7−q8+q12−q13+q14−q15+q19−q20+q21−q22+q24−q25 +q26−q27+q28−q29+q31−q32+q33−q34+q35−q37+q38 −q39+q40−q41+q42−q44+q45−q46+q47−q51+q52−q53 +q54−q58+q59−q65+q66) ×(1−q3+q9−q12+q18−q21+q27−q30+q36−q39+q45−q48 +q54−q57+q63−q66+q72−q75+q81−q87+q90−q96+q99 −q105+q108−q114+q117−q123+q126−q132+q135−q141 +q144−q150+q153−q159+q162). If we apply the same idea as in the proof of Corollary 6, then we see th at [α]qtimes the first factor on the the right-hand side of the above equation is a po lynomial in qwith non-negative integer coefficients, as is [ β]q3times the second factor. Taken together, this establishes the claim. /square Lemma 24. Letαandβbe positive integers with α≥54andβ≥34. Then the expression [α]q[β]q2[140]q[2]q [4]q[7]q[10]q is a polynomial in qwith non-negative integer coefficients.16 C. KRATTENTHALER AND T. W. M ¨ULLER Proof.We have [140]q[2]q [4]q[7]q[10]q = (1−q+q7−q8+q10−q11+q14−q15+q17−q18+q20−q22+q24−q25 +q27−q29+q30−q32+q34−q36+q37−q39+q40−q43+q44−q46 +q47−q53+q54) ×(1−q2+q4−q6+q8−q10+q12−q14+q16−q18+q20−q22+q24 −q26+q28−q30+q32−q34+q36−q38+q40−q42+q44−q46 +q48−q50+q52−q54+q56−q58+q60−q62+q64−q66+q68), If we apply the same idea as in the proof of Corollary 6, then we see th at [α]qtimes the first factor on the the right-hand side of the above equation is a po lynomial in qwith non-negative integer coefficients, as is [ β]q2times the second factor. Taken together, this establishes the claim. /square We are now ready for the proof of the main result of this section. Theorem 25. For all irreducible well-generated complex reflection grou ps and posi- tive integers m, theq-Fuß–Catalan number Catm(W;q)is a polynomial in qwith non- negative integer coefficients. Proof.First, letW=An. In this case, the degrees are 2 ,3,...,n+1, and hence Catm(An;q) =1 [(m+1)n+1]q/bracketleftbigg (m+1)n+1 n/bracketrightbigg q, which, by Proposition 4, is a polynomial in qwith non-negative integer coefficients. Next, letW=G(d,1,n). In this case, the degrees are d,2d,...,nd , and hence Catm(G(d,1,n);q) =/bracketleftbigg (m+1)n n/bracketrightbigg qd, which, by Proposition 3, is a polynomial in qwith non-negative integer coefficients. Now, letW=G(e,e,n). In this case, the degrees are e,2e,...,(n−1)e,n, and hence Catm(G(e,e,n);q) =[m(n−1)e+n]q [n]qn−1/productdisplay i=1[m(n−1)e+ie]q [ie]q =/bracketleftbigg (m+1)(n−1) n−1/bracketrightbigg qe+qn[e]qn/bracketleftbigg (m+1)(n−1) n/bracketrightbigg qe, which, by Proposition 3, is a polynomial in qwith non-negative integer coefficients. It remains to verify the claim for the exceptional groups. ForW=G4, the degrees are 4 ,6, and hence Catm(G4;q) =[6m+4]q[6m+6]q [4]q[6]q=1 [3m+4]q2/bracketleftbigg 3m+4 3/bracketrightbigg q2, which, by Proposition 4, is a polynomial in qwith non-negative integer coefficients.CYCLIC SIEVING FOR GENERALISED NON-CROSSING PARTITIONS 17 ForW=G5, the degrees are 6 ,12, and hence Catm(G5;q) =[12m+6]q[12m+12]q [6]q[12]q=/bracketleftbigg 2m+2 2/bracketrightbigg q6, which, by Proposition 3, is a polynomial in qwith non-negative integer coefficients. ForW=G6, the degrees are 4 ,12, and hence Catm(G6;q) =[12m+4]q[12m+12]q [4]q[12]q= [3m+1]q4[m+1]q12, which is manifestly a polynomial in qwith non-negative integer coefficients. ForW=G8, the degrees are 8 ,12, and hence Catm(G8;q) =[12m+8]q[12m+12]q [8]q[12]q=1 [3m+4]q4/bracketleftbigg 3m+4 3/bracketrightbigg q4, which, by Proposition 4, is a polynomial in qwith non-negative integer coefficients. ForW=G9, the degrees are 8 ,24, and hence Catm(G9;q) =[24m+8]q[24m+24]q [8]q[24]q= [3m+1]q8[m+1]q24, which is manifestly a polynomial in qwith non-negative integer coefficients. ForW=G10, the degrees are 12 ,24, and hence Catm(G10;q) =[24m+12]q[24m+24]q [12]q[24]q=/bracketleftbigg 2m+2 2/bracketrightbigg q12, which, by Proposition 3, is a polynomial in qwith non-negative integer coefficients. ForW=G14, the degrees are 6 ,24, and hence Catm(G14;q) =[24m+6]q[24m+24]q [6]q[24]q= [4m+1]q6[m+1]q24, which is manifestly a polynomial in qwith non-negative integer coefficients. ForW=G16, the degrees are 20 ,30, and hence Catm(G16;q) =[30m+20]q[30m+30]q [20]q[30]q=1 [3m+4]q10/bracketleftbigg 3m+4 3/bracketrightbigg q10, which, by Proposition 4, is a polynomial in qwith non-negative integer coefficients. ForW=G17, the degrees are 20 ,60, and hence Catm(G17;q) =[60m+20]q[60m+60]q [20]q[60]q= [3m+1]q20[m+1]q60, which is manifestly a polynomial in qwith non-negative integer coefficients. ForW=G18, the degrees are 30 ,60, and hence Catm(G18;q) =[60m+30]q[60m+60]q [30]q[60]q=/bracketleftbigg 2m+2 2/bracketrightbigg q30, which, by Proposition 3, is a polynomial in qwith non-negative integer coefficients.18 C. KRATTENTHALER AND T. W. M ¨ULLER ForW=G20, the degrees are 12 ,30, and hence Catm(G20;q) =[30m+12]q[30m+30]q [12]q[30]q =/braceleftBigg/bracketleftbig5m+2 2/bracketrightbig q12[m+1]q30, ifmis even, [5m+2]q6/bracketleftbigm+1 2/bracketrightbig q60[10]q6 [2]q6[5]q6,ifmis odd, which, by Corollary 6, are polynomials in qwith non-negative integer coefficients in both cases. ForW=G21, the degrees are 12 ,60, and hence Catm(G21;q) =[60m+12]q[60m+60]q [12]q[60]q= [5m+1]q12[m+1]q60, which is manifestly a polynomial in qwith non-negative integer coefficients. ForW=G23=H3, the degrees are 2 ,6,10, and hence Catm(H3;q) =[10m+2]q[10m+6]q[10m+10]q [2]q[6]q[10]q =  [5m+2]q2/bracketleftbig5m+3 3/bracketrightbig q6[m+1]q10, ifm≡0 (mod 3),/bracketleftbig5m+1 3/bracketrightbig q6[5m+3]q2[m+1]q10, ifm≡1 (mod 3), [5m+2]q2[5m+3]q2/bracketleftbigm+1 3/bracketrightbig q30[15]q2 [3]q2[5]q2,ifm≡2 (mod 3), which, by Corollary 6, are polynomials in qwith non-negative integer coefficients in all cases. ForW=G24, the degrees are 4 ,6,14, and hence Catm(G24;q) =[14m+4]q[14m+6]q[14m+14]q [4]q[6]q[14]q. We have Catm(G24;q) =  /bracketleftbig7m 2+1/bracketrightbig q4/bracketleftbig14m 6+1/bracketrightbig q6[m+1]q14,ifm≡0 (mod 6),/bracketleftbig7m+2 3/bracketrightbig q6/bracketleftbig7m+3 2/bracketrightbig q4[m+1]q14, ifm≡1 (mod 6), /bracketleftbig7m 2+1/bracketrightbig q4[7m+3]q2/bracketleftbigm+1 3/bracketrightbig q42[21]q2 [3]q2[7]q2,ifm≡2 (mod 6), [7m+2]q2/bracketleftbig7m 3+1/bracketrightbig q6/bracketleftbigm+1 2/bracketrightbig q28[14]q2 [2]q2[7]q2,ifm≡3 (mod 6), /bracketleftbig7m+2 6/bracketrightbig q12[6]q2 [2]q2[3]q2[7m+3]q2[m+1]q14,ifm≡4 (mod 6), [7m+2]q2/bracketleftbig7m+3 2/bracketrightbig q4/bracketleftbigm+1 3/bracketrightbig q42[21]q2 [3]q2[7]q2,ifm≡5 (mod 6), which, by Corollary 6, are polynomials in qwith non-negative integer coefficients in all cases. ForW=G25, the degrees are 6 ,9,12, and hence Catm(G25;q) =[12m+6]q[12m+9]q[12m+12]q [6]q[9]q[12]q=1 [4m+5]q3/bracketleftbigg 4m+5 4/bracketrightbigg q3, which, by Proposition 4, is a polynomial in qwith non-negative integer coefficients.CYCLIC SIEVING FOR GENERALISED NON-CROSSING PARTITIONS 19 ForW=G26, the degrees are 6 ,12,18, and hence Catm(G26;q) =[18m+6]q[18m+12]q[18m+18]q [6]q[12]q[18]q=/bracketleftbigg 3m+3 3/bracketrightbigg q6, which, by Proposition 3, is a polynomial in qwith non-negative integer coefficients. ForW=G27, the degrees are 6 ,12,30, and hence Catm(G27;q) =[30m+6]q[30m+12]q[30m+30]q [6]q[12]q[30]q= [m+1]q30/bracketleftbigg 5m+2 2/bracketrightbigg q6, which, by Proposition 3, is a polynomial in qwith non-negative integer coefficients. ForW=G28=F4, the degrees are 2 ,6,8,12, and hence Catm(F4;q) =[12m+2]q[12m+6]q[12m+8]q[12m+12]q [2]q[6]q[8]q[12]q =  [6m+1]q2[2m+1]q6/bracketleftbig3 2m+1/bracketrightbig q8[m+1]q12, ifmis even, [6m+1]q2[2m+1]q6/bracketleftbigm+1 2/bracketrightbig q24[3m+2]q4[6]q4 [2]q4[3]q4,ifmis odd, which in both cases is a polynomial in qwith non-negative integer coefficients; in the second case this is due to Corollary 6. ForW=G29, the degrees are 4 ,8,12,20, and hence Catm(G29;q) =[20m+4]q[20m+8]q[20m+12]q[20m+20]q [4]q[8]q[12]q[20]q = [m+1]q20/bracketleftbigg 5m+3 3/bracketrightbigg q4, which, by Proposition 3, is a polynomial in qwith non-negative integer coefficients. ForW=G30=H4, the degrees are 2 ,12,20,30, and hence Catm(H4;q) =[30m+2]q[30m+12]q[30m+20]q[30m+30]q [2]q[12]q[20]q[30]q. Ifmis even, then we have Catm(H4;q) = [15m+1]q2/bracketleftbig5 2m+1/bracketrightbig q12/bracketleftbig3 2m+1/bracketrightbig q20[m+1]q30, which is manifestly a polynomial in qwith non-negative integer coefficients. On the other hand, if mis odd, then we may write Catm(H4;q) =/bracketleftbig15m+1 2/bracketrightbig q4[5m+2]q6[3m+2]q10/bracketleftbigm+1 2/bracketrightbig q60[30]q2[2]q2[3]q2[5]q2 [6]q6[10]q2[15]q2, which, by Lemma 16, is a polynomial in qwith non-negative integer coefficients. ForW=G32, the degrees are 12 ,18,24,30, and hence Catm(G32;q) =[30m+12]q[30m+18]q[30m+24]q[30m+30]q [12]q[18]q[24]q[30]q =1 [5m+6]q6/bracketleftbigg 5m+6 5/bracketrightbigg q6, which, by Proposition 4, is a polynomial in qwith non-negative integer coefficients.20 C. KRATTENTHALER AND T. W. M ¨ULLER ForW=G33, the degrees are 4 ,6,10,12,18, and hence Catm(G33;q) =[18m+4]q[18m+6]q[18m+10]q[18m+12]q[18m+18]q [4]q[6]q[10]q[12]q[18]q. Ifm≡0 (mod 10), then we have Catm(G33;q) =/bracketleftbig9 2m+1/bracketrightbig q4[3m+1]q6/bracketleftbig9 5m+1/bracketrightbig q10/bracketleftbig3 2m+1/bracketrightbig q12[m+1]q18, which is manifestly a polynomial in qwith non-negative integer coefficients. If m≡ 1 (mod 10), then we have Catm(G33;q) =/bracketleftbig3m+1 2/bracketrightbig q12/bracketleftbig9m+5 2/bracketrightbig q4/bracketleftbig3m+2 5/bracketrightbig q30[m+1]q18[9m+2]q2[15]q2 [3]q2[5]q2, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡2 (mod 10), then we have Catm(G33;q) =/bracketleftbig9m+2 10/bracketrightbig q20[3m+1]q6/bracketleftbig3 2m+1/bracketrightbig q12[m+1]q18[9m+5]q2[10]q2 [2]q2[5]q2, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡3 (mod 10), then we have Catm(G33;q) =/bracketleftbig3m+1 10/bracketrightbig q60/bracketleftbig9m+5 2/bracketrightbig q4[3m+2]q6[m+1]q18[9m+2]q2[30]q2 [5]q2[6]q2, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡4 (mod 10), then we have Catm(G33;q) =/bracketleftbig9 2m+1/bracketrightbig q4[3m+1]q6/bracketleftbig3 2m+1/bracketrightbig q12/bracketleftbigm+1 5/bracketrightbig q90[9m+5]q2[45]q2 [5]q2[9]q2, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡5 (mod 10), then we have Catm(G33;q) =/bracketleftbig3m+1 2/bracketrightbig q12/bracketleftbig9m+5 10/bracketrightbig q20[3m+2]q6[m+1]q18[9m+2]q2[10]q2 [2]q2[5]q2, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡6 (mod 10), then we have Catm(G33;q) =/bracketleftbig9 2m+1/bracketrightbig q4[3m+1]q6/bracketleftbig3m+2 10/bracketrightbig q60[m+1]q18[9m+5]q2[30]q2 [5]q2[6]q2, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡7 (mod 10), then we have Catm(G33;q) =/bracketleftbig9m+2 5/bracketrightbig q10/bracketleftbig3m+1 2/bracketrightbig q12/bracketleftbig9m+5 2/bracketrightbig q4[3m+2]q6[m+1]q18, which is manifestly a polynomial in qwith non-negative integer coefficients. If m≡ 8 (mod 10), then we have Catm(G33;q) =/bracketleftbig9 2m+1/bracketrightbig q4/bracketleftbig3m+1 5/bracketrightbig q30/bracketleftbig3 2m+1/bracketrightbig q12[m+1]q18[9m+5]q2[15]q2 [3]q2[5]q2,CYCLIC SIEVING FOR GENERALISED NON-CROSSING PARTITIONS 21 which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. Fi- nally, ifm≡9 (mod 10), then we have Catm(G33;q) =/bracketleftbig3m+1 2/bracketrightbig q12/bracketleftbig9m+5 2/bracketrightbig q4[3m+2]q6/bracketleftbigm+1 5/bracketrightbig q90[9m+2]q2[45]q2 [5]q2[9]q2, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. ForW=G34, the degrees are 6 ,12,18,24,30,42, and hence Catm(G34;q) =[42m+6]q[42m+12]q[42m+18]q[42m+24]q[42m+30]q[42m+42]q [6]q[12]q[18]q[24]q[30]q[42]q = [m+1]q42/bracketleftbigg 7m+5 5/bracketrightbigg q6, which, by Proposition 3, is a polynomial in qwith non-negative integer coefficients. ForW=G35=E6, the degrees are 2 ,5,6,8,9,12, and hence Catm(E6;q) =[12m+2]q[12m+5]q[12m+6]q[12m+8]q[12m+9]q[12m+12]q [2]q[5]q[6]q[8]q[9]q[12]q. Ifm≡0 (mod 30),then we have Catm(E6;q) = [6m+1]q2/bracketleftbig12m+5 5/bracketrightbig q5[2m+1]q6/bracketleftbig3m+2 2/bracketrightbig q8/bracketleftbig4m+3 3/bracketrightbig q9[m+1]q12, which is manifestly a polynomial in qwith non-negative integer coefficients. If m≡ 1 (mod 30),then we have Catm(E6;q) = [6m+1]q2/bracketleftbig2m+1 3/bracketrightbig q18[4m+3]q3[6]q3 [2]q3[3]q3 ×/bracketleftbig3m+2 5/bracketrightbig q20[12m+5]q[20]q [4]q[5]q/bracketleftbigm+1 2/bracketrightbig q24[6]q4 [2]q4[3]q4. If one decomposes [6 m+1]q2as [3m+1]q4+q2[3m]q4, then one sees that, by Corollary 6, the above expression is a polynomial in qwith non-negative integer coefficients. If m≡2 (mod 30),then we have Catm(E6;q) = [6m+1]q2[12m+5]q/bracketleftbig2m+1 5/bracketrightbig q30[30]q [5]q[6]q ×/bracketleftbig3m+2 2/bracketrightbig q8[4m+3]q3/bracketleftbigm+1 3/bracketrightbig q36[12]q3 [3]q3[4]q3, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡3 (mod 30),then we have Catm(E6;q) = [6m+1]q2[12m+5]q[2m+1]q6[3m+2]q4 ×/bracketleftbig4m+3 15/bracketrightbig q45[45]q [5]q[9]q/bracketleftbigm+1 2/bracketrightbig q24[6]q4 [2]q4[3]q4,22 C. KRATTENTHALER AND T. W. M ¨ULLER which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡4 (mod 30),then we have Catm(E6;q) =/bracketleftbig12m+2 5/bracketrightbig q5/bracketleftbig12m+5 2/bracketrightbig q2/bracketleftbig2m+1 3/bracketrightbig q18[6]q3 [2]q3[3]q3 ×/bracketleftbig3m+2 2/bracketrightbig q8[4m+3]q3[m+1]q12, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡5 (mod 30),then we have Catm(E6;q) = [6m+1]q2/bracketleftbig12m+5 5/bracketrightbig q5[2m+1]q6 ×[3m+2]q4[4m+3]q3/bracketleftbigm+1 6/bracketrightbig q72[72]q[3]q[4]q [8]q[9]q[12]q, which, by Lemma 7, is a polynomial in qwith non-negative integer coefficients. If m≡6 (mod 30),then we have Catm(E6;q) = [6m+1]q2[12m+5]q[2m+1]q6/bracketleftbig3m+2 10/bracketrightbig q40[40]q [5]q[8]q/bracketleftbig4m+3 3/bracketrightbig q9[m+1]q12, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡7 (mod 30),then we have Catm(E6;q) =/bracketleftbig6m+1 2/bracketrightbig q4[12m+5]q/bracketleftbig2m+1 15/bracketrightbig q90 ×[90]q[3]q[4]q [5]q[6]q[9]q[3m+2]q4[4m+3]q3/bracketleftbigm+1 2/bracketrightbig q24[6]q4 [2]q4[3]q4, which, by Corollary 6 and Lemma 9, is a polynomial in qwith non-negative integer coefficients. If m≡8 (mod 30),then we have Catm(E6;q) = [6m+1]q2[12m+5]q[2m+1]q6/bracketleftbig3m+2 2/bracketrightbig q8 ×/bracketleftbig4m+3 5/bracketrightbig q15[15]q [3]q[5]q/bracketleftbigm+1 3/bracketrightbig q36[12]q3 [3]q3[4]q3, which, by Lemma 11, is a polynomial in qwith non-negative integer coefficients. If m≡9 (mod 30),then we have Catm(E6;q) =/bracketleftbig12m+2 5/bracketrightbig q5/bracketleftbig12m+5 2/bracketrightbig q2[2m+1]q6[3m+2]q4/bracketleftbig4m+3 3/bracketrightbig q9/bracketleftbigm+1 2/bracketrightbig q24[6]q4 [2]q4[3]q4, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡10 (mod 30) ,then we have Catm(E6;q) = [6m+1]q2/bracketleftbig12m+5 5/bracketrightbig q5/bracketleftbig2m+1 3/bracketrightbig q18[6]q3 [2]q3[3]q3/bracketleftbig3m+2 2/bracketrightbig q8[4m+3]q3[m+1]q12,CYCLIC SIEVING FOR GENERALISED NON-CROSSING PARTITIONS 23 which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡11 (mod 30) ,then we have Catm(E6;q) = [6m+1]q2[12m+5]q[2m+1]q6 ×/bracketleftbig3m+2 5/bracketrightbig q20[20]q [4]q[5]q[4m+3]q3/bracketleftbigm+1 6/bracketrightbig q72[72]q[3]q[4]q [8]q[9]q[12]q. If one decomposes [6 m+1]q2as [3m+1]q4+q2[3m]q4, then one sees that, by Corollary 6 and Lemma 7, this is a polynomial in qwith non-negative integer coefficients. If m≡ 12 (mod 30) ,then we have Catm(E6;q) = [6m+1]q2[12m+5]q/bracketleftbig2m+1 5/bracketrightbig q30[30]q [5]q[6]q/bracketleftbig3m+2 2/bracketrightbig q8/bracketleftbig4m+3 3/bracketrightbig q9[m+1]q12, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡13 (mod 30) ,then we have Catm(E6;q) = [6m+1]q2[12m+5]q/bracketleftbig2m+1 3/bracketrightbig q18[6]q3 [2]q3[3]q3 ×[3m+2]q4/bracketleftbig4m+3 5/bracketrightbig q15[15]q [3]q[5]q/bracketleftbigm+1 2/bracketrightbig q24[6]q4 [2]q4[3]q4, which, by Lemma 12, is a polynomial in qwith non-negative integer coefficients. If m≡14 (mod 30) ,then we have Catm(E6;q) =/bracketleftbig12m+2 5/bracketrightbig q5/bracketleftbig12m+5 2/bracketrightbig q2[2m+1]q6/bracketleftbig3m+2 2/bracketrightbig q8[4m+3]q3/bracketleftbigm+1 3/bracketrightbig q36[12]q3 [3]q3[4]q3, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡15 (mod 30) ,then we have Catm(E6;q) = [6m+1]q2/bracketleftbig12m+5 5/bracketrightbig q5[2m+1]q6[3m+2]q4/bracketleftbig4m+3 3/bracketrightbig q9/bracketleftbigm+1 2/bracketrightbig q24[6]q4 [2]q4[3]q4, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡16 (mod 30) ,then we have Catm(E6;q) = [6m+1]q2[12m+5]q/bracketleftbig2m+1 3/bracketrightbig q18[6]q3 [2]q3[3]q3 ×/bracketleftbig3m+2 10/bracketrightbig q40[40]q [5]q[8]q[4m+3]q3[m+1]q12, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡17 (mod 30) ,then we have Catm(E6;q) = [6m+1]q2[12m+5]q/bracketleftbig2m+1 5/bracketrightbig q30[30]q [5]q[6]q ×[3m+2]q4[4m+3]q3/bracketleftbigm+1 6/bracketrightbig q72[72]q[3]q[4]q [8]q[9]q[12]q,24 C. KRATTENTHALER AND T. W. M ¨ULLER which, by Corollary 6 and Lemma 7, is a polynomial in qwith non-negative integer coefficients. If m≡18 (mod 30) ,then we have Catm(E6;q) = [6m+1]q2[12m+5]q[2m+1]q6/bracketleftbig3m+2 2/bracketrightbig q8/bracketleftbig4m+3 15/bracketrightbig q45[45]q [5]q[9]q[m+1]q12, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡19 (mod 30) ,then we have Catm(E6;q) =/bracketleftbig12m+2 5/bracketrightbig q5/bracketleftbig12m+5 2/bracketrightbig q2/bracketleftbig2m+1 3/bracketrightbig q18[6]q3 [2]q3[3]q3 ×[3m+2]q4[4m+3]q3/bracketleftbigm+1 2/bracketrightbig q24[6]q4 [2]q4[3]q4, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡20 (mod 30) ,then we have Catm(E6;q) = [6m+1]q2/bracketleftbig12m+5 5/bracketrightbig q5[2m+1]q6/bracketleftbig3m+2 2/bracketrightbig q8[4m+3]q3/bracketleftbigm+1 3/bracketrightbig q36[12]q3 [3]q3[4]q3, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡21 (mod 30) ,then we have Catm(E6;q) = [6m+1]q2[12m+5]q[2m+1]q6 ×/bracketleftbig3m+2 5/bracketrightbig q20[20]q [4]q[5]q/bracketleftbig4m+3 3/bracketrightbig q9/bracketleftbigm+1 2/bracketrightbig q24[6]q4 [2]q4[3]q4, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡22 (mod 30) ,then we have Catm(E6;q) = [6m+1]q2[12m+5]q/bracketleftbig2m+1 15/bracketrightbig q90[90]q[3]q [5]q[6]q[9]q ×/bracketleftbig3m+2 2/bracketrightbig q8[4m+3]q3[m+1]q12, which, by Lemma 10, is a polynomial in qwith non-negative integer coefficients. If m≡23 (mod 30) ,then we have Catm(E6;q) = [6m+1]q2[12m+5]q[2m+1]q6 ×[3m+2]q4/bracketleftbig4m+3 5/bracketrightbig q15[15]q [3]q[5]q/bracketleftbigm+1 6/bracketrightbig q72[72]q[3]q[4]q [8]q[9]q[12]q, which, by Lemma 8, is a polynomial in qwith non-negative integer coefficients. If m≡24 (mod 30) ,then we have Catm(E6;q) =/bracketleftbig12m+2 5/bracketrightbig q5/bracketleftbig12m+5 2/bracketrightbig q2[2m+1]q6/bracketleftbig3m+2 2/bracketrightbig q8/bracketleftbig4m+3 3/bracketrightbig q9[m+1]q12,CYCLIC SIEVING FOR GENERALISED NON-CROSSING PARTITIONS 25 which is manifestly a polynomial in qwith non-negative integer coefficients. If m≡ 25 (mod 30) ,then we have Catm(E6;q) = [6m+1]q2/bracketleftbig12m+5 5/bracketrightbig q5/bracketleftbig2m+1 3/bracketrightbig q18[6]q3 [2]q3[3]q3 ×[3m+2]q4[4m+3]q3/bracketleftbigm+1 2/bracketrightbig q24[6]q4 [2]q4[3]q4, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡26 (mod 30) ,then we have Catm(E6;q) = [6m+1]q2[12m+5]q[2m+1]q6 ×/bracketleftbig3m+2 10/bracketrightbig q40[40]q [5]q[8]q[4m+3]q3/bracketleftbigm+1 3/bracketrightbig q36[12]q3 [3]q3[4]q3, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡27 (mod 30) ,then we have Catm(E6;q) = [6m+1]q2[12m+5]q/bracketleftbig2m+1 5/bracketrightbig q30[30]q [5]q[6]q ×[3m+2]q4/bracketleftbig4m+3 3/bracketrightbig q9/bracketleftbigm+1 2/bracketrightbig q24[6]q4 [2]q4[3]q4, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡28 (mod 30) ,then we have Catm(E6;q) = [6m+1]q2[12m+5]q/bracketleftbig2m+1 3/bracketrightbig q18[6]q3 [2]q3[3]q3 ×/bracketleftbig3m+2 2/bracketrightbig q8/bracketleftbig4m+3 5/bracketrightbig q15[15]q [3]q[5]q[m+1]q12, which, by Lemma 12, is a polynomial in qwith non-negative integer coefficients. If m≡29 (mod 30) ,then we have Catm(E6;q) =/bracketleftbig6m+1 5/bracketrightbig q10[10]q [2]q[5]q[12m+5]q[2m+1]q6 ×[3m+2]q4[4m+3]q3/bracketleftbigm+1 6/bracketrightbig q72[72]q[3]q[4]q [8]q[9]q[12]q, which, by Corollary 6 and Lemma 7, is a polynomial in qwith non-negative integer coefficients. ForW=G36=E7, the degrees are 2 ,6,8,10,12,14,18, and hence Catm(E7;q) =[18m+2]q[18m+6]q[18m+8]q[18m+10]q [2]q[6]q[8]q[10]q ×[18m+12]q[18m+14]q[18m+18]q [12]q[14]q[18]q.26 C. KRATTENTHALER AND T. W. M ¨ULLER Ifm≡0 (mod 140) ,then we have Catm(E7;q) = [9m+1]q2[3m+1]q6/bracketleftbig9m+4 4/bracketrightbig q8/bracketleftbig9m+5 5/bracketrightbig q10/bracketleftbig3m+2 2/bracketrightbig q12/bracketleftbig9m+7 7/bracketrightbig q14[m+1]q18, which is manifestly a polynomial in qwith non-negative integer coefficients. If m≡ 1 (mod 140) ,then we have Catm(E7;q) =/bracketleftbig9m+1 5/bracketrightbig q10/bracketleftbig3m+1 2/bracketrightbig q12[9m+4]q2/bracketleftbig9m+5 7/bracketrightbig q14[3m+2]q6/bracketleftbig9m+7 4/bracketrightbig q8[m+1]q18, which is manifestly a polynomial in qwith non-negative integer coefficients. If m≡ 2 (mod 140) ,then we have Catm(E7;q) = [9m+1]q2/bracketleftbig3m+1 7/bracketrightbig q42[21]q2 [3]q2[7]q2/bracketleftbig9m+4 2/bracketrightbig q4[9m+5]q2 ×/bracketleftbig3m+2 4/bracketrightbig q24[6]q4 [2]q4[3]q4/bracketleftbig9m+7 5/bracketrightbig q10[m+1]q18, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡3 (mod 140) ,then we have Catm(E7;q) =/bracketleftbig9m+1 7/bracketrightbig q14/bracketleftbig3m+1 10/bracketrightbig q60[30]q2 [5]q2[6]q2[9m+4]q2 ×/bracketleftbig9m+5 4/bracketrightbig q8[3m+2]q6[9m+7]q2[m+1]q18, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡4 (mod 140) ,then we have Catm(E7;q) = [9m+1]q2[3m+1]q6/bracketleftbig9m+4 20/bracketrightbig q40[20]q2 [4]q2[5]q2[9m+5]q2 ×/bracketleftbig3m+2 14/bracketrightbig q84[42]q2 [6]q2[7]q2[9m+7]q2[m+1]q18, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡5 (mod 140) ,then we have Catm(E7;q) = [9m+1]q2/bracketleftbig3m+1 2/bracketrightbig q12/bracketleftbig9m+4 7/bracketrightbig q14/bracketleftbig9m+5 5/bracketrightbig q10[3m+2]q6/bracketleftbig9m+7 4/bracketrightbig q8[m+1]q18, which is manifestly a polynomial in qwith non-negative integer coefficients. If m≡ 6 (mod 140) ,then we have Catm(E7;q) =/bracketleftbig9m+1 5/bracketrightbig q10[3m+1]q6/bracketleftbig9m+4 2/bracketrightbig q4[9m+5]q2 ×/bracketleftbig3m+2 4/bracketrightbig q24[6]q4 [2]q4[3]q4[9m+7]q2/bracketleftbigm+1 7/bracketrightbig q126[63]q2 [7]q2[9]q2, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡7 (mod 140) ,then we have Catm(E7;q) =/bracketleftbig9m+1 4/bracketrightbig q8/bracketleftbig3m+1 2/bracketrightbig q12[9m+4]q2 ×[9m+5]q2[3m+2]q6/bracketleftbig9m+7 35/bracketrightbig q70[35]q2 [5]q2[7]q2[m+1]q18,CYCLIC SIEVING FOR GENERALISED NON-CROSSING PARTITIONS 27 which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡8 (mod 140) ,then we have Catm(E7;q) = [9m+1]q2/bracketleftbig3m+1 5/bracketrightbig q30[15]q2 [3]q2[5]q2/bracketleftbig9m+4 4/bracketrightbig q8 ×/bracketleftbig9m+5 7/bracketrightbig q14/bracketleftbig3m+2 2/bracketrightbig q12[9m+7]q2[m+1]q18, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡9 (mod 140) ,then we have Catm(E7;q) = [9m+1]q2/bracketleftbig3m+1 14/bracketrightbig q84[42]q2 [6]q2[7]q2/bracketleftbig9m+4 5/bracketrightbig q10 ×[9m+5]q2[3m+2]q6/bracketleftbig9m+7 4/bracketrightbig q8[m+1]q18, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡10 (mod 140) ,then we have Catm(E7;q) =/bracketleftbig9m+1 7/bracketrightbig q14[3m+1]q6/bracketleftbig9m+4 2/bracketrightbig q4/bracketleftbig9m+5 5/bracketrightbig q10 ×/bracketleftbig3m+2 4/bracketrightbig q24[6]q4 [2]q4[3]q4[9m+7]q2[m+1]q18, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡11 (mod 140) ,then we have Catm(E7;q) =/bracketleftbig9m+1 5/bracketrightbig q10/bracketleftbig3m+1 2/bracketrightbig q12[9m+4]q2/bracketleftbig9m+5 4/bracketrightbig q8 ×/bracketleftbig3m+2 7/bracketrightbig q42[21]q2 [3]q2[7]q2[9m+7]q2[m+1]q18, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡12 (mod 140) ,then we have Catm(E7;q) = [9m+1]q2[3m+1]q6/bracketleftbig9m+4 28/bracketrightbig q56[28]q2 [4]q2[7]q2[9m+5]q2 ×/bracketleftbig3m+2 2/bracketrightbig q12/bracketleftbig9m+7 5/bracketrightbig q10[m+1]q18, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡13 (mod 140) ,then we have Catm(E7;q) = [9m+1]q2/bracketleftbig3m+1 10/bracketrightbig q60[30]q2 [5]q2[6]q2[9m+4]q2 ×[9m+5]q2[3m+2]q6/bracketleftbig9m+7 4/bracketrightbig q8/bracketleftbigm+1 7/bracketrightbig q126[63]q2 [7]q2[9]q2,28 C. KRATTENTHALER AND T. W. M ¨ULLER which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡14 (mod 140) ,then we have Catm(E7;q) = [9m+1]q2[3m+1]q6/bracketleftbig9m+4 10/bracketrightbig q20[10]q2 [2]q2[5]q2 ×[9m+5]q2/bracketleftbig3m+2 4/bracketrightbig q24[6]q4 [2]q4[3]q4/bracketleftbig9m+7 7/bracketrightbig q14[m+1]q18. Ifonedecomposes[9 m+5]q2as[9m 2+3]q4+q2[9m 2+2]q4, thenoneseesthat,byCorollary6, this is a polynomial in qwith non-negative integer coefficients. If m≡15 (mod 140) , then we have Catm(E7;q) =/bracketleftbig9m+1 4/bracketrightbig q8/bracketleftbig3m+1 2/bracketrightbig q12[9m+4]q2 ×/bracketleftbig9m+5 35/bracketrightbig q70[35]q2 [5]q2[7]q2[3m+2]q6[9m+7]q2[m+1]q18, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡16 (mod 140) ,then we have Catm(E7;q) =/bracketleftbig9m+1 5/bracketrightbig q10/bracketleftbig3m+1 7/bracketrightbig q42[21]q2 [3]q2[7]q2/bracketleftbig9m+4 4/bracketrightbig q8[9m+5]q2 ×/bracketleftbig3m+2 2/bracketrightbig q12[9m+7]q2[m+1]q18, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡17 (mod 140) ,then we have Catm(E7;q) =/bracketleftbig9m+1 7/bracketrightbig q14/bracketleftbig3m+1 4/bracketrightbig q24[6]q4 [2]q4[3]q4[9m+4]q2/bracketleftbig9m+5 2/bracketrightbig q4 ×[3m+2]q6/bracketleftbig9m+7 5/bracketrightbig q10[m+1]q18, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡18 (mod 140) ,then we have Catm(E7;q) = [9m+1]q2/bracketleftbig3m+1 5/bracketrightbig q30[15]q2 [3]q2[5]q2 ×/bracketleftbig9m+4 2/bracketrightbig q4[9m+5]q2/bracketleftbig3m+2 28/bracketrightbig q168[84]q2[2]q2 [4]q2[6]q2[7]q2[9m+7]q2[m+1]q18, which, by Corollary 6 and Lemma 13, is a polynomial in qwith non-negative integer coefficients. If m≡19 (mod 140) ,then we have Catm(E7;q) =/bracketleftbig9m+1 4/bracketrightbig q8/bracketleftbig3m+1 2/bracketrightbig q12/bracketleftbig9m+4 35/bracketrightbig q70[35]q2 [5]q2[7]q2[9m+5]q2 ×[3m+2]q6[9m+7]q2[m+1]q18,CYCLIC SIEVING FOR GENERALISED NON-CROSSING PARTITIONS 29 which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡20 (mod 140) ,then we have Catm(E7;q) = [9m+1]q2[3m+1]q6/bracketleftbig9m+4 4/bracketrightbig q8/bracketleftbig9m+5 5/bracketrightbig q10/bracketleftbig3m+2 2/bracketrightbig q12 ×[9m+7]q2/bracketleftbigm+1 7/bracketrightbig q126[63]q2 [7]q2[9]q2, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡21 (mod 140) ,then we have Catm(E7;q) =/bracketleftbig9m+1 5/bracketrightbig q10/bracketleftbig3m+1 4/bracketrightbig q24[6]q4 [2]q4[3]q4[9m+4]q2/bracketleftbig9m+5 2/bracketrightbig q4 ×[3m+2]q6/bracketleftbig9m+7 7/bracketrightbig q14[m+1]q18, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡22 (mod 140) ,then we have Catm(E7;q) = [9m+1]q2[3m+1]q6/bracketleftbig9m+4 2/bracketrightbig q4/bracketleftbig9m+5 7/bracketrightbig q14/bracketleftbig3m+2 4/bracketrightbig q24[6]q4 [2]q4[3]q4 ×/bracketleftbig9m+7 5/bracketrightbig q10[m+1]q18, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡23 (mod 140) ,then we have Catm(E7;q) =/bracketleftbig9m+1 4/bracketrightbig q8/bracketleftbig3m+1 35/bracketrightbig q210[105]q2 [3]q2[5]q2[7]q2[9m+4]q2[9m+5]q2 ×[3m+2]q6[9m+7]q2/bracketleftbigm+1 2/bracketrightbig q36[6]q6 [2]q6[3]q6, which, by Corollary 6 and Lemma 14, is a polynomial in qwith non-negative integer coefficients. If m≡24 (mod 140) ,then we have Catm(E7;q) =/bracketleftbig9m+1 7/bracketrightbig q14[3m+1]q6/bracketleftbig9m+4 20/bracketrightbig q40[20]q2 [4]q2[5]q2[9m+5]q2 ×/bracketleftbig3m+2 2/bracketrightbig q12[9m+7]q2[m+1]q18, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡25 (mod 140) ,then we have Catm(E7;q) = [9m+1]q2/bracketleftbig3m+1 2/bracketrightbig q12 ×[9m+4]q2/bracketleftbig9m+5 5/bracketrightbig q10/bracketleftbig3m+2 7/bracketrightbig q42[21]q2 [3]q2[7]q2/bracketleftbig9m+7 4/bracketrightbig q8[m+1]q18,30 C. KRATTENTHALER AND T. W. M ¨ULLER which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡26 (mod 140) ,then we have Catm(E7;q) =/bracketleftbig9m+1 5/bracketrightbig q10[3m+1]q6/bracketleftbig9m+4 14/bracketrightbig q28[14]q2 [2]q2[7]q2[9m+5]q2 ×/bracketleftbig3m+2 4/bracketrightbig q24[6]q4 [2]q4[3]q4[9m+7]q2[m+1]q18. Ifonedecomposes[9 m+7]q2as[9m 2+4]q4+q2[9m 2+3]q4, thenoneseesthat,byCorollary6, this is a polynomial in qwith non-negative integer coefficients. If m≡27 (mod 140) , then we have Catm(E7;q) =/bracketleftbig9m+1 4/bracketrightbig q8/bracketleftbig3m+1 2/bracketrightbig q12[9m+4]q2 ×[9m+5]q2[3m+2]q6/bracketleftbig9m+7 5/bracketrightbig q10/bracketleftbigm+1 7/bracketrightbig q126[63]q2 [7]q2[9]q2, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡28 (mod 140) ,then we have Catm(E7;q) = [9m+1]q2/bracketleftbig3m+1 5/bracketrightbig q30[15]q2 [3]q2[5]q2/bracketleftbig9m+4 4/bracketrightbig q8[9m+5]q2 ×/bracketleftbig3m+2 2/bracketrightbig q12/bracketleftbig9m+7 7/bracketrightbig q14[m+1]q18, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡29 (mod 140) ,then we have Catm(E7;q) = [9m+1]q2/bracketleftbig3m+1 2/bracketrightbig q12/bracketleftbig9m+4 5/bracketrightbig q10/bracketleftbig9m+5 7/bracketrightbig q14[3m+2]q6/bracketleftbig9m+7 4/bracketrightbig q8[m+1]q18, which is manifestly a polynomial in qwith non-negative integer coefficients. If m≡ 30 (mod 140) ,then we have Catm(E7;q) = [9m+1]q2/bracketleftbig3m+1 7/bracketrightbig q42[21]q2 [3]q2[7]q2/bracketleftbig9m+4 2/bracketrightbig q4/bracketleftbig9m+5 5/bracketrightbig q10 ×/bracketleftbig3m+2 4/bracketrightbig q24[6]q4 [2]q4[3]q4[9m+7]q2[m+1]q18, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡31 (mod 140) ,then we have Catm(E7;q) =/bracketleftbig9m+1 35/bracketrightbig q70[35]q2 [5]q2[7]q2/bracketleftbig3m+1 2/bracketrightbig q12[9m+4]q2 ×/bracketleftbig9m+5 4/bracketrightbig q8[3m+2]q6[9m+7]q2[m+1]q18,CYCLIC SIEVING FOR GENERALISED NON-CROSSING PARTITIONS 31 which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡32 (mod 140) ,then we have Catm(E7;q) = [9m+1]q2[3m+1]q6/bracketleftbig9m+4 4/bracketrightbig q8 ×[9m+5]q2/bracketleftbig3m+2 14/bracketrightbig q84[42]q2 [6]q2[7]q2/bracketleftbig9m+7 5/bracketrightbig q10[m+1]q18, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡33 (mod 140) ,then we have Catm(E7;q) = [9m+1]q2/bracketleftbig3m+1 10/bracketrightbig q60[30]q2 [5]q2[6]q2/bracketleftbig9m+4 7/bracketrightbig q14 ×[9m+5]q2[3m+2]q6/bracketleftbig9m+7 4/bracketrightbig q8[m+1]q18, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡34 (mod 140) ,then we have Catm(E7;q) = [9m+1]q2[3m+1]q6/bracketleftbig9m+4 10/bracketrightbig q20[10]q2 [2]q2[5]q2[9m+5]q2 ×/bracketleftbig3m+2 4/bracketrightbig q24[6]q4 [2]q4[3]q4[9m+7]q2/bracketleftbigm+1 7/bracketrightbig q126[63]q2 [7]q2[9]q2. Ifonedecomposes[9 m+7]q2as[9m 2+4]q4+q2[9m 2+3]q4, thenoneseesthat,byCorollary6, this is a polynomial in qwith non-negative integer coefficients. If m≡35 (mod 140) , then we have Catm(E7;q) =/bracketleftbig9m+1 4/bracketrightbig q8/bracketleftbig3m+1 2/bracketrightbig q12[9m+4]q2/bracketleftbig9m+5 5/bracketrightbig q10[3m+2]q6/bracketleftbig9m+7 7/bracketrightbig q14[m+1]q18, which is manifestly a polynomial in qwith non-negative integer coefficients. If m≡ 36 (mod 140) ,then we have Catm(E7;q) =/bracketleftbig9m+1 5/bracketrightbig q10[3m+1]q6/bracketleftbig9m+4 4/bracketrightbig q8/bracketleftbig9m+5 7/bracketrightbig q14/bracketleftbig3m+2 2/bracketrightbig q12[9m+7]q2[m+1]q18, which is manifestly a polynomial in qwith non-negative integer coefficients. If m≡ 37 (mod 140) ,then we have Catm(E7;q) = [9m+1]q2/bracketleftbig3m+1 14/bracketrightbig q84[42]q2 [6]q2[7]q2[9m+4]q2 ×[9m+5]q2[3m+2]q6/bracketleftbig9m+7 20/bracketrightbig q40[20]q2 [4]q2[5]q2[m+1]q18, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡38 (mod 140) ,then we have Catm(E7;q) =/bracketleftbig9m+1 7/bracketrightbig q14/bracketleftbig3m+1 5/bracketrightbig q30[15]q2 [3]q2[5]q2/bracketleftbig9m+4 2/bracketrightbig q4[9m+5]q2 ×/bracketleftbig3m+2 4/bracketrightbig q24[6]q4 [2]q4[3]q4[9m+7]q2[m+1]q18,32 C. KRATTENTHALER AND T. W. M ¨ULLER which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡39 (mod 140) ,then we have Catm(E7;q) =/bracketleftbig9m+1 4/bracketrightbig q8/bracketleftbig3m+1 2/bracketrightbig q12/bracketleftbig9m+4 5/bracketrightbig q10[9m+5]q2 ×/bracketleftbig3m+2 7/bracketrightbig q42[21]q2 [3]q2[7]q2[9m+7]q2[m+1]q18, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡40 (mod 140) ,then we have Catm(E7;q) = [9m+1]q2[3m+1]q6/bracketleftbig9m+4 28/bracketrightbig q56[28]q2 [4]q2[7]q2 ×/bracketleftbig9m+5 5/bracketrightbig q10/bracketleftbig3m+2 2/bracketrightbig q12[9m+7]q2[m+1]q18, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡41 (mod 140) ,then we have Catm(E7;q) =/bracketleftbig9m+1 5/bracketrightbig q10/bracketleftbig3m+1 2/bracketrightbig q12[9m+4]q2 ×[9m+5]q2[3m+2]q6/bracketleftbig9m+7 4/bracketrightbig q8/bracketleftbigm+1 7/bracketrightbig q126[63]q2 [7]q2[9]q2, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡42 (mod 140) ,then we have Catm(E7;q) = [9m+1]q2[3m+1]q6 ×/bracketleftbig9m+4 2/bracketrightbig q4[9m+5]q2/bracketleftbig3m+2 4/bracketrightbig q24[6]q4 [2]q4[3]q4/bracketleftbig9m+7 35/bracketrightbig q70[35]q2 [5]q2[7]q2[m+1]q18, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡43 (mod 140) ,then we have Catm(E7;q) =/bracketleftbig9m+1 4/bracketrightbig q8/bracketleftbig3m+1 10/bracketrightbig q60[30]q2 [5]q2[6]q2[9m+4]q2 ×/bracketleftbig9m+5 7/bracketrightbig q14[3m+2]q6[9m+7]q2[m+1]q18, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡44 (mod 140) ,then we have Catm(E7;q) = [9m+1]q2/bracketleftbig3m+1 7/bracketrightbig q42[21]q2 [3]q2[7]q2/bracketleftbig9m+4 20/bracketrightbig q40[20]q2 [4]q2[5]q2[9m+5]q2 ×/bracketleftbig3m+2 2/bracketrightbig q12[9m+7]q2[m+1]q18, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡45 (mod 140) ,then we have Catm(E7;q) =/bracketleftbig9m+1 7/bracketrightbig q14/bracketleftbig3m+1 2/bracketrightbig q12[9m+4]q2/bracketleftbig9m+5 5/bracketrightbig q10[3m+2]q6/bracketleftbig9m+7 4/bracketrightbig q8[m+1]q18,CYCLIC SIEVING FOR GENERALISED NON-CROSSING PARTITIONS 33 which is manifestly a polynomial in qwith non-negative integer coefficients. If m≡ 46 (mod 140) ,then we have Catm(E7;q) =/bracketleftbig9m+1 5/bracketrightbig q10[3m+1]q6 ×/bracketleftbig9m+4 2/bracketrightbig q4[9m+5]q2/bracketleftbig3m+2 28/bracketrightbig q168[84]q2[2]q2 [4]q2[6]q2[7]q2[9m+7]q2[m+1]q18, which, by Lemma 13, is a polynomial in qwith non-negative integer coefficients. If m≡47 (mod 140) ,then we have Catm(E7;q) =/bracketleftbig9m+1 4/bracketrightbig q8/bracketleftbig3m+1 2/bracketrightbig q12/bracketleftbig9m+4 7/bracketrightbig q14[9m+5]q2[3m+2]q6/bracketleftbig9m+7 5/bracketrightbig q10[m+1]q18, which is manifestly a polynomial in qwith non-negative integer coefficients. If m≡ 48 (mod 140) ,then we have Catm(E7;q) = [9m+1]q2/bracketleftbig3m+1 5/bracketrightbig q30[15]q2 [3]q2[5]q2/bracketleftbig9m+4 4/bracketrightbig q8[9m+5]q2 ×/bracketleftbig3m+2 2/bracketrightbig q12[9m+7]q2/bracketleftbigm+1 7/bracketrightbig q126[63]q2 [7]q2[9]q2, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡49 (mod 140) ,then we have Catm(E7;q) = [9m+1]q2/bracketleftbig3m+1 4/bracketrightbig q24[6]q4 [2]q4[3]q4/bracketleftbig9m+4 5/bracketrightbig q10/bracketleftbig9m+5 2/bracketrightbig q4 ×[3m+2]q6/bracketleftbig9m+7 7/bracketrightbig q14[m+1]q18, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡50 (mod 140) ,then we have Catm(E7;q) = [9m+1]q2[3m+1]q6/bracketleftbig9m+4 2/bracketrightbig q4 ×/bracketleftbig9m+5 35/bracketrightbig q70[35]q2 [5]q2[7]q2/bracketleftbig3m+2 4/bracketrightbig q24[6]q4 [2]q4[3]q4[9m+7]q2[m+1]q18, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡51 (mod 140) ,then we have Catm(E7;q) =/bracketleftbig9m+1 5/bracketrightbig q10/bracketleftbig3m+1 14/bracketrightbig q84[42]q2 [6]q2[7]q2[9m+4]q2 ×/bracketleftbig9m+5 4/bracketrightbig q8[3m+2]q6[9m+7]q2[m+1]q18, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡52 (mod 140) ,then we have Catm(E7;q) =/bracketleftbig9m+1 7/bracketrightbig q14[3m+1]q6/bracketleftbig9m+4 4/bracketrightbig q8[9m+5]q2/bracketleftbig3m+2 2/bracketrightbig q12/bracketleftbig9m+7 5/bracketrightbig q10[m+1]q18,34 C. KRATTENTHALER AND T. W. M ¨ULLER which is manifestly a polynomial in qwith non-negative integer coefficients. If m≡ 53 (mod 140) ,then we have Catm(E7;q) = [9m+1]q2/bracketleftbig3m+1 10/bracketrightbig q60[30]q2 [5]q2[6]q2[9m+4]q2 ×[9m+5]q2/bracketleftbig3m+2 7/bracketrightbig q42[21]q2 [3]q2[7]q2/bracketleftbig9m+7 4/bracketrightbig q8[m+1]q18, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡54 (mod 140) ,then we have Catm(E7;q) = [9m+1]q2[3m+1]q6/bracketleftbig9m+4 70/bracketrightbig q140[70]q2 [2]q2[5]q2[7]q2[9m+5]q2 ×/bracketleftbig3m+2 4/bracketrightbig q24[6]q4 [2]q4[3]q4[9m+7]q2[m+1]q18. Ifonedecomposes[9 m+7]q2as[9m 2+4]q4+q2[9m 2+3]q4, thenoneseesthat, byCorollary6 and Lemma 15, this is a polynomial in qwith non-negative integer coefficients. If m≡55 (mod 140) ,then we have Catm(E7;q) =/bracketleftbig9m+1 4/bracketrightbig q8/bracketleftbig3m+1 2/bracketrightbig q12[9m+4]q2/bracketleftbig9m+5 5/bracketrightbig q10[3m+2]q6 ×[9m+7]q2/bracketleftbigm+1 7/bracketrightbig q126[63]q2 [7]q2[9]q2, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡56 (mod 140) ,then we have Catm(E7;q) =/bracketleftbig9m+1 5/bracketrightbig q10[3m+1]q6/bracketleftbig9m+4 4/bracketrightbig q8[9m+5]q2/bracketleftbig3m+2 2/bracketrightbig q12/bracketleftbig9m+7 7/bracketrightbig q14[m+1]q18, which is manifestly a polynomial in qwith non-negative integer coefficients. If m≡ 57 (mod 140) ,then we have Catm(E7;q) =/bracketleftbig9m+1 2/bracketrightbig q4/bracketleftbig3m+1 4/bracketrightbig q24[6]q4 [2]q4[3]q4 ×[9m+4]q2/bracketleftbig9m+5 7/bracketrightbig q14[3m+2]q6/bracketleftbig9m+7 5/bracketrightbig q10[m+1]q18, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡58 (mod 140) ,then we have Catm(E7;q) = [9m+1]q2/bracketleftbig3m+1 35/bracketrightbig q210[105]q2 [3]q2[5]q2[7]q2/bracketleftbig9m+4 2/bracketrightbig q4[9m+5]q2 ×/bracketleftbig3m+2 4/bracketrightbig q24[6]q4 [2]q4[3]q4[9m+7]q2[m+1]q18, which, by Corollary 6 and Lemma 14, is a polynomial in qwith non-negative integer coefficients. If m≡59 (mod 140) ,then we have Catm(E7;q) =/bracketleftbig9m+1 7/bracketrightbig q14/bracketleftbig3m+1 2/bracketrightbig q12/bracketleftbig9m+4 5/bracketrightbig q10/bracketleftbig9m+5 4/bracketrightbig q8[3m+2]q6[9m+7]q2[m+1]q18,CYCLIC SIEVING FOR GENERALISED NON-CROSSING PARTITIONS 35 which is manifestly a polynomial in qwith non-negative integer coefficients. If m≡ 60 (mod 140) ,then we have Catm(E7;q) = [9m+1]q2[3m+1]q6/bracketleftbig9m+4 4/bracketrightbig q8/bracketleftbig9m+5 5/bracketrightbig q10 ×/bracketleftbig3m+2 14/bracketrightbig q84[42]q2 [6]q2[7]q2[9m+7]q2[m+1]q18, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡61 (mod 140) ,then we have Catm(E7;q) =/bracketleftbig9m+1 5/bracketrightbig q10/bracketleftbig3m+1 2/bracketrightbig q12/bracketleftbig9m+4 7/bracketrightbig q14[9m+5]q2[3m+2]q6/bracketleftbig9m+7 4/bracketrightbig q8[m+1]q18, which is manifestly a polynomial in qwith non-negative integer coefficients. If m≡ 62 (mod 140) ,then we have Catm(E7;q) = [9m+1]q2[3m+1]q6 ×/bracketleftbig9m+4 2/bracketrightbig q4[9m+5]q2/bracketleftbig3m+2 4/bracketrightbig q24[6]q4 [2]q4[3]q4/bracketleftbig9m+7 5/bracketrightbig q10/bracketleftbigm+1 7/bracketrightbig q126[63]q2 [7]q2[9]q2, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡63 (mod 140) ,then we have Catm(E7;q) =/bracketleftbig9m+1 4/bracketrightbig q8/bracketleftbig3m+1 10/bracketrightbig q60[30]q2 [5]q2[6]q2[9m+4]q2 ×[9m+5]q2[3m+2]q6/bracketleftbig9m+7 7/bracketrightbig q14[m+1]q18, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡64 (mod 140) ,then we have Catm(E7;q) = [9m+1]q2[3m+1]q6/bracketleftbig9m+4 20/bracketrightbig q40[20]q2 [4]q2[5]q2 ×/bracketleftbig9m+5 7/bracketrightbig q14/bracketleftbig3m+2 2/bracketrightbig q12[9m+7]q2[m+1]q18, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡65 (mod 140) ,then we have Catm(E7;q) = [9m+1]q2/bracketleftbig3m+1 14/bracketrightbig q84[42]q2 [6]q2[7]q2[9m+4]q2 ×/bracketleftbig9m+5 5/bracketrightbig q10[3m+2]q6/bracketleftbig9m+7 4/bracketrightbig q8[m+1]q18, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡66 (mod 140) ,then we have Catm(E7;q) =/bracketleftbig9m+1 35/bracketrightbig q70[35]q2 [5]q2[7]q2[3m+1]q6/bracketleftbig9m+4 2/bracketrightbig q4[9m+5]q2 ×/bracketleftbig3m+2 4/bracketrightbig q24[6]q4 [2]q4[3]q4[9m+7]q2[m+1]q18,36 C. KRATTENTHALER AND T. W. M ¨ULLER which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡67 (mod 140) ,then we have Catm(E7;q) =/bracketleftbig9m+1 4/bracketrightbig q8/bracketleftbig3m+1 2/bracketrightbig q12[9m+4]q2 ×[9m+5]q2/bracketleftbig3m+2 7/bracketrightbig q42[21]q2 [3]q2[7]q2/bracketleftbig9m+7 5/bracketrightbig q10[m+1]q18, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡68 (mod 140) ,then we have Catm(E7;q) = [9m+1]q2/bracketleftbig3m+1 5/bracketrightbig q30[15]q2 [3]q2[5]q2 ×/bracketleftbig9m+4 28/bracketrightbig q56[28]q2 [4]q2[7]q2[9m+5]q2/bracketleftbig3m+2 2/bracketrightbig q12[9m+7]q2[m+1]q18, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡69 (mod 140) ,then we have Catm(E7;q) = [9m+1]q2/bracketleftbig3m+1 2/bracketrightbig q12/bracketleftbig9m+4 5/bracketrightbig q10 ×[9m+5]q2[3m+2]q6/bracketleftbig9m+7 4/bracketrightbig q8/bracketleftbigm+1 7/bracketrightbig q126[63]q2 [7]q2[9]q2, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡70 (mod 140) ,then we have Catm(E7;q) = [9m+1]q2[3m+1]q6/bracketleftbig9m+4 2/bracketrightbig q4/bracketleftbig9m+5 5/bracketrightbig q10/bracketleftbig3m+2 4/bracketrightbig q24[6]q4 [2]q4[3]q4 ×/bracketleftbig9m+7 7/bracketrightbig q14[m+1]q18, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡71 (mod 140) ,then we have Catm(E7;q) =/bracketleftbig9m+1 20/bracketrightbig q40[20]q2 [4]q2[5]q2/bracketleftbig3m+1 2/bracketrightbig q12[9m+4]q2 ×/bracketleftbig9m+5 7/bracketrightbig q14[3m+2]q6[9m+7]q2[m+1]q18, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡72 (mod 140) ,then we have Catm(E7;q) = [9m+1]q2/bracketleftbig3m+1 7/bracketrightbig q42[21]q2 [3]q2[7]q2/bracketleftbig9m+4 4/bracketrightbig q8 ×[9m+5]q2/bracketleftbig3m+2 2/bracketrightbig q12/bracketleftbig9m+7 5/bracketrightbig q10[m+1]q18,CYCLIC SIEVING FOR GENERALISED NON-CROSSING PARTITIONS 37 which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡73 (mod 140) ,then we have Catm(E7;q) =/bracketleftbig9m+1 7/bracketrightbig q14/bracketleftbig3m+1 10/bracketrightbig q60[30]q2 [5]q2[6]q2[9m+4]q2 ×[9m+5]q2[3m+2]q6/bracketleftbig9m+7 4/bracketrightbig q8[m+1]q18, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡74 (mod 140) ,then we have Catm(E7;q) = [9m+1]q2[3m+1]q6/bracketleftbig9m+4 10/bracketrightbig q20[10]q2 [2]q2[5]q2 ×[9m+5]q2/bracketleftbig3m+2 28/bracketrightbig q168[84]q2[2]q2 [4]q2[6]q2[7]q2[9m+7]q2[m+1]q18. Ifonedecomposes[9 m+7]q2as[9m 2+4]q4+q2[9m 2+3]q4, thenoneseesthat, byCorollary6 and Lemma 13, this is a polynomial in qwith non-negative integer coefficients. If m≡75 (mod 140) ,then we have Catm(E7;q) =/bracketleftbig9m+1 4/bracketrightbig q8/bracketleftbig3m+1 2/bracketrightbig q12/bracketleftbig9m+4 7/bracketrightbig q14/bracketleftbig9m+5 5/bracketrightbig q10[3m+2]q6[9m+7]q2[m+1]q18, which is manifestly a polynomial in qwith non-negative integer coefficients. If m≡ 76 (mod 140) ,then we have Catm(E7;q) =/bracketleftbig9m+1 5/bracketrightbig q10[3m+1]q6/bracketleftbig9m+4 4/bracketrightbig q8[9m+5]q2 ×/bracketleftbig3m+2 2/bracketrightbig q12[9m+7]q2/bracketleftbigm+1 7/bracketrightbig q126[63]q2 [7]q2[9]q2, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡77 (mod 140) ,then we have Catm(E7;q) = [9m+1]q2/bracketleftbig3m+1 4/bracketrightbig q24[6]q4 [2]q4[3]q4[9m+4]q2 ×/bracketleftbig9m+5 2/bracketrightbig q4[3m+2]q6/bracketleftbig9m+7 35/bracketrightbig q70[35]q2 [5]q2[7]q2[m+1]q18, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡78 (mod 140) ,then we have Catm(E7;q) = [9m+1]q2/bracketleftbig3m+1 5/bracketrightbig q30[15]q2 [3]q2[5]q2 ×/bracketleftbig9m+4 2/bracketrightbig q4/bracketleftbig9m+5 7/bracketrightbig q14/bracketleftbig3m+2 4/bracketrightbig q24[6]q4 [2]q4[3]q4[9m+7]q2[m+1]q18,38 C. KRATTENTHALER AND T. W. M ¨ULLER which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡79 (mod 140) ,then we have Catm(E7;q) =/bracketleftbig9m+1 4/bracketrightbig q8/bracketleftbig3m+1 14/bracketrightbig q84[42]q2 [6]q2[7]q2/bracketleftbig9m+4 5/bracketrightbig q10[9m+5]q2 ×[3m+2]q6[9m+7]q2[m+1]q18, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡80 (mod 140) ,then we have Catm(E7;q) =/bracketleftbig9m+1 7/bracketrightbig q14[3m+1]q6/bracketleftbig9m+4 4/bracketrightbig q8/bracketleftbig9m+5 5/bracketrightbig q10/bracketleftbig3m+2 2/bracketrightbig q12[9m+7]q2[m+1]q18, which is manifestly a polynomial in qwith non-negative integer coefficients. If m≡ 81 (mod 140) ,then we have Catm(E7;q) =/bracketleftbig9m+1 5/bracketrightbig q10/bracketleftbig3m+1 2/bracketrightbig q12[9m+4]q2 ×[9m+5]q2/bracketleftbig3m+2 7/bracketrightbig q42[21]q2 [3]q2[7]q2/bracketleftbig9m+7 4/bracketrightbig q8[m+1]q18, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡82 (mod 140) ,then we have Catm(E7;q) = [9m+1]q2[3m+1]q6/bracketleftbig9m+4 14/bracketrightbig q28[14]q2 [2]q2[7]q2 ×[9m+5]q2/bracketleftbig3m+2 4/bracketrightbig q24[6]q4 [2]q4[3]q4/bracketleftbig9m+7 5/bracketrightbig q10[m+1]q18. Ifonedecomposes[9 m+5]q2as[9m 2+4]q4+q2[9m 2+2]q4, thenoneseesthat,byCorollary6, this is a polynomial in qwith non-negative integer coefficients. If m≡83 (mod 140) , then we have Catm(E7;q) =/bracketleftbig9m+1 4/bracketrightbig q8/bracketleftbig3m+1 10/bracketrightbig q60[30]q2 [5]q2[6]q2[9m+4]q2[9m+5]q ×[3m+2]q6[9m+7]q2/bracketleftbigm+1 7/bracketrightbig q126[63]q2 [7]q2[9]q2, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡84 (mod 140) ,then we have Catm(E7;q) = [9m+1]q2[3m+1]q6/bracketleftbig9m+4 20/bracketrightbig q40[20]q2 [4]q2[5]q2 ×[9m+5]q2/bracketleftbig3m+2 2/bracketrightbig q12/bracketleftbig9m+7 7/bracketrightbig q14[m+1]q18, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡85 (mod 140) ,then we have Catm(E7;q) = [9m+1]q2/bracketleftbig3m+1 2/bracketrightbig q12 ×[9m+4]q2/bracketleftbig9m+5 35/bracketrightbig q70[35]q2 [5]q2[7]q2[3m+2]q6/bracketleftbig9m+7 4/bracketrightbig q8[m+1]q18,CYCLIC SIEVING FOR GENERALISED NON-CROSSING PARTITIONS 39 which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡86 (mod 140) ,then we have Catm(E7;q) =/bracketleftbig9m+1 5/bracketrightbig q10/bracketleftbig3m+1 7/bracketrightbig q42[21]q2 [3]q2[7]q2/bracketleftbig9m+4 2/bracketrightbig q4[9m+5]q2 ×/bracketleftbig3m+2 4/bracketrightbig q24[6]q4 [2]q4[3]q4[9m+7]q2[m+1]q18, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡87 (mod 140) ,then we have Catm(E7;q) =/bracketleftbig9m+1 7/bracketrightbig q14/bracketleftbig3m+1 2/bracketrightbig q12[9m+4]q2/bracketleftbig9m+5 4/bracketrightbig q8[3m+2]q6/bracketleftbig9m+7 5/bracketrightbig q10[m+1]q18, which is manifestly a polynomial in qwith non-negative integer coefficients. If m≡ 88 (mod 140) ,then we have Catm(E7;q) = [9m+1]q2/bracketleftbig3m+1 5/bracketrightbig q30[15]q2 [3]q2[5]q2/bracketleftbig9m+4 4/bracketrightbig q8[9m+5]q2 ×/bracketleftbig3m+2 14/bracketrightbig q84[42]q2 [6]q2[7]q2[9m+7]q2[m+1]q18, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡89 (mod 140) ,then we have Catm(E7;q) = [9m+1]q2/bracketleftbig3m+1 2/bracketrightbig q12/bracketleftbig9m+4 35/bracketrightbig q70[35]q2 [5]q2[7]q2 ×[9m+5]q2[3m+2]q6/bracketleftbig9m+7 4/bracketrightbig q8[m+1]q18, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡90 (mod 140) ,then we have Catm(E7;q) = [9m+1]q2[3m+1]q6/bracketleftbig9m+4 2/bracketrightbig q4/bracketleftbig9m+5 5/bracketrightbig q10/bracketleftbig3m+2 4/bracketrightbig q24[6]q4 [2]q4[3]q4 ×[9m+7]q2/bracketleftbigm+1 7/bracketrightbig q126[63]q2 [7]q2[9]q2, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡91 (mod 140) ,then we have Catm(E7;q) =/bracketleftbig9m+1 5/bracketrightbig q10/bracketleftbig3m+1 2/bracketrightbig q12[9m+4]q2/bracketleftbig9m+5 4/bracketrightbig q8[3m+2]q6/bracketleftbig9m+7 7/bracketrightbig q14[m+1]q18, which is manifestly a polynomial in qwith non-negative integer coefficients. If m≡ 92 (mod 140) ,then we have Catm(E7;q) = [9m+1]q2[3m+1]q6/bracketleftbig9m+4 4/bracketrightbig q8/bracketleftbig9m+5 7/bracketrightbig q14/bracketleftbig3m+2 2/bracketrightbig q12/bracketleftbig9m+7 5/bracketrightbig q10[m+1]q18,40 C. KRATTENTHALER AND T. W. M ¨ULLER which is manifestly a polynomial in qwith non-negative integer coefficients. If m≡ 93 (mod 140) ,then we have Catm(E7;q) = [9m+1]q2/bracketleftbig3m+1 35/bracketrightbig q210[105]q2 [3]q2[5]q2[7]q2[9m+4]q2 ×[9m+5]q2[3m+2]q6/bracketleftbig9m+7 4/bracketrightbig q8/bracketleftbigm+1 2/bracketrightbig q36[6]q6 [2]q6[3]q6, which, by Corollary 6 and Lemma 14, is a polynomial in qwith non-negative integer coefficients. If m≡94 (mod 140) ,then we have Catm(E7;q) =/bracketleftbig9m+1 7/bracketrightbig q14[3m+1]q6/bracketleftbig9m+4 10/bracketrightbig q20[10]q2 [2]q2[5]q2[9m+5]q2 ×/bracketleftbig3m+2 4/bracketrightbig q24[6]q4 [2]q4[3]q4[9m+7]q2[m+1]q18. Ifonedecomposes[9 m+7]q2as[9m 2+4]q4+q2[9m 2+3]q4, thenoneseesthat,byCorollary6, this is a polynomial in qwith non-negative integer coefficients. If m≡95 (mod 140) , then we have Catm(E7;q) =/bracketleftbig9m+1 4/bracketrightbig q8/bracketleftbig3m+1 2/bracketrightbig q12[9m+4]q2/bracketleftbig9m+5 5/bracketrightbig q10 ×/bracketleftbig3m+2 7/bracketrightbig q42[21]q2 [3]q2[7]q2[9m+7]q2[m+1]q18, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡96 (mod 140) ,then we have Catm(E7;q) =/bracketleftbig9m+1 5/bracketrightbig q10[3m+1]q6/bracketleftbig9m+4 28/bracketrightbig q56[28]q2 [4]q2[7]q2[9m+5]q2 ×/bracketleftbig3m+2 2/bracketrightbig q12[9m+7]q2[m+1]q18, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡97 (mod 140) ,then we have Catm(E7;q) = [9m+1]q2/bracketleftbig3m+1 4/bracketrightbig q24[6]q4 [2]q4[3]q4[9m+4]q2/bracketleftbig9m+5 2/bracketrightbig q4 ×[3m+2]q6/bracketleftbig9m+7 5/bracketrightbig q10/bracketleftbigm+1 7/bracketrightbig q126[63]q2 [7]q2[9]q2, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡98 (mod 140) ,then we have Catm(E7;q) = [9m+1]q2/bracketleftbig3m+1 5/bracketrightbig q30[15]q2 [3]q2[5]q2 ×/bracketleftbig9m+4 2/bracketrightbig q4[9m+5]q2/bracketleftbig3m+2 4/bracketrightbig q24[6]q4 [2]q4[3]q4/bracketleftbig9m+7 7/bracketrightbig q14[m+1]q18,CYCLIC SIEVING FOR GENERALISED NON-CROSSING PARTITIONS 41 which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡99 (mod 140) ,then we have Catm(E7;q) =/bracketleftbig9m+1 4/bracketrightbig q8/bracketleftbig3m+1 2/bracketrightbig q12/bracketleftbig9m+4 5/bracketrightbig q10/bracketleftbig9m+5 7/bracketrightbig q14[3m+2]q6[9m+7]q2[m+1]q18, which is manifestly a polynomial in qwith non-negative integer coefficients. If m≡ 100 (mod 140) ,then we have Catm(E7;q) = [9m+1]q2/bracketleftbig3m+1 7/bracketrightbig q42[21]q2 [3]q2[7]q2/bracketleftbig9m+4 4/bracketrightbig q8 ×/bracketleftbig9m+5 5/bracketrightbig q10/bracketleftbig3m+2 2/bracketrightbig q12[9m+7]q2[m+1]q18, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡101 (mod 140) ,then we have Catm(E7;q) =/bracketleftbig9m+1 35/bracketrightbig q70[35]q2 [5]q2[7]q2/bracketleftbig3m+1 2/bracketrightbig q12[9m+4]q2 ×[9m+5]q2[3m+2]q6/bracketleftbig9m+7 4/bracketrightbig q8[m+1]q18, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡102 (mod 140) ,then we have Catm(E7;q) = [9m+1]q2[3m+1]q6 ×/bracketleftbig9m+4 2/bracketrightbig q4[9m+5]q2/bracketleftbig3m+2 28/bracketrightbig q168[84]q2[2]q2 [4]q2[6]q2[7]q2/bracketleftbig9m+7 5/bracketrightbig q10[m+1]q18, which, by Lemma 13, is a polynomial in qwith non-negative integer coefficients. If m≡103 (mod 140) ,then we have Catm(E7;q) =/bracketleftbig9m+1 4/bracketrightbig q8/bracketleftbig3m+1 10/bracketrightbig q60[30]q2 [5]q2[6]q2/bracketleftbig9m+4 7/bracketrightbig q14[9m+5]q2 ×[3m+2]q6[9m+7]q2[m+1]q18, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡104 (mod 140) ,then we have Catm(E7;q) = [9m+1]q2[3m+1]q6/bracketleftbig9m+4 20/bracketrightbig q40[20]q2 [4]q2[5]q2[9m+5]q2 ×/bracketleftbig3m+2 2/bracketrightbig q12[9m+7]q2/bracketleftbigm+1 7/bracketrightbig q126[63]q2 [7]q2[9]q2, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡105 (mod 140) ,then we have Catm(E7;q) = [9m+1]q2/bracketleftbig3m+1 4/bracketrightbig q24[6]q4 [2]q4[3]q4 ×[9m+4]q2/bracketleftbig9m+5 10/bracketrightbig q20[10]q2 [2]q2[5]q2[3m+2]q6/bracketleftbig9m+7 7/bracketrightbig q14[m+1]q18.42 C. KRATTENTHALER AND T. W. M ¨ULLER If one decomposes [9 m+1]q2as [9m+1 2]q4+q2[9m+1 2]q4, then one sees that, by Corollary 6, this is a polynomial in qwith non-negative integer coefficients. If m≡106 (mod 140) , then we have Catm(E7;q) =/bracketleftbig9m+1 5/bracketrightbig q10[3m+1]q6/bracketleftbig9m+4 2/bracketrightbig q4/bracketleftbig9m+5 7/bracketrightbig q14/bracketleftbig3m+2 4/bracketrightbig q24[6]q4 [2]q4[3]q4 ×[9m+7]q2[m+1]q18, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡107 (mod 140) ,then we have Catm(E7;q) =/bracketleftbig9m+1 4/bracketrightbig q8/bracketleftbig3m+1 14/bracketrightbig q84[42]q2 [6]q2[7]q2[9m+4]q2 ×[9m+5]q2[3m+2]q6/bracketleftbig9m+7 5/bracketrightbig q10[m+1]q18, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡108 (mod 140) ,then we have Catm(E7;q) =/bracketleftbig9m+1 7/bracketrightbig q14/bracketleftbig3m+1 5/bracketrightbig q30[15]q2 [3]q2[5]q2/bracketleftbig9m+4 4/bracketrightbig q8[9m+5]q2 ×/bracketleftbig3m+2 2/bracketrightbig q12[9m+7]q2[m+1]q18, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡109 (mod 140) ,then we have Catm(E7;q) = [9m+1]q2/bracketleftbig3m+1 2/bracketrightbig q12/bracketleftbig9m+4 5/bracketrightbig q10 ×[9m+5]q2/bracketleftbig3m+2 7/bracketrightbig q42[21]q2 [3]q2[7]q2/bracketleftbig9m+7 4/bracketrightbig q8[m+1]q18, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡110 (mod 140) ,then we have Catm(E7;q) = [9m+1]q2[3m+1]q6/bracketleftbig9m+4 14/bracketrightbig q28[14]q2 [2]q2[7]q2 ×/bracketleftbig9m+5 5/bracketrightbig q10/bracketleftbig3m+2 4/bracketrightbig q24[6]q4 [2]q4[3]q4[9m+7]q2[m+1]q18. Ifonedecomposes[9 m+7]q2as[9m 2+4]q4+q2[9m 2+3]q4, thenoneseesthat,byCorollary6, this is a polynomial in qwith non-negative integer coefficients. If m≡111 (mod 140) , then we have Catm(E7;q) =/bracketleftbig9m+1 5/bracketrightbig q10/bracketleftbig3m+1 2/bracketrightbig q12[9m+4]q2/bracketleftbig9m+5 4/bracketrightbig q8[3m+2]q6 ×[9m+7]q2/bracketleftbigm+1 7/bracketrightbig q126[63]q2 [7]q2[9]q2,CYCLIC SIEVING FOR GENERALISED NON-CROSSING PARTITIONS 43 which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡112 (mod 140) ,then we have Catm(E7;q) = [9m+1]q2[3m+1]q6/bracketleftbig9m+4 4/bracketrightbig q8 ×[9m+5]q2/bracketleftbig3m+2 2/bracketrightbig q12/bracketleftbig9m+7 35/bracketrightbig q70[35]q2 [5]q2[7]q2[m+1]q18, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡113 (mod 140) ,then we have Catm(E7;q) = [9m+1]q2/bracketleftbig3m+1 10/bracketrightbig q60[30]q2 [5]q2[6]q2[9m+4]q2 ×/bracketleftbig9m+5 7/bracketrightbig q14[3m+2]q6/bracketleftbig9m+7 4/bracketrightbig q8[m+1]q18, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡114 (mod 140) ,then we have Catm(E7;q) = [9m+1]q2/bracketleftbig3m+1 7/bracketrightbig q42[21]q2 [3]q2[7]q2/bracketleftbig9m+4 10/bracketrightbig q20[10]q2 [2]q2[5]q2[9m+5]q2 ×/bracketleftbig3m+2 4/bracketrightbig q24[6]q4 [2]q4[3]q4[9m+7]q2[m+1]q18. Ifonedecomposes[9 m+7]q2as[9m 2+4]q4+q2[9m 2+3]q4, thenoneseesthat,byCorollary6, this is a polynomial in qwith non-negative integer coefficients. If m≡115 (mod 140) , then we have Catm(E7;q) =/bracketleftbig9m+1 28/bracketrightbig q56[28]q2 [4]q2[7]q2/bracketleftbig3m+1 2/bracketrightbig q12[9m+4]q2 ×/bracketleftbig9m+5 5/bracketrightbig q10[3m+2]q6[9m+7]q[m+1]q18, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡116 (mod 140) ,then we have Catm(E7;q) =/bracketleftbig9m+1 5/bracketrightbig q10[3m+1]q6/bracketleftbig9m+4 4/bracketrightbig q8[9m+5]q2 ×/bracketleftbig3m+2 14/bracketrightbig q84[42]q2 [6]q2[7]q2[9m+7]q2[m+1]q18, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡117 (mod 140) ,then we have Catm(E7;q) = [9m+1]q2/bracketleftbig3m+1 4/bracketrightbig q24[6]q4 [2]q4[3]q4/bracketleftbig9m+4 7/bracketrightbig q14/bracketleftbig9m+5 2/bracketrightbig q4 ×[3m+2]q6/bracketleftbig9m+7 5/bracketrightbig q10[m+1]q18,44 C. KRATTENTHALER AND T. W. M ¨ULLER which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡118 (mod 140) ,then we have Catm(E7;q) = [9m+1]q2/bracketleftbig3m+1 5/bracketrightbig q30[15]q2 [3]q2[5]q2 ×/bracketleftbig9m+4 2/bracketrightbig q4[9m+5]q2/bracketleftbig3m+2 4/bracketrightbig q24[6]q4 [2]q4[3]q4[9m+7]q2/bracketleftbigm+1 7/bracketrightbig q126[63]q2 [7]q2[9]q2, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡119 (mod 140) ,then we have Catm(E7;q) =/bracketleftbig9m+1 4/bracketrightbig q8/bracketleftbig3m+1 2/bracketrightbig q12/bracketleftbig9m+4 5/bracketrightbig q10[9m+5]q2[3m+2]q6/bracketleftbig9m+7 7/bracketrightbig q14[m+1]q18, which is manifestly a polynomial in qwith non-negative integer coefficients. If m≡ 120 (mod 140) ,then we have Catm(E7;q) = [9m+1]q2[3m+1]q6/bracketleftbig9m+4 4/bracketrightbig q8 ×/bracketleftbig9m+5 35/bracketrightbig q70[35]q2 [5]q2[7]q2/bracketleftbig3m+2 2/bracketrightbig q12[9m+7]q2[m+1]q18, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡121 (mod 140) ,then we have Catm(E7;q) =/bracketleftbig9m+1 5/bracketrightbig q10/bracketleftbig3m+1 14/bracketrightbig q84[42]q2 [6]q2[7]q2[9m+4]q2 ×[9m+5]q2[3m+2]q6/bracketleftbig9m+7 4/bracketrightbig q8[m+1]q18, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡122 (mod 140) ,then we have Catm(E7;q) =/bracketleftbig9m+1 7/bracketrightbig q14[3m+1]q6 ×/bracketleftbig9m+4 2/bracketrightbig q4[9m+5]q2/bracketleftbig3m+2 4/bracketrightbig q24[6]q4 [2]q4[3]q4/bracketleftbig9m+7 5/bracketrightbig q10[m+1]q18, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡123 (mod 140) ,then we have Catm(E7;q) =/bracketleftbig9m+1 4/bracketrightbig q8/bracketleftbig3m+1 10/bracketrightbig q60[30]q2 [5]q2[6]q2[9m+4]q2[9m+5]q2 ×/bracketleftbig3m+2 7/bracketrightbig q42[21]q2 [3]q2[7]q2[9m+7]q2[m+1]q18,CYCLIC SIEVING FOR GENERALISED NON-CROSSING PARTITIONS 45 which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡124 (mod 140) ,then we have Catm(E7;q) = [9m+1]q2[3m+1]q6/bracketleftbig9m+4 28/bracketrightbig q56[28]q2 [4]q2[7]q2[9m+5]q2 ×/bracketleftbig3m+2 2/bracketrightbig q12[9m+7]q2/bracketleftbigm+1 5/bracketrightbig q90[45]q2 [5]q2[9]q2, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡125 (mod 140) ,then we have Catm(E7;q) = [9m+1]q2/bracketleftbig3m+1 2/bracketrightbig q12 ×[9m+4]q2/bracketleftbig9m+5 5/bracketrightbig q10[3m+2]q6/bracketleftbig9m+7 4/bracketrightbig q8/bracketleftbigm+1 7/bracketrightbig q126[63]q2 [7]q2[9]q2, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡126 (mod 140) ,then we have Catm(E7;q) =/bracketleftbig9m+1 5/bracketrightbig q10[3m+1]q6 ×/bracketleftbig9m+4 2/bracketrightbig q4[9m+5]q2/bracketleftbig3m+2 4/bracketrightbig q24[6]q4 [2]q4[3]q4/bracketleftbig9m+7 7/bracketrightbig q14[m+1]q18, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡127 (mod 140) ,then we have Catm(E7;q) =/bracketleftbig9m+1 4/bracketrightbig q8/bracketleftbig3m+1 2/bracketrightbig q12[9m+4]q2/bracketleftbig9m+5 7/bracketrightbig q14[3m+2]q6/bracketleftbig9m+7 5/bracketrightbig q10[m+1]q18, which is manifestly a polynomial in qwith non-negative integer coefficients. If m≡ 128 (mod 140) ,then we have Catm(E7;q) = [9m+1]q2/bracketleftbig3m+1 35/bracketrightbig q210[105]q2 [3]q2[5]q2[7]q2/bracketleftbig9m+4 4/bracketrightbig q8[9m+5]q2 ×/bracketleftbig3m+2 2/bracketrightbig q12[9m+7]q2[m+1]q18, which, by Lemma 14, is a polynomial in qwith non-negative integer coefficients. If m≡129 (mod 140) ,then we have Catm(E7;q) =/bracketleftbig9m+1 7/bracketrightbig q14/bracketleftbig3m+1 2/bracketrightbig q12/bracketleftbig9m+4 5/bracketrightbig q10[9m+5]q2[3m+2]q6/bracketleftbig9m+7 4/bracketrightbig q8[m+1]q18, which is manifestly a polynomial in qwith non-negative integer coefficients. If m≡ 130 (mod 140) ,then we have Catm(E7;q) = [9m+1]q2[3m+1]q6 ×/bracketleftbig9m+4 2/bracketrightbig q4/bracketleftbig9m+5 5/bracketrightbig q10/bracketleftbig3m+2 28/bracketrightbig q168[84]q2[2]q2 [4]q2[6]q2[7]q2[9m+7]q2[m+1]q18, which, by Lemma 13, is a polynomial in qwith non-negative integer coefficients. If m≡131 (mod 140) ,then we have Catm(E7;q) =/bracketleftbig9m+1 5/bracketrightbig q10/bracketleftbig3m+1 2/bracketrightbig q12/bracketleftbig9m+4 7/bracketrightbig q14/bracketleftbig9m+5 4/bracketrightbig q8[3m+2]q6[9m+7]q2[m+1]q18,46 C. KRATTENTHALER AND T. W. M ¨ULLER which is manifestly a polynomial in qwith non-negative integer coefficients. If m≡ 132 (mod 140) ,then we have Catm(E7;q) = [9m+1]q2[3m+1]q6/bracketleftbig9m+4 4/bracketrightbig q8 ×[9m+5]q2/bracketleftbig3m+2 2/bracketrightbig q12/bracketleftbig9m+7 5/bracketrightbig q10/bracketleftbigm+1 7/bracketrightbig q126[63]q2 [7]q2[9]q2, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡133 (mod 140) ,then we have Catm(E7;q) = [9m+1]q2/bracketleftbig3m+1 10/bracketrightbig q60[30]q2 [5]q2[6]q2[9m+4]q2 ×[9m+5]q2[3m+2]q6/bracketleftbig9m+7 28/bracketrightbig q56[28]q2 [4]q2[7]q2[m+1]q18, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡134 (mod 140) ,then we have Catm(E7;q) = [9m+1]q2[3m+1]q6/bracketleftbig9m+4 10/bracketrightbig q20[10]q2 [2]q2[5]q2 ×/bracketleftbig9m+5 7/bracketrightbig q14/bracketleftbig3m+2 4/bracketrightbig q24[6]q4 [2]q4[3]q4[9m+7]q2[m+1]q18. Ifonedecomposes[9 m+7]q2as[9m 2+4]q4+q2[9m 2+3]q4, thenoneseesthat,byCorollary6, this is a polynomial in qwith non-negative integer coefficients. If m≡135 (mod 140) , then we have Catm(E7;q) =/bracketleftbig9m+1 4/bracketrightbig q8/bracketleftbig3m+1 14/bracketrightbig q84[42]q2 [6]q2[7]q2[9m+4]q2 ×/bracketleftbig9m+5 5/bracketrightbig q10[3m+2]q6[9m+7]q2[m+1]q18, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡136 (mod 140) ,then we have Catm(E7;q) =/bracketleftbig9m+1 35/bracketrightbig q70[35]q2 [5]q2[7]q2[3m+1]q6/bracketleftbig9m+4 4/bracketrightbig q8[9m+5]q2 ×/bracketleftbig3m+2 2/bracketrightbig q12[9m+7]q2[m+1]q18, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡137 (mod 140) ,then we have Catm(E7;q) = [9m+1]q2/bracketleftbig3m+1 4/bracketrightbig q24[6]q4 [2]q4[3]q4[9m+4]q2/bracketleftbig9m+5 2/bracketrightbig q4 ×/bracketleftbig3m+2 7/bracketrightbig q42[21]q2 [3]q2[7]q2/bracketleftbig9m+7 5/bracketrightbig q10[m+1]q18,CYCLIC SIEVING FOR GENERALISED NON-CROSSING PARTITIONS 47 which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡138 (mod 140) ,then we have Catm(E7;q) = [9m+1]q2/bracketleftbig3m+1 5/bracketrightbig q30[15]q2 [3]q2[5]q2/bracketleftbig9m+4 14/bracketrightbig q28[14]q2 [2]q2[7]q2[9m+5]q2 ×/bracketleftbig3m+2 4/bracketrightbig q24[6]q4 [2]q4[3]q4[9m+7]q2[m+1]q18. Ifonedecomposes[9 m+7]q2as[9m 2+4]q4+q2[9m 2+3]q4, thenoneseesthat,byCorollary6, this is a polynomial in qwith non-negative integer coefficients. If m≡139 (mod 140) , then we have Catm(E7;q) =/bracketleftbig9m+1 4/bracketrightbig q8/bracketleftbig3m+1 2/bracketrightbig q12/bracketleftbig9m+4 5/bracketrightbig q10[9m+5]q2 ×[3m+2]q6[9m+7]q2/bracketleftbigm+1 7/bracketrightbig q126[63]q2 [7]q2[9]q2, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. ForW=G37=E8, the degrees are 2 ,8,12,14,18,20,24,30, and hence Catm(E7;q) =[30m+2]q[30m+8]q[30m+12]q[30m+14]q [2]q[8]q[12]q[14]q ×[30m+18]q[30m+20]q[30m+24]q[30m+30]q [18]q[20]q[24]q[30]q. Ifm≡0 (mod 84),then we have Catm(E8;q) = [15m+1]q2/bracketleftbig15m+4 4/bracketrightbig q8/bracketleftbig5m+2 2/bracketrightbig q12/bracketleftbig15m+7 7/bracketrightbig q14 ×/bracketleftbig5m+3 3/bracketrightbig q18/bracketleftbig3m+2 2/bracketrightbig q20/bracketleftbig5m+4 4/bracketrightbig q24[m+1]q30, which is manifestly a polynomial in qwith non-negative integer coefficients. If m≡ 1 (mod 84),then we have Catm(E8;q) =/bracketleftbig15m+1 4/bracketrightbig q8[15m+4]q2/bracketleftbig5m+2 7/bracketrightbig q42[21]q2 [3]q2[7]q2/bracketleftbig15m+7 2/bracketrightbig q4/bracketleftbig5m+3 4/bracketrightbig q24 ×[3m+2]q10/bracketleftbig5m+4 3/bracketrightbig q18/bracketleftbigm+1 2/bracketrightbig q60[30]q2[2]q2[3]q2[5]q2 [6]q2[10]q2[15]q2, which, by Corollary 6 and Lemma 16, is a polynomial in qwith non-negative integer coefficients. If m≡2 (mod 84),then we have Catm(E8;q) = [15m+1]q2/bracketleftbig15m+4 2/bracketrightbig q4/bracketleftbig5m+2 12/bracketrightbig q72[12]q6 [3]q6[4]q6[15m+7]q2[5m+3]q6 ×/bracketleftbig3m+2 4/bracketrightbig q40[10]q4 [2]q4[5]q4/bracketleftbig5m+4 14/bracketrightbig q84[42]q2 [6]q2[7]q2[m+1]q30,48 C. KRATTENTHALER AND T. W. M ¨ULLER which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡3 (mod 84),then we have Catm(E8;q) =/bracketleftbig15m+1 2/bracketrightbig q4/bracketleftbig15m+4 7/bracketrightbig q14[5m+2]q6/bracketleftbig15m+7 4/bracketrightbig q8/bracketleftbig5m+3 6/bracketrightbig q36[6]q6 [2]q6[3]q6 ×[3m+2]q10[5m+4]q6/bracketleftbigm+1 4/bracketrightbig q120[60]q2[2]q2[3]q2[5]q2 [10]q2[12]q2[15]q2, which, by Corollary 6 and Lemma 19, is a polynomial in qwith non-negative integer coefficients. If m≡4 (mod 84),then we have Catm(E8;q) = [15m+1]q2/bracketleftbig15m+4 4/bracketrightbig q8/bracketleftbig5m+2 2/bracketrightbig q12 ×[15m+7]q2[5m+3]q6/bracketleftbig3m+2 14/bracketrightbig q140[70]q2 [7]q2[10]q2/bracketleftbig5m+4 12/bracketrightbig q72[12]q6 [3]q6[4]q6[m+1]q30, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡5 (mod 84),then we have Catm(E8;q) =/bracketleftbig15m+1 4/bracketrightbig q8[15m+4]q2/bracketleftbig5m+2 3/bracketrightbig q18/bracketleftbig15m+7 2/bracketrightbig q4/bracketleftbig5m+3 28/bracketrightbig q168[84]q2 [7]q2[12]q2 ×[3m+2]q10[5m+4]q6/bracketleftbigm+1 2/bracketrightbig q60[30]q2[2]q2[3]q2[5]q2 [6]q2[10]q2[15]q2, which, by Corollary 6 and Lemma 16, is a polynomial in qwith non-negative integer coefficients. If m≡6 (mod 84),then we have Catm(E8;q) =/bracketleftbig15m+1 7/bracketrightbig q14/bracketleftbig15m+4 2/bracketrightbig q4/bracketleftbig5m+2 4/bracketrightbig q24[15m+7]q2 ×/bracketleftbig5m+3 3/bracketrightbig q18/bracketleftbig3m+2 4/bracketrightbig q40[10]q4 [2]q4[5]q4/bracketleftbig5m+4 2/bracketrightbig q12[m+1]q30, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡7 (mod 84),then we have Catm(E8;q) =/bracketleftbig15m+1 2/bracketrightbig q4[15m+4]q2[5m+2]q6/bracketleftbig15m+7 28/bracketrightbig q56[28]q2 [4]q2[7]q2/bracketleftbig5m+3 2/bracketrightbig q12 ×[3m+2]q10/bracketleftbig5m+4 3/bracketrightbig q18/bracketleftbigm+1 4/bracketrightbig q120[60]q2[2]q2[3]q2[5]q2 [10]q2[12]q2[15]q2, which, by Corollary 6 and Lemma 19, is a polynomial in qwith non-negative integer coefficients. If m≡8 (mod 84),then we have Catm(E8;q) = [15m+1]q2/bracketleftbig15m+4 4/bracketrightbig q8/bracketleftbig5m+2 42/bracketrightbig q252[126]q2[3]q2 [6]q2[7]q2[9]q2[15m+7]q2[5m+3]q6 ×/bracketleftbig3m+2 2/bracketrightbig q20/bracketleftbig5m+4 4/bracketrightbig q24[m+1]q30,CYCLIC SIEVING FOR GENERALISED NON-CROSSING PARTITIONS 49 which, by Lemma 22, is a polynomial in qwith non-negative integer coefficients. If m≡9 (mod 84),then we have Catm(E8;q) =/bracketleftbig15m+1 4/bracketrightbig q8[15m+4]q2[5m+2]q6/bracketleftbig15m+7 2/bracketrightbig q4/bracketleftbig5m+3 12/bracketrightbig q72[12]q6 [3]q6[4]q6 ×[3m+2]q10/bracketleftbig5m+4 7/bracketrightbig q42[21]q2 [3]q2[7]q2/bracketleftbigm+1 2/bracketrightbig q60[30]q2[2]q2[3]q2[5]q2 [6]q2[10]q2[15]q2, which, by Corollary 6 and Lemma 16, is a polynomial in qwith non-negative integer coefficients. If m≡10 (mod 84) ,then we have Catm(E8;q) = [15m+1]q2/bracketleftbig15m+4 14/bracketrightbig q28[14]q2 [2]q2[7]q2/bracketleftbig5m+2 4/bracketrightbig q24 ×[15m+7]q2[5m+3]q6/bracketleftbig3m+2 4/bracketrightbig q40[10]q4 [2]q4[5]q4/bracketleftbig5m+4 6/bracketrightbig q36[6]q6 [2]q6[3]q6[m+1]q30. If one decomposes [15 m+ 7]q2as [15m 2+ 4]q4+q2[15m 2+ 3]q4, then one sees that, by Corollary 6, this is a polynomial in qwith non-negative integer coefficients. If m≡ 11 (mod 84) ,then we have Catm(E8;q) =/bracketleftbig15m+1 2/bracketrightbig q4[15m+4]q2/bracketleftbig5m+2 3/bracketrightbig q18/bracketleftbig15m+7 4/bracketrightbig q8/bracketleftbig5m+3 2/bracketrightbig q12 ×/bracketleftbig3m+2 7/bracketrightbig q70[35]q2 [5]q2[7]q2[5m+4]q6/bracketleftbigm+1 4/bracketrightbig q120[60]q2[2]q2[3]q2[5]q2 [10]q2[12]q2[15]q2, which, by Corollary 6 and Lemma 20, is a polynomial in qwith non-negative integer coefficients. If m≡12 (mod 84) ,then we have Catm(E8;q) = [15m+1]q2/bracketleftbig15m+4 4/bracketrightbig q8/bracketleftbig5m+2 2/bracketrightbig q12 ×[15m+7]q2/bracketleftbig5m+3 21/bracketrightbig q126[63]q2 [7]q2[9]q2/bracketleftbig3m+2 2/bracketrightbig q20/bracketleftbig5m+4 4/bracketrightbig q24[m+1]q30, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡13 (mod 84) ,then we have Catm(E8;q) =/bracketleftbig15m+1 28/bracketrightbig q56[28]q2 [4]q2[7]q2[15m+4]q2[5m+2]q6/bracketleftbig15m+7 2/bracketrightbig q4/bracketleftbig5m+3 4/bracketrightbig q24 ×[3m+2]q10/bracketleftbig5m+4 3/bracketrightbig q18/bracketleftbigm+1 2/bracketrightbig q60[30]q2[2]q2[3]q2[5]q2 [6]q2[10]q2[15]q2, which, by Corollary 6 and Lemma 16, is a polynomial in qwith non-negative integer coefficients. If m≡14 (mod 84) ,then we have Catm(E8;q) = [15m+1]q2/bracketleftbig15m+4 2/bracketrightbig q4/bracketleftbig5m+2 12/bracketrightbig q72[12]q6 [3]q6[4]q6/bracketleftbig15m+7 7/bracketrightbig q14 ×[5m+3]q6/bracketleftbig3m+2 4/bracketrightbig q40[10]q4 [2]q4[5]q4/bracketleftbig5m+4 2/bracketrightbig q12[m+1]q30,50 C. KRATTENTHALER AND T. W. M ¨ULLER which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡15 (mod 84) ,then we have Catm(E8;q) =/bracketleftbig15m+1 2/bracketrightbig q4[15m+4]q2/bracketleftbig5m+2 7/bracketrightbig q42[21]q2 [3]q2[7]q2/bracketleftbig15m+7 4/bracketrightbig q8/bracketleftbig5m+3 6/bracketrightbig q36[6]q6 [2]q6[3]q6 ×[3m+2]q10[5m+4]q6/bracketleftbigm+1 4/bracketrightbig q120[60]q2[2]q2[3]q2[5]q2 [10]q2[12]q2[15]q2, which, by Corollary 6 and Lemma 19, is a polynomial in qwith non-negative integer coefficients. If m≡16 (mod 84) ,then we have Catm(E8;q) = [15m+1]q2/bracketleftbig15m+4 4/bracketrightbig q8/bracketleftbig5m+2 2/bracketrightbig q12[15m+7]q2[5m+3]q6 ×/bracketleftbig3m+2 2/bracketrightbig q20/bracketleftbig5m+4 84/bracketrightbig q504[252]q2[3]q2 [7]q2[9]q2[12]q2[m+1]q30, which, by Lemma 23, is a polynomial in qwith non-negative integer coefficients. If m≡17 (mod 84) ,then we have Catm(E8;q) =/bracketleftbig15m+1 4/bracketrightbig q8/bracketleftbig15m+4 7/bracketrightbig q14/bracketleftbig5m+2 3/bracketrightbig q18/bracketleftbig15m+7 2/bracketrightbig q4/bracketleftbig5m+3 4/bracketrightbig q24 ×[3m+2]q10[5m+4]q6/bracketleftbigm+1 2/bracketrightbig q60[30]q2[2]q2[3]q2[5]q2 [6]q2[10]q2[15]q2, which, by Lemma 16, is a polynomial in qwith non-negative integer coefficients. If m≡18 (mod 84) ,then we have Catm(E8;q) = [15m+1]q2/bracketleftbig15m+4 2/bracketrightbig q4/bracketleftbig5m+2 4/bracketrightbig q24[15m+7]q2/bracketleftbig5m+3 3/bracketrightbig q18 /bracketleftbig3m+2 28/bracketrightbig q280[140]q2[2]q2 [4]q2[7]q2[10]q2/bracketleftbig5m+4 2/bracketrightbig q12[m+1]q30, which, by Lemma 24, is a polynomial in qwith non-negative integer coefficients. If m≡19 (mod 84) ,then we have Catm(E8;q) =/bracketleftbig15m+1 2/bracketrightbig q4[15m+4]q2[5m+2]q6/bracketleftbig15m+7 4/bracketrightbig q8/bracketleftbig5m+3 14/bracketrightbig q84[42]q2 [6]q2[7]q2 ×[3m+2]q10/bracketleftbig5m+4 3/bracketrightbig q18/bracketleftbigm+1 4/bracketrightbig q120[60]q2[2]q2[3]q2[5]q2 [10]q2[12]q2[15]q2, which, by Corollary 6 and Lemma 19, is a polynomial in qwith non-negative integer coefficients. If m≡20 (mod 84) ,then we have Catm(E8;q) =/bracketleftbig15m+1 7/bracketrightbig q14/bracketleftbig15m+4 4/bracketrightbig q8/bracketleftbig5m+2 6/bracketrightbig q36[6]q6 [2]q6[3]q6[15m+7]q2[5m+3]q6 ×/bracketleftbig3m+2 2/bracketrightbig q20/bracketleftbig5m+4 4/bracketrightbig q24[m+1]q30,CYCLIC SIEVING FOR GENERALISED NON-CROSSING PARTITIONS 51 which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡21 (mod 84) ,then we have Catm(E8;q) =/bracketleftbig15m+1 4/bracketrightbig q8[15m+4]q2[5m+2]q6/bracketleftbig15m+7 14/bracketrightbig q28[14]q2 [2]q2[7]q2/bracketleftbig5m+3 12/bracketrightbig q72[12]q6 [3]q6[4]q6 ×[3m+2]q10[5m+4]q6/bracketleftbigm+1 2/bracketrightbig q60[30]q2[2]q2[3]q2[5]q2 [6]q2[10]q2[15]q2, which, by Corollary 6 and Lemma 17, is a polynomial in qwith non-negative integer coefficients. If m≡22 (mod 84) ,then we have Catm(E8;q) = [15m+1]q2/bracketleftbig15m+4 2/bracketrightbig q4/bracketleftbig5m+2 28/bracketrightbig q168[84]q2 [7]q2[12]q2[15m+7]q2 ×[5m+3]q6/bracketleftbig3m+2 4/bracketrightbig q40[10]q4 [2]q4[5]q4/bracketleftbig5m+4 6/bracketrightbig q36[6]q6 [2]q6[3]q6[m+1]q30, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡23 (mod 84) ,then we have Catm(E8;q) =/bracketleftbig15m+1 2/bracketrightbig q4[15m+4]q2/bracketleftbig5m+2 3/bracketrightbig q18/bracketleftbig15m+7 4/bracketrightbig q8/bracketleftbig5m+3 2/bracketrightbig q12 ×[3m+2]q10/bracketleftbig5m+4 7/bracketrightbig q42[21]q2 [3]q2[7]q2/bracketleftbigm+1 4/bracketrightbig q120[60]q2[2]q2[3]q2[5]q2 [10]q2[12]q2[15]q2, which, by Corollary 6 and Lemma 19, is a polynomial in qwith non-negative integer coefficients. If m≡24 (mod 84) ,then we have Catm(E8;q) = [15m+1]q2/bracketleftbig15m+4 28/bracketrightbig q56[28]q2 [4]q2[7]q2/bracketleftbig5m+2 2/bracketrightbig q12[15m+7]q2 ×/bracketleftbig5m+3 3/bracketrightbig q18/bracketleftbig3m+2 2/bracketrightbig q20/bracketleftbig5m+4 4/bracketrightbig q24[m+1]q30, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡25 (mod 84) ,then we have Catm(E8;q) =/bracketleftbig15m+1 4/bracketrightbig q8[15m+4]q2[5m+2]q6/bracketleftbig15m+7 2/bracketrightbig q4/bracketleftbig5m+3 4/bracketrightbig q24 ×/bracketleftbig3m+2 7/bracketrightbig q70[35]q2 [5]q2[7]q2/bracketleftbig5m+4 3/bracketrightbig q18/bracketleftbigm+1 2/bracketrightbig q60[30]q2[2]q2[3]q2[5]q2 [6]q2[10]q2[15]q2, which, by Lemma 18, is a polynomial in qwith non-negative integer coefficients. If m≡26 (mod 84) ,then we have Catm(E8;q) = [15m+1]q2/bracketleftbig15m+4 2/bracketrightbig q4/bracketleftbig5m+2 12/bracketrightbig q72[12]q6 [3]q6[4]q6[15m+7]q2/bracketleftbig5m+3 7/bracketrightbig q42[21]q2 [3]q2[7]q2 ×/bracketleftbig3m+2 4/bracketrightbig q40[10]q4 [2]q4[5]q4/bracketleftbig5m+4 2/bracketrightbig q12[m+1]q30. If one decomposes [15 m+1]q2as [5m+1]q6+q2[5m]q6+q4[5m]q6, then one sees that, by Corollary 6, this is a polynomial in qwith non-negative integer coefficients. If52 C. KRATTENTHALER AND T. W. M ¨ULLER m≡27 (mod 84) ,then we have Catm(E8;q) =/bracketleftbig15m+1 14/bracketrightbig q28[14]q2 [2]q2[7]q2[15m+4]q2[5m+2]q6/bracketleftbig15m+7 4/bracketrightbig q8/bracketleftbig5m+3 6/bracketrightbig q36[6]q6 [2]q6[3]q6 ×[3m+2]q10[5m+4]q6/bracketleftbigm+1 4/bracketrightbig q120[60]q2[2]q2[3]q2[5]q2 [10]q2[12]q2[15]q2, which, by Corollary 6 and Lemma 21, is a polynomial in qwith non-negative integer coefficients. If m≡28 (mod 84) ,then we have Catm(E8;q) = [15m+1]q2/bracketleftbig15m+4 4/bracketrightbig q8/bracketleftbig5m+2 2/bracketrightbig q12/bracketleftbig15m+7 7/bracketrightbig q14 ×[5m+3]q6/bracketleftbig3m+2 2/bracketrightbig q20/bracketleftbig5m+4 12/bracketrightbig q72[12]q6 [3]q6[4]q6[m+1]q30, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡29 (mod 84) ,then we have Catm(E8;q) =/bracketleftbig15m+1 4/bracketrightbig q8[15m+4]q2/bracketleftbig5m+2 21/bracketrightbig q126[63]q2 [7]q2[9]q2/bracketleftbig15m+7 2/bracketrightbig q4/bracketleftbig5m+3 4/bracketrightbig q24 ×[3m+2]q10[5m+4]q6/bracketleftbigm+1 2/bracketrightbig q60[30]q2[2]q2[3]q2[5]q2 [6]q2[10]q2[15]q2, which, by Corollary 6 and Lemma 16, is a polynomial in qwith non-negative integer coefficients. If m≡30 (mod 84) ,then we have Catm(E8;q) = [15m+1]q2/bracketleftbig15m+4 2/bracketrightbig q4/bracketleftbig5m+2 4/bracketrightbig q24[15m+7]q2/bracketleftbig5m+3 3/bracketrightbig q18 /bracketleftbig3m+2 4/bracketrightbig q40[10]q4 [2]q4[5]q4/bracketleftbig5m+4 14/bracketrightbig q84[42]q2 [6]q2[7]q2[m+1]q30, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡31 (mod 84) ,then we have Catm(E8;q) =/bracketleftbig15m+1 2/bracketrightbig q4/bracketleftbig15m+4 7/bracketrightbig q14[5m+2]q6/bracketleftbig15m+7 4/bracketrightbig q8/bracketleftbig5m+3 2/bracketrightbig q12 ×[3m+2]q10/bracketleftbig5m+4 3/bracketrightbig q18/bracketleftbigm+1 4/bracketrightbig q120[60]q2[2]q2[3]q2[5]q2 [10]q2[12]q2[15]q2, which, by Lemma 19, is a polynomial in qwith non-negative integer coefficients. If m≡32 (mod 84) ,then we have Catm(E8;q) = [15m+1]q2/bracketleftbig15m+4 4/bracketrightbig q8/bracketleftbig5m+2 6/bracketrightbig q36[6]q6 [2]q6[3]q6[15m+7]q2[5m+3]q6 ×/bracketleftbig3m+2 14/bracketrightbig q140[70]q2 [7]q2[10]q2/bracketleftbig5m+4 4/bracketrightbig q24[m+1]q30,CYCLIC SIEVING FOR GENERALISED NON-CROSSING PARTITIONS 53 which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡33 (mod 84) ,then we have Catm(E8;q) =/bracketleftbig15m+1 4/bracketrightbig q8[15m+4]q2[5m+2]q6/bracketleftbig15m+7 2/bracketrightbig q4/bracketleftbig5m+3 84/bracketrightbig q504[252]q2[3]q2 [7]q2[9]q2[12]q2 ×[3m+2]q10[5m+4]q6/bracketleftbigm+1 2/bracketrightbig q60[30]q2[2]q2[3]q2[5]q2 [6]q2[10]q2[15]q2, which, by Lemmas 16and23, is apolynomial in qwithnon-negative integer coefficients. Ifm≡34 (mod 84) ,then we have Catm(E8;q) =/bracketleftbig15m+1 7/bracketrightbig q14/bracketleftbig15m+4 2/bracketrightbig q4/bracketleftbig5m+2 4/bracketrightbig q24[15m+7]q2 [5m+3]q6/bracketleftbig3m+2 4/bracketrightbig q40[10]q4 [2]q4[5]q4/bracketleftbig5m+4 6/bracketrightbig q36[6]q6 [2]q6[3]q6[m+1]q30, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡35 (mod 84) ,then we have Catm(E8;q) =/bracketleftbig15m+1 2/bracketrightbig q4[15m+4]q2/bracketleftbig5m+2 3/bracketrightbig q18/bracketleftbig15m+7 28/bracketrightbig q56[28]q2 [4]q2[7]q2/bracketleftbig5m+3 2/bracketrightbig q12 ×[3m+2]q10[5m+4]q6/bracketleftbigm+1 4/bracketrightbig q120[60]q2[2]q2[3]q2[5]q2 [10]q2[12]q2[15]q2, which, by Corollary 6 and Lemma 19, is a polynomial in qwith non-negative integer coefficients. If m≡36 (mod 84) ,then we have Catm(E8;q) = [15m+1]q2/bracketleftbig15m+4 4/bracketrightbig q8/bracketleftbig5m+2 14/bracketrightbig q84[42]q2 [6]q2[7]q2[15m+7]q2/bracketleftbig5m+3 3/bracketrightbig q18 ×/bracketleftbig3m+2 2/bracketrightbig q20/bracketleftbig5m+4 4/bracketrightbig q24[m+1]q30, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡37 (mod 84) ,then we have Catm(E8;q) =/bracketleftbig15m+1 4/bracketrightbig q8[15m+4]q2[5m+2]q6/bracketleftbig15m+7 2/bracketrightbig q4/bracketleftbig5m+3 4/bracketrightbig q24 ×[3m+2]q10/bracketleftbigg5m+4 21/bracketrightbigg q126[63]q2 [7]q2[9]q2/bracketleftbigm+1 2/bracketrightbig q60[30]q2[2]q2[3]q2[5]q2 [6]q2[10]q2[15]q2, which, by Corollary 6 and Lemma 16, is a polynomial in qwith non-negative integer coefficients. If m≡38 (mod 84) ,then we have Catm(E8;q) = [15m+1]q2/bracketleftbig15m+4 14/bracketrightbig q28[14]q2 [2]q2[7]q2/bracketleftbig5m+2 12/bracketrightbig q72[12]q6 [3]q6[4]q6 ×[15m+7]q2[5m+3]q6/bracketleftbig3m+2 4/bracketrightbig q40[10]q4 [2]q4[5]q4/bracketleftbig5m+4 2/bracketrightbig q12[m+1]q30. If one decomposes [15 m+ 7]q2as [15m 2+ 4]q4+q2[15m 2+ 3]q4, then one sees that, by Corollary 6, this is a polynomial in qwith non-negative integer coefficients. If m≡54 C. KRATTENTHALER AND T. W. M ¨ULLER 39 (mod 84) ,then we have Catm(E8;q) =/bracketleftbig15m+1 2/bracketrightbig q4[15m+4]q2[5m+2]q6/bracketleftbig15m+7 4/bracketrightbig q8/bracketleftbig5m+3 6/bracketrightbig q36[6]q6 [2]q6[3]q6 ×/bracketleftbig3m+2 7/bracketrightbig q70[35]q2 [5]q2[7]q2[5m+4]q6/bracketleftbigm+1 4/bracketrightbig q120[60]q2[2]q2[3]q2[5]q2 [10]q2[12]q2[15]q2, which, by Corollary 6 and Lemma 20, is a polynomial in qwith non-negative integer coefficients. If m≡40 (mod 84) ,then we have Catm(E8;q) = [15m+1]q2/bracketleftbig15m+4 4/bracketrightbig q8/bracketleftbig5m+2 2/bracketrightbig q12[15m+7]q2/bracketleftbig5m+3 7/bracketrightbig q42[21]q2 [3]q2[7]q2 ×/bracketleftbig3m+2 2/bracketrightbig q20/bracketleftbig5m+4 12/bracketrightbig q72[12]q6 [3]q6[4]q6[m+1]q30. If one decomposes [15 m+7]q2as [5m+1]q6+q2[5m]q6+q4[5m]q6, then one sees that, by Corollary 6, this is a polynomial in qwith non-negative integer coefficients. If m≡41 (mod 84) ,then we have Catm(E8;q) =/bracketleftbig15m+1 28/bracketrightbig q56[28]q2 [4]q2[7]q2[15m+4]q2/bracketleftbig5m+2 3/bracketrightbig q18/bracketleftbig15m+7 2/bracketrightbig q4/bracketleftbig5m+3 4/bracketrightbig q24 ×[3m+2]q10[5m+4]q6/bracketleftbigm+1 2/bracketrightbig q60[30]q2[2]q2[3]q2[5]q2 [6]q2[10]q2[15]q2, which, by Corollary 6 and Lemma 16, is a polynomial in qwith non-negative integer coefficients. If m≡42 (mod 84) ,then we have Catm(E8;q) = [15m+1]q2/bracketleftbig15m+4 2/bracketrightbig q4/bracketleftbig5m+2 4/bracketrightbig q24/bracketleftbig15m+7 7/bracketrightbig q14 ×/bracketleftbig5m+3 3/bracketrightbig q18/bracketleftbig3m+2 4/bracketrightbig q40[10]q4 [2]q4[5]q4/bracketleftbig5m+4 2/bracketrightbig q12[m+1]q30, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡43 (mod 84) ,then we have Catm(E8;q) =/bracketleftbig15m+1 2/bracketrightbig q4[15m+4]q2/bracketleftbig5m+2 7/bracketrightbig q42[21]q2 [3]q2[7]q2/bracketleftbig15m+7 4/bracketrightbig q8/bracketleftbig5m+3 2/bracketrightbig q12 ×[3m+2]q10/bracketleftbig5m+4 3/bracketrightbig q18/bracketleftbigm+1 4/bracketrightbig q120[60]q2[2]q2[3]q2[5]q2 [10]q2[12]q2[15]q2, which, by Corollary 6 and Lemma 19, is a polynomial in qwith non-negative integer coefficients. If m≡44 (mod 84) ,then we have Catm(E8;q) = [15m+1]q2/bracketleftbig15m+4 4/bracketrightbig q8/bracketleftbig5m+2 6/bracketrightbig q36[6]q6 [2]q6[3]q6[15m+7]q2[5m+3]q6 ×/bracketleftbig3m+2 2/bracketrightbig q20/bracketleftbig5m+4 28/bracketrightbig q168[84]q2 [7]q2[12]q2[m+1]q30,CYCLIC SIEVING FOR GENERALISED NON-CROSSING PARTITIONS 55 which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡45 (mod 84) ,then we have Catm(E8;q) =/bracketleftbig15m+1 4/bracketrightbig q8/bracketleftbig15m+4 7/bracketrightbig q14[5m+2]q6/bracketleftbig15m+7 2/bracketrightbig q4/bracketleftbig5m+3 12/bracketrightbig q72[12]q6 [3]q6[4]q6 ×[3m+2]q10[5m+4]q6/bracketleftbigm+1 2/bracketrightbig q60[30]q2[2]q2[3]q2[5]q2 [6]q2[10]q2[15]q2, which, by Corollary 6 and Lemma 16, is a polynomial in qwith non-negative integer coefficients. If m≡46 (mod 84) ,then we have Catm(E8;q) = [15m+1]q2/bracketleftbig15m+4 2/bracketrightbig q4/bracketleftbig5m+2 4/bracketrightbig q24[15m+7]q2 [5m+3]q6/bracketleftbig3m+2 28/bracketrightbig q280[140]q2[2]q2 [4]q2[7]q2[10]q2/bracketleftbig5m+4 6/bracketrightbig q36[6]q6 [2]q6[3]q6[m+1]q30, which, by Corollary 6 and Lemma 24, is a polynomial in qwith non-negative integer coefficients. If m≡47 (mod 84) ,then we have Catm(E8;q) =/bracketleftbig15m+1 2/bracketrightbig q4[15m+4]q2/bracketleftbig5m+2 3/bracketrightbig q18/bracketleftbig15m+7 4/bracketrightbig q8/bracketleftbig5m+3 14/bracketrightbig q84[42]q2 [6]q2[7]q2 ×[3m+2]q10[5m+4]q6/bracketleftbigm+1 4/bracketrightbig q120[60]q2[2]q2[3]q2[5]q2 [10]q2[12]q2[15]q2, which, by Corollary 6 and Lemma 19, is a polynomial in qwith non-negative integer coefficients. If m≡48 (mod 84) ,then we have Catm(E8;q) =/bracketleftbig15m+1 7/bracketrightbig q14/bracketleftbig15m+4 4/bracketrightbig q8/bracketleftbig5m+2 2/bracketrightbig q12[15m+7]q2 /bracketleftbig5m+3 3/bracketrightbig q18/bracketleftbig3m+2 2/bracketrightbig q20/bracketleftbig5m+4 4/bracketrightbig q24[m+1]q30, which is manifestly a polynomial in qwith non-negative integer coefficients. If m≡ 49 (mod 84) ,then we have Catm(E8;q) =/bracketleftbig15m+1 4/bracketrightbig q8[15m+4]q2[5m+2]q6/bracketleftbig15m+7 14/bracketrightbig q28[14]q2 [2]q2[7]q2/bracketleftbig5m+3 4/bracketrightbig q24 ×[3m+2]q10/bracketleftbig5m+4 3/bracketrightbig q18/bracketleftbigm+1 2/bracketrightbig q60[30]q2[2]q2[3]q2[5]q2 [6]q2[10]q2[15]q2, which, by Corollary 6 and Lemma 16, is a polynomial in qwith non-negative integer coefficients. If m≡50 (mod 84) ,then we have Catm(E8;q) = [15m+1]q2/bracketleftbig15m+4 2/bracketrightbig q4/bracketleftbig5m+2 28/bracketrightbig q168[84]q2 [7]q2[12]q2[15m+7]q2 ×[5m+3]q6/bracketleftbig3m+2 4/bracketrightbig q40[10]q4 [2]q4[5]q4/bracketleftbig5m+4 2/bracketrightbig q12/bracketleftbigm+1 3/bracketrightbig q90[15]q6 [3]q6[5]q6,56 C. KRATTENTHALER AND T. W. M ¨ULLER which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡51 (mod 84) ,then we have Catm(E8;q) =/bracketleftbig15m+1 2/bracketrightbig q4[15m+4]q2[5m+2]q6/bracketleftbig15m+7 4/bracketrightbig q8/bracketleftbig5m+3 6/bracketrightbig q36[6]q6 [2]q6[3]q6 ×[3m+2]q10/bracketleftbig5m+4 7/bracketrightbig q42[21]q2 [3]q2[7]q2/bracketleftbigm+1 4/bracketrightbig q120[60]q2[2]q2[3]q2[5]q2 [10]q2[12]q2[15]q2, which, by Corollary 6 and Lemma 19, is a polynomial in qwith non-negative integer coefficients. If m≡52 (mod 84) ,then we have Catm(E8;q) = [15m+1]q2/bracketleftbig15m+4 28/bracketrightbig q56[28]q2 [4]q2[7]q2/bracketleftbig5m+2 2/bracketrightbig q12[15m+7]q2 ×[5m+3]q6/bracketleftbig3m+2 2/bracketrightbig q20/bracketleftbig5m+4 12/bracketrightbig q72[12]q6 [3]q6[4]q6[m+1]q30, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡53 (mod 84) ,then we have Catm(E8;q) =/bracketleftbig15m+1 4/bracketrightbig q8[15m+4]q2/bracketleftbig5m+2 3/bracketrightbig q18/bracketleftbig15m+7 2/bracketrightbig q4/bracketleftbig5m+3 4/bracketrightbig q24 ×/bracketleftbig3m+2 7/bracketrightbig q70[35]q2 [5]q2[7]q2[5m+4]q6/bracketleftbigm+1 2/bracketrightbig q60[30]q2[2]q2[3]q2[5]q2 [6]q2[10]q2[15]q2, which, by Lemma 18, is a polynomial in qwith non-negative integer coefficients. If m≡54 (mod 84) ,then we have Catm(E8;q) = [15m+1]q2/bracketleftbig15m+4 2/bracketrightbig q4/bracketleftbig5m+2 4/bracketrightbig q24[15m+7]q2 /bracketleftbig5m+3 21/bracketrightbig q126[63]q2 [7]q2[9]q2/bracketleftbig3m+2 4/bracketrightbig q40[10]q4 [2]q4[5]q4/bracketleftbig5m+4 2/bracketrightbig q12[m+1]q30, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡55 (mod 84) ,then we have Catm(E8;q) =/bracketleftbig15m+1 14/bracketrightbig q28[14]q2 [2]q2[7]q2[15m+4]q2[5m+2]q6/bracketleftbig15m+7 4/bracketrightbig q8/bracketleftbig5m+3 2/bracketrightbig q12 ×[3m+2]q10/bracketleftbig5m+4 3/bracketrightbig q18/bracketleftbigm+1 4/bracketrightbig q120[60]q2[2]q2[3]q2[5]q2 [10]q2[12]q2[15]q2, which, by Lemma 21, is a polynomial in qwith non-negative integer coefficients. If m≡56 (mod 84) ,then we have Catm(E8;q) = [15m+1]q2/bracketleftbig15m+4 4/bracketrightbig q8/bracketleftbig5m+2 6/bracketrightbig q36[6]q6 [2]q6[3]q6/bracketleftbig15m+7 7/bracketrightbig q14[5m+3]q6 ×/bracketleftbig3m+2 2/bracketrightbig q20/bracketleftbig5m+4 4/bracketrightbig q24[m+1]q30,CYCLIC SIEVING FOR GENERALISED NON-CROSSING PARTITIONS 57 which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡57 (mod 84) ,then we have Catm(E8;q) =/bracketleftbig15m+1 4/bracketrightbig q8[15m+4]q2/bracketleftbig5m+2 7/bracketrightbig q42[21]q2 [3]q2[7]q2/bracketleftbig15m+7 2/bracketrightbig q4/bracketleftbig5m+3 12/bracketrightbig q72[12]q6 [3]q6[4]q6 ×[3m+2]q10[5m+4]q6/bracketleftbigm+1 2/bracketrightbig q60[30]q2[2]q2[3]q2[5]q2 [6]q2[10]q2[15]q2, which, by Corollary 6 and Lemma 16, is a polynomial in qwith non-negative integer coefficients. If m≡58 (mod 84) ,then we have Catm(E8;q) = [15m+1]q2/bracketleftbigm+4 2/bracketrightbig q4/bracketleftbig5m+2 4/bracketrightbig q24[15m+7]q2[5m+3]q6 /bracketleftbig3m+2 4/bracketrightbig q40[10]q4 [2]q4[5]q4/bracketleftbig5m+4 42/bracketrightbig q252[126]q2[3]q2 [6]q2[7]q2[9]q2[m+1]q30, which, by Lemma 22, is a polynomial in qwith non-negative integer coefficients. If m≡59 (mod 84) ,then we have Catm(E8;q) =/bracketleftbig15m+1 2/bracketrightbig q4/bracketleftbig15m+4 7/bracketrightbig q14/bracketleftbig5m+2 3/bracketrightbig q18/bracketleftbig15m+7 4/bracketrightbig q8/bracketleftbig5m+3 2/bracketrightbig q12 ×[3m+2]q10[5m+4]q6/bracketleftbigm+1 4/bracketrightbig q120[60]q2[2]q2[3]q2[5]q2 [10]q2[12]q2[15]q2, which, by Lemma 19, is a polynomial in qwith non-negative integer coefficients. If m≡60 (mod 84) ,then we have Catm(E8;q) = [15m+1]q2/bracketleftbig15m+4 4/bracketrightbig q8/bracketleftbig5m+2 2/bracketrightbig q12[15m+7]q2/bracketleftbig5m+3 3/bracketrightbig q18 ×/bracketleftbig3m+2 14/bracketrightbig q140[70]q2 [7]q2[10]q2/bracketleftbig5m+4 4/bracketrightbig q24[m+1]q30, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡61 (mod 84) ,then we have Catm(E8;q) =/bracketleftbig15m+1 4/bracketrightbig q8[15m+4]q2[5m+2]q6/bracketleftbig15m+7 2/bracketrightbig q4/bracketleftbig5m+3 28/bracketrightbig q168[84]q2 [7]q2[12]q2 ×[3m+2]q10/bracketleftbig5m+4 3/bracketrightbig q18/bracketleftbigm+1 2/bracketrightbig q60[30]q2[2]q2[3]q2[5]q2 [6]q2[10]q2[15]q2, which, by Corollary 6 and Lemma 16, is a polynomial in qwith non-negative integer coefficients. If m≡62 (mod 84) ,then we have Catm(E8;q) =/bracketleftbig15m+1 7/bracketrightbig q14/bracketleftbig15m+4 2/bracketrightbig q4/bracketleftbig5m+2 12/bracketrightbig q72[12]q6 [3]q6[4]q6[15m+7]q2[5m+3]q6 ×/bracketleftbig3m+2 4/bracketrightbig q40[10]q4 [2]q4[5]q4/bracketleftbig5m+4 2/bracketrightbig q12[m+1]q30,58 C. KRATTENTHALER AND T. W. M ¨ULLER which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡63 (mod 84) ,then we have Catm(E8;q) =/bracketleftbig15m+1 2/bracketrightbig q4[15m+4]q2[5m+2]q6/bracketleftbig15m+7 28/bracketrightbig q56[28]q2 [4]q2[7]q2/bracketleftbig5m+3 6/bracketrightbig q36[6]q6 [2]q6[3]q6 ×[3m+2]q10[5m+4]q6/bracketleftbigm+1 4/bracketrightbig q120[60]q2[2]q2[3]q2[5]q2 [10]q2[12]q2[15]q2, which, by Corollary 6 and Lemma 19, is a polynomial in qwith non-negative integer coefficients. If m≡64 (mod 84) ,then we have Catm(E8;q) = [15m+1]q2/bracketleftbig15m+4 4/bracketrightbig q8/bracketleftbig5m+2 14/bracketrightbig q84[42]q2 [6]q2[7]q2[15m+7]q2 ×[5m+3]q6/bracketleftbig3m+2 2/bracketrightbig q20/bracketleftbig5m+4 12/bracketrightbig q72[12]q6 [3]q6[4]q6[m+1]q30, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡65 (mod 84) ,then we have Catm(E8;q) =/bracketleftbig15m+1 4/bracketrightbig q8[15m+4]q2/bracketleftbig5m+2 3/bracketrightbig q18/bracketleftbig15m+7 2/bracketrightbig q4/bracketleftbig5m+3 4/bracketrightbig q24 ×[3m+2]q10/bracketleftbig5m+4 7/bracketrightbig q42[21]q2 [3]q2[7]q2/bracketleftbigm+1 2/bracketrightbig q60[30]q2[2]q2[3]q2[5]q2 [6]q2[10]q2[15]q2, which, by Corollary 6 and Lemma 16, is a polynomial in qwith non-negative integer coefficients. If m≡66 (mod 84) ,then we have Catm(E8;q) = [15m+1]q2/bracketleftbig15m+4 14/bracketrightbig q28[14]q2 [2]q2[7]q2/bracketleftbig5m+2 4/bracketrightbig q24[15m+7]q2 ×/bracketleftbig5m+3 3/bracketrightbig q18/bracketleftbig3m+2 4/bracketrightbig q40[10]q4 [2]q4[5]q4/bracketleftbig5m+4 2/bracketrightbig q12[m+1]q30, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡67 (mod 84) ,then we have Catm(E8;q) =/bracketleftbig15m+1 2/bracketrightbig q4[15m+4]q2[5m+2]q6/bracketleftbig15m+7 4/bracketrightbig q8/bracketleftbig5m+3 2/bracketrightbig q12 ×/bracketleftbig3m+2 7/bracketrightbig q70[35]q2 [5]q2[7]q2/bracketleftbig5m+4 3/bracketrightbig q18/bracketleftbigm+1 4/bracketrightbig q120[60]q2[2]q2[3]q2[5]q2 [10]q2[12]q2[15]q2, which, by Lemma 20, is a polynomial in qwith non-negative integer coefficients. If m≡68 (mod 84) ,then we have Catm(E8;q) = [15m+1]q2/bracketleftbig15m+4 4/bracketrightbig q8/bracketleftbig5m+2 6/bracketrightbig q36[6]q6 [2]q6[3]q6 ×[15m+7]q2/bracketleftbig5m+3 7/bracketrightbig q42[21]q2 [3]q2[7]q2/bracketleftbig3m+2 2/bracketrightbig q20/bracketleftbig5m+4 4/bracketrightbig q24[m+1]q30.CYCLIC SIEVING FOR GENERALISED NON-CROSSING PARTITIONS 59 If one decomposes [15 m+1]q2as [5m+1]q6+q2[5m]q6+q4[5m]q6, then one sees that, by Corollary 6, this is a polynomial in qwith non-negative integer coefficients. If m≡69 (mod 84) ,then we have Catm(E8;q) =/bracketleftbig15m+1 28/bracketrightbig q56[28]q2 [4]q2[7]q2[15m+4]q2[5m+2]q6/bracketleftbig15m+7 2/bracketrightbig q4/bracketleftbig5m+3 12/bracketrightbig q72[12]q6 [3]q6[4]q6 ×[3m+2]q10[5m+4]q6/bracketleftbigm+1 2/bracketrightbig q60[30]q2[2]q2[3]q2[5]q2 [6]q2[10]q2[15]q2, which, by Corollary 6 and Lemma 16, is a polynomial in qwith non-negative integer coefficients. If m≡70 (mod 84) ,then we have Catm(E8;q) = [15m+1]q2/bracketleftbig15m+4 2/bracketrightbig q4/bracketleftbig5m+2 4/bracketrightbig q24/bracketleftbig15m+7 7/bracketrightbig q14 ×[5m+3]q6/bracketleftbig3m+2 4/bracketrightbig q40[10]q4 [2]q4[5]q4/bracketleftbig5m+4 6/bracketrightbig q36[6]q6 [2]q6[3]q6[m+1]q30, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡71 (mod 84) ,then we have Catm(E8;q) =/bracketleftbig15m+1 2/bracketrightbig q4[15m+4]q2/bracketleftbig5m+2 21/bracketrightbig q126[63]q2 [7]q2[9]q2/bracketleftbig15m+7 4/bracketrightbig q8/bracketleftbig5m+3 2/bracketrightbig q12 ×[3m+2]q10[5m+4]q6/bracketleftbigm+1 4/bracketrightbig q120[60]q2[2]q2[3]q2[5]q2 [10]q2[12]q2[15]q2, which, by Corollary 6 and Lemma 19, is a polynomial in qwith non-negative integer coefficients. If m≡72 (mod 84) ,then we have Catm(E8;q) = [15m+1]q2/bracketleftbig15m+4 4/bracketrightbig q8/bracketleftbig5m+2 2/bracketrightbig q12[15m+7]q2/bracketleftbig5m+3 3/bracketrightbig q18/bracketleftbig3m+2 2/bracketrightbig q20 /bracketleftbig5m+4 28/bracketrightbig q168[84]q2 [7]q2[12]q2[m+1]q30, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡73 (mod 84) ,then we have Catm(E8;q) =/bracketleftbig15m+1 4/bracketrightbig q8/bracketleftbig15m+4 7/bracketrightbig q14[5m+2]q6/bracketleftbig15m+7 2/bracketrightbig q4/bracketleftbig5m+3 4/bracketrightbig q24 ×[3m+2]q10/bracketleftbig5m+4 3/bracketrightbig q18/bracketleftbigm+1 2/bracketrightbig q60[30]q2[2]q2[3]q2[5]q2 [6]q2[10]q2[15]q2, which, by Lemma 16, is a polynomial in qwith non-negative integer coefficients. If m≡74 (mod 84) ,then we have Catm(E8;q) = [15m+1]q2/bracketleftbig15m+4 2/bracketrightbig q4/bracketleftbig5m+2 4/bracketrightbig q24[15m+7]q2[5m+3]q6 ×/bracketleftbig3m+2 28/bracketrightbig q280[140]q2[2]q2 [4]q2[7]q2[10]q2/bracketleftbig5m+4 2/bracketrightbig q12/bracketleftbigm+1 3/bracketrightbig q90[15]q6 [3]q6[5]q6,60 C. KRATTENTHALER AND T. W. M ¨ULLER which, by Corollary 6 and Lemma 24, is a polynomial in qwith non-negative integer coefficients. If m≡75 (mod 84) ,then we have Catm(E8;q) =/bracketleftbig15m+1 2/bracketrightbig q4[15m+4]q2[5m+2]q6/bracketleftbig15m+7 4/bracketrightbig q8/bracketleftbig5m+3 42/bracketrightbig q252[126]q2[3]q2 [6]q2[7]q2[9]q2 ×[3m+2]q10[5m+4]q6/bracketleftbigm+1 4/bracketrightbig q120[60]q2[2]q2[3]q2[5]q2 [10]q2[12]q2[15]q2, which, by Lemmas 19and22, is apolynomial in qwithnon-negative integer coefficients. Ifm≡76 (mod 84) ,then we have Catm(E8;q) =/bracketleftbig15m+1 7/bracketrightbig q14/bracketleftbig15m+4 4/bracketrightbig q8/bracketleftbig5m+2 2/bracketrightbig q12[15m+7]q2 ×[5m+3]q6/bracketleftbig3m+2 2/bracketrightbig q20/bracketleftbig5m+4 12/bracketrightbig q72[12]q6 [3]q6[4]q6[m+1]q30, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡77 (mod 84) ,then we have Catm(E8;q) =/bracketleftbig15m+1 4/bracketrightbig q8[15m+4]q2/bracketleftbig5m+2 3/bracketrightbig q18/bracketleftbig15m+7 14/bracketrightbig q28[14]q2 [2]q2[7]q2/bracketleftbig5m+3 4/bracketrightbig q24 ×[3m+2]q10[5m+4]q6/bracketleftbigm+1 2/bracketrightbig q60[30]q2[2]q2[3]q2[5]q2 [6]q2[10]q2[15]q2, which, by Corollary 6 and Lemma 16, is a polynomial in qwith non-negative integer coefficients. If m≡78 (mod 84) ,then we have Catm(E8;q) = [15m+1]q2/bracketleftbig15m+4 2/bracketrightbig q4/bracketleftbig5m+2 28/bracketrightbig q168[84]q2 [7]q2[12]q2[15m+7]q2/bracketleftbig5m+3 3/bracketrightbig q18 ×/bracketleftbig3m+2 4/bracketrightbig q40[10]q4 [2]q4[5]q4/bracketleftbig5m+4 2/bracketrightbig q12[m+1]q30, which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡79 (mod 84) ,then we have Catm(E8;q) =/bracketleftbig15m+1 2/bracketrightbig q4[15m+4]q2[5m+2]q6/bracketleftbig15m+7 4/bracketrightbig q8/bracketleftbig5m+3 2/bracketrightbig q12 ×[3m+2]q10/bracketleftbig5m+4 21/bracketrightbig q126[63]q2 [7]q2[9]q2/bracketleftbigm+1 4/bracketrightbig q120[60]q2[2]q2[3]q2[5]q2 [10]q2[12]q2[15]q2, which, by Corollary 6 and Lemma 19, is a polynomial in qwith non-negative integer coefficients. If m≡80 (mod 84) ,then we have Catm(E8;q) = [15m+1]q2/bracketleftbig15m+4 28/bracketrightbig q56[28]q2 [4]q2[7]q2/bracketleftbig5m+2 6/bracketrightbig q36[6]q6 [2]q6[3]q6[15m+7]q2[5m+3]q6 ×/bracketleftbig3m+2 2/bracketrightbig q20/bracketleftbig5m+4 4/bracketrightbig q24[m+1]q30,CYCLIC SIEVING FOR GENERALISED NON-CROSSING PARTITIONS 61 which, by Corollary 6, is a polynomial in qwith non-negative integer coefficients. If m≡81 (mod 84) ,then we have Catm(E8;q) =/bracketleftbig15m+1 4/bracketrightbig q8[15m+4]q2[5m+2]q6/bracketleftbig15m+7 2/bracketrightbig q4/bracketleftbig5m+3 12/bracketrightbig q72[12]q6 [3]q6[4]q6 ×/bracketleftbig3m+2 7/bracketrightbig q70[35]q2 [5]q2[7]q2[5m+4]q6/bracketleftbigm+1 2/bracketrightbig q60[30]q2[2]q2[3]q2[5]q2 [6]q2[10]q2[15]q2, which, by Corollary 6 and Lemma 18, is a polynomial in qwith non-negative integer coefficients. If m≡82 (mod 84) ,then we have Catm(E8;q) = [15m+1]q2/bracketleftbig15m+4 2/bracketrightbig q4/bracketleftbig5m+2 4/bracketrightbig q24[15m+7]q2/bracketleftbig5m+3 7/bracketrightbig q42[21]q2 [3]q2[7]q2 /bracketleftbig3m+2 4/bracketrightbig q40[10]q4 [2]q4[5]q4/bracketleftbig5m+4 6/bracketrightbig q36[6]q6 [2]q6[3]q6[m+1]q30. If one decomposes [15 m+1]q2as [5m+1]q6+q2[5m]q6+q4[5m]q6, then one sees that, by Corollary 6, this is a polynomial in qwith non-negative integer coefficients. If m≡83 (mod 84) ,then we have Catm(E8;q) =/bracketleftbig15m+1 14/bracketrightbig q28[14]q2 [2]q2[7]q2[15m+4]q2/bracketleftbig5m+2 3/bracketrightbig q18/bracketleftbig15m+7 4/bracketrightbig q8/bracketleftbig5m+3 2/bracketrightbig q12 ×[3m+2]q10[5m+4]q6/bracketleftbigm+1 4/bracketrightbig q120[60]q2[2]q2[3]q2[5]q2 [10]q2[12]q2[15]q2, which, by Corollary 6 and Lemma 21, is a polynomial in qwith non-negative integer coefficients. /square 5.Auxiliary results I This section collects several auxiliary results which allow us to reduce the problem of proving Theorem 2, or the equivalent statement (3.3), for the 2 6 exceptional groups listed in Section 2 to a finite problem. While Lemmas 27 and 28 cover spec ial choices of the parameters, Lemmas 26 and 30 afford an inductive procedur e. More precisely, if we assume that we have already verified Theorem 2 for all groups o f smaller rank, then Lemmas 26 and 30, together with Lemmas 27 and 31, reduce th e verification of Theorem 2 for the group that we are currently considering to a finit e problem; see Remark 3. The final lemma of this section, Lemma 32, disposes of com plex reflection groups with a special property satisfied by their degrees. Letp=am+b, 0≤bnthen FixNCm(W)(φp) =/braceleftbig (c;ε,...,ε)/bracerightbig . Proof.Let us suppose that ( w0;w1,...,w m)∈FixNCm(W)(φp) and that there exists a j≥1 such that wj/ne}ationslash=ε. By (5.8), it then follows for such a jthat alsowk/ne}ationslash=εfor allk≡j−lm1b(modm), where, as before, bis defined as the unique integer with h1=am2+band 0≤b < m 2. Since, by assumption, gcd( b,m2) = 1, there are exactlym2suchk’s which are distinct mod m. However, this implies that the sum of the absolute lengths of the wi’s, 0≤i≤m, is at least m2> n, a contradiction to Remark 1.(2). /square Remark 3.(1) If we put ourselves in the situation of the assumptions of Lemma 30, then we may conclude that equation (3.3) only needs to be checked f or pairs (m2,h2) subject to the following restrictions: m2≥2,gcd(h1,m2) = 1,andh2divides all degrees of W. (5.9) Indeed, Lemmas 27 and 30 together imply that equation (3.3) is alway s satisfied in all other cases. (2) Still putting ourselves in the situation of Lemma 30, if m2>nandm2h2does not divide any of the degrees of W, then equation (3.3) is satisfied. Indeed, Lemma 31 says thatinthiscasetheleft-handsideof (3.3)equals1,whileastraightf orwardcomputation using (5.4) shows that in this case the right-hand side of (3.3) equals 1 as well. (3)It shouldbeobserved that thisleaves afinitenumber of choices form2to consider, whence a finite number of choices for ( m1,m2,h1,h2). Altogether, there remains a finite number of choices for p=h1m1to be checked. Lemma 32. LetWbe an irreducible well-generated complex reflection group o f rankn with the property that di|hfori= 1,2,...,n. Then Theorem 2is true for this group W. Proof.By Lemma 26, we may restrict ourselves to divisors pofmh.CYCLIC SIEVING FOR GENERALISED NON-CROSSING PARTITIONS 65 Suppose that e2πip/mhis adi-th rootof unity for some i. In other words, mh/pdivides di. Sincediis a divisor of hby assumption, the integer mh/palso divides h. But this is equivalent to saying that mdividesp, and equation (3.3) holds by Lemma 27. Now assume that mh/pdoes not divide any of the di’s. Then, by (5.4), the right- hand side of (3.3) equals 1. On the other hand, ( c;ε,...,ε) is always an element of FixNCm(W)(φp). To see that there are no others, we make appeal to the classific a- tion of all irreducible well-generated complex reflection groups, whic h we recalled in Section 2. Inspection reveals that all groups satisfying the hypot heses of the lemma have rank n≤2. Except for the groups contained in the infinite series G(d,1,n) andG(e,e,n) for which Theorem 2 has been established in [19], these are the grou ps G5,G6,G9,G10,G14,G17,G18,G21. We now discuss these groups case by case, keeping the notation of Lemma 30. In order to simplify the argument, we not e that Lemma 31 implies that equation (3.3) holds if m2>2, so that in the following arguments we always may assume that m2= 2. CaseG5. The degrees are 6 ,12, and therefore Remark 3.(1) implies that equa- tion (3.3) is always satisfied. CaseG6. The degrees are 4 ,12, and therefore, according to Remark 3.(1), we need only consider the casewhere h2= 4andm2= 2, that is, p= 3m/2. Then (5.8) becomes φp/parenleftbig (w0;w1,...,w m)/parenrightbig = (∗;c2wm 2+1c−2,c2wm 2+2c−2,...,c2wmc−2,cw1c−1,...,cw m 2c−1/parenrightbig . (5.10) If (w0;w1,...,w m) isfixed by φpandnot equal to ( c;ε,...,ε), there must exist an iwith 1≤i≤m 2such thatℓT(wi) =ℓT(wm 2+i) = 1,wm 2+i=cwic−1,wiwm 2+i=wicwic−1=c, and allwj, withj/ne}ationslash=i,m 2+i, equalε. However, with the help of the GAPpackage CHEVIE[14, 28], one verifies that there is no wiinG6such that ℓT(wi) = 1 and wicwic−1=c are simultaneously satisfied. Hence, the left-hand side of (3.3) is eq ual to 1, as required. CaseG9. The degrees are 8 ,24, and therefore, according to Remark 3.(1), we need only consider the case where h2= 8 andm2= 2, that is, p= 3m/2. This is the same p as forG6. Again, CHEVIEfinds no solution. Hence, the left-hand side of (3.3) is equal to 1, as required. CaseG10. The degrees are 12 ,24, and therefore Remark 3.(1) implies that equa- tion (3.3) is always satisfied. CaseG14. The degrees are 6 ,24, and therefore Remark 3.(1) implies that equa- tion (3.3) is always satisfied. CaseG17. The degrees are 20 ,60, and therefore, according to Remark 3.(1), we need only consider the cases where h2= 20 orh2= 4. In the first case, p= 3m/2, which is the samepas forG6. Again,CHEVIEfinds no solution. In the second case, p= 15m/2. Then (5.8) becomes φp/parenleftbig (w0;w1,...,w m)/parenrightbig = (∗;c8wm 2+1c−8,c8wm 2+2c−8,...,c8wmc−8,c7w1c−7,...,c7wm 2c−7/parenrightbig .(5.11) By Lemma 29, every element of NC(W) is fixed under conjugation by c3, and, thus, on elements fixed by φp, the above action of φpreduces to the one in (5.10). This action66 C. KRATTENTHALER AND T. W. M ¨ULLER was already discussed in the first case. Hence, in both cases, the le ft-hand side of (3.3) is equal to 1, as required. CaseG18. The degrees are 30 ,60, and therefore Remark 3.(1) implies that equa- tion (3.3) is always satisfied. CaseG21. The degrees are 12 ,60, and therefore, according to Remark 3.(1), we need only consider the cases where h2= 12 orh2= 4. In the first case, p= 5m/2, so that (5.8) becomes φp/parenleftbig (w0;w1,...,w m)/parenrightbig = (∗;c3wm 2+1c−3,c3wm 2+2c−3,...,c3wmc−3,c2w1c−2,...,c2wm 2c−2/parenrightbig .(5.12) If (w0;w1,...,w m) is fixed by φpand not equal to ( c;ε,...,ε), there must exist an i with 1≤i≤m 2such thatℓT(wi) = 1 andwic2wic−2=c. However, with the help of theGAPpackageCHEVIE[14, 28], one verifies that there is no such solution to this equation. In the second case, p= 15m/2. Then (5.8) becomes the action in (5.11). By Lemma 29, every element of NC(W) is fixed under conjugation by c5, and, thus, on elements fixed by φp, the action of φpin (5.11) reduces to the one in the first case. Hence, in both cases, the left-hand side of (3.3) is equal to 1, as re quired. This completes the proof of the lemma. /square 6.Case-by-case verification of Theorem 2 In the sequel we write ζdfor a primitive d-th root of unity. CaseG4.The degrees are 4 ,6, and hence we have Catm(G4;q) =[6m+6]q[6m+4]q [6]q[4]q. Letζbe a 6m-th root of unity. In what follows, we abbreviate the assertion tha t “ζis a primitive d-th root of unity” as “ ζ=ζd.” The following cases on the right-hand side of (3.3) occur: lim q→ζCatm(G4;q) =m+1,ifζ=ζ6,ζ3, (6.1a) lim q→ζCatm(G4;q) =3m+2 2,ifζ=ζ4,2|m, (6.1b) lim q→ζCatm(G4;q) = Catm(G4),ifζ=−1 orζ= 1, (6.1c) lim q→ζCatm(G4;q) = 1,otherwise. (6.1d) We must now prove that the left-hand side of (3.3) in each case agre es with the values exhibited in (6.1). The only cases not covered by Lemmas 27 an d 28 are the ones in (6.1b) and (6.1d). On the other hand, the only case left to co nsider according to Remark 3 is the case where h2=m2= 2, that is the case (6.1b) where p= 3m/2. In particular, mmust be divisible by 2. The action of φpis the same as the one in (5.10). With the help of CHEVIE, one finds that each of the 3 (complex) reflections in G4which are less than the (chosen) Coxeter element is a valid choice for wi, and each of these choices gives rise to m/2 elements in NCm(G4) since the index iranges from 1 to m/2. Hence, in total, we obtain 1+3m 2=3m+2 2elements in Fix NCm(G4)(φp), which agrees with the limit in (6.1b).CYCLIC SIEVING FOR GENERALISED NON-CROSSING PARTITIONS 67 CaseG8.The degrees are 8 ,12, and hence we have Catm(G8;q) =[12m+12]q[12m+8]q [12]q[8]q. Letζbe a 12m-th root of unity. The following cases on the right-hand side of (3.3) occur: lim q→ζCatm(G8;q) =m+1,ifζ=ζ12,ζ6,ζ3, (6.2a) lim q→ζCatm(G8;q) =3m+2 2,ifζ=ζ8,2|m, (6.2b) lim q→ζCatm(G8;q) = Catm(G8),ifζ=ζ4,−1,1, (6.2c) lim q→ζCatm(G8;q) = 1,otherwise. (6.2d) We must now prove that the left-hand side of (3.3) in each case agre es with the values exhibited in (6.2). The only cases not covered by Lemmas 27 an d 28 are the ones in (6.2b) and (6.2d). On the other hand, the only case left to co nsider according to Remark 3 is the case where h2= 4 andm2= 2, that is the case (6.2b) where p= 3m/2. In particular, mmust be divisible by 2. The action of φpis the same as the one in (5.10). With the help of CHEVIE, one finds that each of the 3 (complex) reflections in G8which are less than the (chosen) Coxeter element is a valid choice for wi, and each of these choices gives rise to m/2 elements in NCm(G8) since the index iranges from 1 tom/2. Hence, in total, we obtain 1+3m 2=3m+2 2elements in Fix NCm(G8)(φp), which agrees with the limit in (6.2b). CaseG16.The degrees are 20 ,30, and hence we have Catm(G16;q) =[30m+30]q[30m+20]q [30]q[20]q. Letζbe a 30m-th root of unity. The following cases on the right-hand side of (3.3) occur: lim q→ζCatm(G16;q) =m+1,ifζ=ζ30,ζ15,ζ6,ζ3, (6.3a) lim q→ζCatm(G16;q) =3m+2 2,ifζ=ζ20,ζ4,2|m, (6.3b) lim q→ζCatm(G16;q) = Catm(G16),ifζ=ζ10,ζ5,−1,1, (6.3c) lim q→ζCatm(G16;q) = 1,otherwise. (6.3d) We must now prove that the left-hand side of (3.3) in each case agre es with the values exhibited in (6.3). The only cases not covered by Lemmas 27 an d 28 are the ones in (6.3b) and (6.3d). On the other hand, the only cases left to c onsider according to Remark 3 are the cases where h2= 10 andm2= 2, respectively h2=m2= 2. Both cases belong to (6.3b). In the first case, we have p= 3m/2, while in the second case we havep= 15m/2. In particular, mmust be divisible by 2. In the first case, the action ofφpis the same as the one in (5.10). With the help of CHEVIE, one finds that each of the 3 (complex) reflections in G16which are less than the (chosen) Coxeter element is a valid choice for wi, and each of these choices gives rise to m/2 elements in NCm(G16)68 C. KRATTENTHALER AND T. W. M ¨ULLER since the index iranges from 1 to m/2. On the other hand, if p= 15m/2, then the action ofφpis the same as the one in (5.11). By Lemma 29, every element of NC(W) is fixed under conjugation by c3, and, thus, on elements fixed by φp, the action of φp reduces to the one in the first case. Hence, in total, we obtain 1+3m 2=3m+2 2elements in Fix NCm(G16)(φp), which agrees with the limit in (6.3b). CaseG20.The degrees are 12 ,30, and hence we have Catm(G20;q) =[30m+30]q[30m+12]q [30]q[12]q. Letζbe a 30m-th root of unity. The following cases on the right-hand side of (3.3) occur: lim q→ζCatm(G20;q) =m+1,ifζ=ζ30,ζ15,ζ10,ζ5, (6.4a) lim q→ζCatm(G20;q) =5m+2 2,ifζ=ζ12,ζ4,2|m, (6.4b) lim q→ζCatm(G20;q) = Catm(G20),ifζ=ζ6,ζ3,−1,1, (6.4c) lim q→ζCatm(G20;q) = 1,otherwise. (6.4d) We must now prove that the left-hand side of (3.3) in each case agre es with the values exhibited in (6.4). The only cases not covered by Lemmas 27 an d 28 are the ones in (6.4b) and (6.4d). On the other hand, the only cases left to c onsider according to Remark 3 are the cases where h2= 6 andm2= 2, respectively h2=m2= 2. Both cases belong to (6.4b). In the first case, we have p= 5m/2, while in the second case we havep= 15m/2. In particular, mmust be divisible by 2. In the first case, the action ofφpis the same as the one in (5.12). With the help of CHEVIE, one finds that each of the 5 (complex) reflections in G20which are less than the (chosen) Coxeter element is a valid choice for wi, and each of these choices gives rise to m/2 elements in NCm(G20) since the index iranges from 1 to m/2. On the other hand, if p= 15m/2, then the action ofφpis the same as the one in (5.11). By Lemma 29, every element of NC(W) is fixed under conjugation by c5, and, thus, on elements fixed by φp, the action of φp reduces to the one in the first case. Hence, in total, we obtain 1+5m 2=5m+2 2elements in Fix NCm(G20)(φp), which agrees with the limit in (6.4b). CaseG23=H3.The degrees are 2 ,6,10, and hence we have Catm(H3;q) =[10m+10]q[10m+6]q[10m+2]q [10]q[6]q[2]q.CYCLIC SIEVING FOR GENERALISED NON-CROSSING PARTITIONS 69 Letζbe a 10m-th root of unity. The following cases on the right-hand side of (3.3) occur: lim q→ζCatm(H3;q) =m+1,ifζ=ζ10,ζ5, (6.5a) lim q→ζCatm(H3;q) =10m+6 6,ifζ=ζ6,ζ3,3|m, (6.5b) lim q→ζCatm(H3;q) = Catm(H3),ifζ=−1 orζ= 1, (6.5c) lim q→ζCatm(H3;q) = 1,otherwise. (6.5d) We must now prove that the left-handside of (3.3) in each case agre es with the values exhibited in (6.5). The only cases not covered by Lemmas 27 and 28 ar e the ones in (6.5b) and (6.5d). By Lemma 26, we are free to choose p= 5m/3 ifζ=ζ6, respectively p= 10m/3 ifζ=ζ3. In both cases, mmust be divisible by 3. We start with the case that p= 5m/3. From (5.1), we infer φp/parenleftbig (w0;w1,...,w m)/parenrightbig = (∗;c2wm 3+1c−2,c2wm 3+2c−2,...,c2wmc−2,cw1c−1,...,cw m 3c−1/parenrightbig . (6.6) Supposing that ( w0;w1,...,w m) is fixed by φp, we obtain the system of equations wi=c2wm 3+ic−2, i= 1,2,...,2m 3, (6.7a) wi=cwi−2m 3c−1, i=2m 3+1,2m 3+2,...,m. (6.7b) There are two distinct possibilities for choosing the wi’s, 1≤i≤m: either all the wi’s are equal to ε, or there is an iwith 1≤i≤m 3such that ℓT(wi) =ℓT(wi+m 3) =ℓT(wi+2m 3) = 1. Writingt1,t2,t3forwi,wi+m 3,wi+2m 3, respectively, the equations (6.7) reduce to t1=c2t2c−2, (6.8a) t2=c2t3c−2, (6.8b) t3=ct1c−1. (6.8c) One of these equations is in fact superfluous: if we substitute (6.8b ) and (6.8c) in (6.8a), then we obtain t1=c5t1c−5which is automatically satisfied due to Lemma 29 withd= 2. Since (w0;w1,...,w m)∈NCm(H3), we must have t1t2t3=c. Combining this with (6.8), we infer that t1(c−2t1c2)(ct1c−1) =c. (6.9) With the help of Stembridge’s Maplepackagecoxeter [36], one obtains five solutions fort1in this equation: t1∈/braceleftbig [2],[3],[2,1,2],[1,2,3,2,1],[1,3,2,1,2,1,3]/bracerightbig . (6.10) Here we have used the short notation of coxeter: if{s1,s2,s3}is a simple system of generators of H3, corresponding to the Dynkin diagram displayed in Figure 1, then [j1,j2,...,j k] stands for the element sj1sj2...sjk. We claim that each of the above five solutions gives rise to m/3 elements of FixNCm(H3)(φp). Indeed, given t1, the elements t2andt3can be computed by (6.8a) and (6.8c), and there are m/3 possibilities to choose the index iforwi.70 C. KRATTENTHALER AND T. W. M ¨ULLER • • •1 2 35 Figure 1. The Dynkin diagram for H3 In total, we obtain 1 + 5m 3=10m+6 6elements in Fix NCm(H3)(φp), which agrees with the limit in (6.5b). In the case that p= 10m/3, we infer from (5.1) that φp/parenleftbig (w0;w1,...,w m)/parenrightbig = (∗;c4w2m 3+1c−4,c4w2m 3+2c−4,...,c4wmc−4,c3w1c−3,...,c3w2m 3c−3/parenrightbig .(6.11) Supposing that ( w0;w1,...,w m) is fixed by φp, we obtain the system of equations wi=c4w2m 3+ic−4, i= 1,2,...,m 3, (6.12a) wi=c3wi−m 3c−3, i=m 3+1,m 3+2,...,m. (6.12b) There are two distinct possibilities for choosing the wi’s, 1≤i≤m: either all the wi’s are equal to ε, or there is an iwith 1≤i≤m 3such that ℓT(wi) =ℓT(wi+m 3) =ℓT(wi+2m 3) = 1. Writingt1,t2,t3forwi,wi+m 3,wi+2m 3, respectively, the equations (6.12) reduce to t1=c4t3c−4, (6.13a) t2=c3t1c−3, (6.13b) t3=c3t2c−3. (6.13c) One of these equations is in fact superfluous: if we substitute (6.13 b) and (6.13c) in (6.13a), then we obtain t1=c10t1c−10which is automatically satisfied since c10=ε. Since (w0;w1,...,w m)∈NCm(H3), we must have t1t2t3=c. Combining this with (6.13), we infer that t1(c3t1c−3)(c−4t1c4) =c. (6.14) Using that c5t1c−5=t1, due to Lemma 29 with d= 2, we see that this equation is equivalent with (6.9). Therefore, we are facing exactly the same en umeration problem here as forp= 5m/3, and, consequently, the number of solutions to (6.14) is the same , namely5m+3 3, as required. Finally, we turn to (6.5d). By Remark 3, the only choices for h2andm2to be considered are h2= 1 andm2= 3,h2=m2= 2, respectively h2= 2 andm2= 3. These correspond to the choices p= 10m/3,p= 5m/2, respectively p= 5m/3, out of which only p= 5m/2 has not yet been discussed and belongs to the current case. The corresponding action of φpis given by (5.12). A computation with Stembridge’s Maple packagecoxeter [36] finds no solution. Hence, the left-hand side of (3.3) is equal to 1 , as required.CYCLIC SIEVING FOR GENERALISED NON-CROSSING PARTITIONS 71 CaseG24.The degrees are 4 ,6,14, and hence we have Catm(G24;q) =[14m+14]q[14m+6]q[14m+4]q [14]q[6]q[4]q. Letζbe a 14m-th root of unity. The following cases on the right-hand side of (3.3) occur: lim q→ζCatm(G24;q) =m+1,ifζ=ζ14,ζ7, (6.15a) lim q→ζCatm(G24;q) =7m+3 3,ifζ=ζ6,ζ3,3|m, (6.15b) lim q→ζCatm(G24;q) =7m+2 2,ifζ=ζ4,2|m, (6.15c) lim q→ζCatm(G24;q) = Catm(G24),ifζ=−1 orζ= 1, (6.15d) lim q→ζCatm(G24;q) = 1,otherwise. (6.15e) We must now prove that the left-handside of (3.3) in each case agre es with the values exhibited in (6.15). The only cases not covered by Lemmas 27 and 28 a re the ones in (6.15b), (6.15c), and (6.15e). We first consider (6.15b). By Lemma 26, we are free to choose p= 7m/3 ifζ=ζ6, respectively p= 14m/3 ifζ=ζ3. In both cases, mmust be divisible by 3. We start with the case that p= 7m/3. From (5.1), we infer φp/parenleftbig (w0;w1,...,w m)/parenrightbig = (∗;c3w2m 3+1c−3,c3w2m 3+2c−3,...,c3wmc−3,c2w1c−2,...,c2w2m 3c−2/parenrightbig . Supposing that ( w0;w1,...,w m) is fixed by φp, we obtain the system of equations wi=c3w2m 3+ic−3, i= 1,2,...,m 3, (6.16a) wi=c2wi−m 3c−2, i=m 3+1,m 3+2,...,m. (6.16b) There are two distinct possibilities for choosing the wi’s, 1≤i≤m: either all the wi’s are equal to ε, or there is an iwith 1≤i≤m 3such that ℓT(wi) =ℓT(wi+m 3) =ℓT(wi+2m 3) = 1. Writingt1,t2,t3forwi,wi+m 3,wi+2m 3, respectively, the equations (6.16) reduce to t1=c3t3c−3, (6.17a) t2=c2t1c−2, (6.17b) t3=c2t2c−2. (6.17c) One of these equations is in fact superfluous: if we substitute (6.17 b) and (6.17c) in (6.17a), then we obtain t1=c7t1c−7which is automatically satisfied due to Lemma 29 withd= 2. Since (w0;w1,...,w m)∈NCm(G24), we must have t1t2t3=c. Combining this with (6.17), we infer that t1(c2t1c−2)(c4t1c−4) =c. (6.18) With the help of CHEVIE, one obtains 7 solutions for t1in this equation: t1∈/braceleftbig [1],[2],[3],[15],[16],[19],[21]/bracerightbig , (6.19)72 C. KRATTENTHALER AND T. W. M ¨ULLER each of them giving rise to m/3 elements of Fix NCm(G24)(φp) sinceiranges from 1 to m/3. Here we have used the short notation of CHEVIE: [j1,j2,...,j k] stands for the elementrj1rj2...rjk, whereriis thei-th (complex) reflection corresponding to the i-th root in the internal ordering of the roots of G24inCHEVIE. In total, we obtain 1 + 7m 3=7m+3 3elements in Fix NCm(G24)(φp), which agrees with the limit in (6.15b). In the case that p= 14m/3, we infer from (5.1) that φp/parenleftbig (w0;w1,...,w m)/parenrightbig = (∗;c5wm 3+1c−5,c5wm 3+2c−5,...,c5wmc−5,c4w1c−4,...,c4wm 3c−4/parenrightbig . Supposing that ( w0;w1,...,w m) is fixed by φp, we obtain the system of equations wi=c5wm 3+ic−5, i= 1,2,...,2m 3, (6.20a) wi=c4wi−2m 3c−4, i=2m 3+1,2m 3+2,...,m. (6.20b) There are two distinct possibilities for choosing the wi’s, 1≤i≤m: either all the wi’s are equal to ε, or there is an iwith 1≤i≤m 3such that ℓT(wi) =ℓT(wi+m 3) =ℓT(wi+2m 3) = 1. Writingt1,t2,t3forwi,wi+m 3,wi+2m 3, respectively, the equations (6.20) reduce to t1=c5t2c−5, (6.21a) t2=c5t3c−5, (6.21b) t3=c4t1c−4. (6.21c) One of these equations is in fact superfluous: if we substitute (6.21 b) and (6.21c) in (6.21a), then we obtain t1=c14t1c−14which is automatically satisfied since c14=ε. Since (w0;w1,...,w m)∈NCm(G24), we must have t1t2t3=c. Combining this with (6.21), we infer that t1(c9t1c−9)(c−4t1c4) =c. (6.22) Using that c7t1c−7=t1, due to Lemma 29 with d= 2, we see that this equation is equivalent with (6.18). Therefore, we are facing exactly the same e numeration problem here as forp= 7m/3, and, consequently, the number of solutions to (6.22) is the same , namely7m+3 3, as required. Ournextcaseis(6.15c). ByLemma26, wearefreetochoose p= 7m/2. Inparticular, mmust be divisible by 2. From (5.1), we infer φp/parenleftbig (w0;w1,...,w m)/parenrightbig = (∗;c4wm 2+1c−4,c4wm 2+2c−4,...,c4wmc−4,c3w1c−3,...,c3wm 2c−3/parenrightbig . Supposing that ( w0;w1,...,w m) is fixed by φp, we obtain the system of equations wi=c4wm 2+ic−4, i= 1,2,...,m 2, (6.23a) wi=c3wi−m 2c−3, i=m 2+1,m 2+2,...,m. (6.23b) There are two distinct possibilities for choosing the wi’s, 1≤i≤m: either all the wi’s are equal to ε, or there is an iwith 1≤i≤m 2such that ℓT(wi) =ℓT(wi+m 2) = 1.CYCLIC SIEVING FOR GENERALISED NON-CROSSING PARTITIONS 73 Writingt1,t2forwi,wi+m 2, respectively, the equations (6.23) reduce to t1=c4t2c−4, (6.24a) t2=c3t1c−3. (6.24b) One of these equations is in fact superfluous: if we substitute (6.24 b) in (6.24a), then we obtaint1=c7t1c−7which is automatically satisfied due to Lemma 29 with d= 2. Since (w0;w1,...,w m)∈NCm(G24), we must have t1t2≤Tc, where≤Tis the partial order defined in (2.1). Combining this with (6.24), we infer that t1(c3t1c−3)≤Tc. (6.25) With the help of CHEVIE, one obtains 7 solutions for t1in this relation: t1∈/braceleftbig [5],[6],[7],[9],[12],[29],[32]/bracerightbig , (6.26) each of them giving rise to m/2 elements of Fix NCm(G24)(φp) sinceiranges from 1 to m/2. Here we have used again the short notation of CHEVIEreferring to the internal ordering of the roots of G24inCHEVIE. In total, we obtain 1 + 7m 2=7m+2 2elements in Fix NCm(G24)(φp), which agrees with the limit in (6.15c). Finally, we turn to (6.15e). By Remark 3, the only choices for h2andm2to be con- sidered are h2= 1 andm2= 3,h2=m2= 2, andh2= 2 andm2= 3. These correspond to the choices p= 14m/3,p= 7m/2, respectively p= 7m/3, all of which have already been discussed as they do not belong to (6.15e). Hence, (3.3) must necessarily hold, as required. CaseG25.The degrees are 6 ,9,12, and hence we have Catm(G25;q) =[12m+12]q[12m+9]q[12m+6]q [12]q[9]q[6]q. Letζbe a 12m-th root of unity. The following cases on the right-hand side of (3.3) occur: lim q→ζCatm(G25;q) =m+1,ifζ=ζ12,ζ4, (6.27a) lim q→ζCatm(G25;q) =4m+3 3,ifζ=ζ9,3|m, (6.27b) lim q→ζCatm(G25;q) = (m+1)(2m+1),ifζ=ζ6,−1 (6.27c) lim q→ζCatm(G25;q) = Catm(G25),ifζ=ζ3,1, (6.27d) lim q→ζCatm(G25;q) = 1,otherwise. (6.27e) We must now prove that the left-handside of (3.3) in each case agre es with the values exhibited in (6.27). The only cases not covered by Lemmas 27 and 28 a re the ones in (6.27b) and (6.27e). We first consider (6.27b). By Lemma 26, we are free to choose p= 4m/3. In particular, mmust be divisible by 3. From (5.1), we infer φp/parenleftbig (w0;w1,...,w m)/parenrightbig = (∗;c2w2m 3+1c−2,c2w2m 3+2c−2,...,c2wmc−2,cw1c−1,...,cw 2m 3c−1/parenrightbig .74 C. KRATTENTHALER AND T. W. M ¨ULLER Supposing that ( w0;w1,...,w m) is fixed by φp, we obtain the system of equations wi=c2w2m 3+ic−2, i= 1,2,...,m 3, (6.28a) wi=cwi−m 3c−1, i=m 3+1,m 3+2,...,m. (6.28b) There are two distinct possibilities for choosing the wi’s, 1≤i≤m: either all the wi’s are equal to ε, or there is an iwith 1≤i≤m 3such that ℓT(wi) =ℓT(wi+m 3) =ℓT(wi+2m 3) = 1. Writingt1,t2,t3forwi,wi+m 3,wi+2m 3, respectively, the equations (6.28) reduce to t1=c2t3c−2, (6.29a) t2=ct1c−1, (6.29b) t3=ct2c−1. (6.29c) One of these equations is in fact superfluous: if we substitute (6.29 b) and (6.29c) in (6.29a), then we obtain t1=c4t1c−4which is automatically satisfied due to Lemma 29 withd= 3. Since (w0;w1,...,w m)∈NCm(G25), we must have t1t2t3=c. Combining this with (6.29), we infer that t1(ct1c−1)(c2t1c−2) =c. (6.30) With the help of CHEVIE, one obtains four solutions for t1in this equation: t1∈/braceleftbig [1],[2],[3],[14]/bracerightbig , (6.31) each of them giving rise to m/3 elements of Fix NCm(G25)(φp) sinceiranges from 1 to m/3. Here we have used again the short notation of CHEVIEreferring to the internal ordering of the roots of G25inCHEVIE. In total, we obtain 1 + 4m 3=4m+3 3elements in Fix NCm(G25)(φp), which agrees with the limit in (6.27b). Finally, we turn to (6.27e). By Remark 3, the only choice for h2andm2to be considered are h2=m2= 3. This corresponds to the choice p= 4m/3, which has already been discussed as they do not belong to (6.27e). Hence, (3 .3) must necessarily hold, as required. CaseG26.The degrees are 6 ,12,18, and hence we have Catm(G26;q) =[18m+18]q[18m+12]q[18m+6]q [18]q[12]q[6]q. Letζbe a 14m-th root of unity. The following cases on the right-hand side of (3.3) occur: lim q→ζCatm(G26;q) =m+1,ifζ=ζ18,ζ9, (6.32a) lim q→ζCatm(G26;q) =3m+2 2,ifζ=ζ12,ζ4,2|m, (6.32b) lim q→ζCatm(G26;q) = Catm(G26),ifζ=ζ6,ζ3,−1,1, (6.32c) lim q→ζCatm(G26;q) = 1,otherwise. (6.32d)CYCLIC SIEVING FOR GENERALISED NON-CROSSING PARTITIONS 75 We must now prove that the left-handside of (3.3) in each case agre es with the values exhibited in (6.32). The only cases not covered by Lemmas 27 and 28 a re the ones in (6.32b) and (6.32d). We first consider (6.32b). By Lemma 26, we are free to choose p= 3m/2 ifζ=ζ12, respectively p= 9m/2 ifζ=ζ4. In both cases, mmust be divisible by 2. We start with the case that p= 3m/2. From (5.1), we infer φp/parenleftbig (w0;w1,...,w m)/parenrightbig = (∗;c2wm 2+1c−2,c2wm 2+2c−2,...,c2wmc−2,cw1c−1,...,cw m 2c−1/parenrightbig . Supposing that ( w0;w1,...,w m) is fixed by φp, we obtain the system of equations wi=c2wm 2+ic−2, i= 1,2,...,m 2, (6.33a) wi=cwi−m 2c−1, i=m 2+1,m 2+2,...,m. (6.33b) There are two distinct possibilities for choosing the wi’s, 1≤i≤m: either all the wi’s are equal to ε, or there is an iwith 1≤i≤m 2such that ℓT(wi) =ℓT(wi+m 2) = 1. Writingt1,t2forwi,wi+m 2, respectively, the equations (6.33) reduce to t1=c2t2c−2, (6.34a) t2=c1t1c−1. (6.34b) One of these equations is in fact superfluous: if we substitute (6.34 b) in (6.34a), then we obtaint1=c3t1c−3which is automatically satisfied due to Lemma 29 with d= 6. Since (w0;w1,...,w m)∈NCm(G26), we must have t1t2≤Tc. Combining this with (6.34), we infer that t1(ct1c−1)≤Tc. (6.35) With the help of CHEVIE, one obtains three solutions for t1in this equation: t1∈/braceleftbig [2],[3],[12]/bracerightbig , each of them giving rise to m/2 elements of Fix NCm(G26)(φp) sinceiranges from 1 to m/2. Here we have again used the short notation of CHEVIEreferring to the internal ordering of the roots of G26inCHEVIE. In total, we obtain 1 + 3m 2=3m+2 2elements in Fix NCm(G26)(φp), which agrees with the limit in (6.32b). In the case that p= 9m/2, we infer from (5.1) that φp/parenleftbig (w0;w1,...,w m)/parenrightbig = (∗;c5wm 2+1c−5,c5wm 2+2c−5,...,c5wmc−5,c4w1c−4,...,c4wm 2c−4/parenrightbig . Supposing that ( w0;w1,...,w m) is fixed by φp, we obtain the system of equations wi=c5wm 2+ic−5, i= 1,2,...,m 2, (6.36a) wi=c4wi−m 2c−4, i=m 2+1,m 2+2,...,m. (6.36b) There are two distinct possibilities for choosing the wi’s, 1≤i≤m: either all the wi’s are equal to ε, or there is an iwith 1≤i≤m 3such that ℓT(wi) =ℓT(wi+m 2) = 1.76 C. KRATTENTHALER AND T. W. M ¨ULLER Writingt1,t2forwi,wi+m 2, respectively, the equations (6.36) reduce to t1=c5t2c−5, (6.37a) t2=c4t1c−4. (6.37b) One of these equations is in fact superfluous: if we substitute (6.37 b) in (6.37a), then we obtaint1=c9t1c−9which is automatically satisfied due to Lemma 29 with d= 2. Since (w0;w1,...,w m)∈NCm(G26), we must have t1t2≤Tc. Combining this with (6.37), we infer that t1(c4t1c−4)≤Tc. (6.38) Using that c3t1c−3=t1, due to Lemma 29 with d= 6, we see that this equation is equivalent with (6.35). Therefore, we are facing exactly the same e numeration problem here as forp= 3m/2, and, consequently, the number of solutions to (6.38) is the same , namely3m+2 2, as required. Finally, we turn to (6.32d). By Remark 3, the only choices for h2andm2to be considered are h2= 6 andm2= 2, respectively h2=m2= 2. These correspond to the choicesp= 3m/2, respectively p= 9m/2, all of which have already been discussed as they do not belong to (6.32d). Hence, (3.3) must necessarily hold, a s required. CaseG27.The degrees are 6 ,12,30, and hence we have Catm(G27;q) =[30m+30]q[30m+12]q[30m+6]q [30]q[12]q[6]q. Letζbe a 14m-th root of unity. The following cases on the right-hand side of (3.3) occur: lim q→ζCatm(G27;q) =m+1,ifζ=ζ30,ζ15,ζ10,ζ5, (6.39a) lim q→ζCatm(G27;q) =5m+2 2,ifζ=ζ12,ζ4,2|m, (6.39b) lim q→ζCatm(G27;q) = Catm(G27),ifζ=ζ6,ζ3,−1,1, (6.39c) lim q→ζCatm(G27;q) = 1,otherwise. (6.39d) We must now prove that the left-handside of (3.3) in each case agre es with the values exhibited in (6.39). The only cases not covered by Lemmas 27 and 28 a re the ones in (6.39b) and (6.39d). We first consider (6.39b). By Lemma 26, we are free to choose p= 5m/2 ifζ=ζ12, respectively p= 15m/2 ifζ=ζ4. In both cases, mmust be divisible by 2. We start with the case that p= 5m/2. From (5.1), we infer φp/parenleftbig (w0;w1,...,w m)/parenrightbig = (∗;c3wm 2+1c−3,c3wm 2+2c−3,...,c3wmc−3,c2w1c−2,...,c2wm 2c−2/parenrightbig . Supposing that ( w0;w1,...,w m) is fixed by φp, we obtain the system of equations wi=c3wm 2+ic−3, i= 1,2,...,m 2, (6.40a) wi=c2wi−m 2c−2, i=m 2+1,m 2+2,...,m. (6.40b)CYCLIC SIEVING FOR GENERALISED NON-CROSSING PARTITIONS 77 There are two distinct possibilities for choosing the wi’s, 1≤i≤m: either all the wi’s are equal to ε, or there is an iwith 1≤i≤m 2such that ℓT(wi) =ℓT(wi+m 2) = 1. Writingt1,t2forwi,wi+m 2, respectively, the equations (6.40) reduce to t1=c3t2c−3, (6.41a) t2=c2t1c−2. (6.41b) One of these equations is in fact superfluous: if we substitute (6.41 b) in (6.41a), then we obtaint1=c5t1c−5which is automatically satisfied due to Lemma 29 with d= 6. Since (w0;w1,...,w m)∈NCm(G27), we must have t1t2≤Tc. Combining this with (6.41), we infer that t1(c2t1c−2)≤Tc. (6.42) With the help of CHEVIE, one obtains five solutions for t1in this equation: t1∈/braceleftbig [1],[2],[15],[16],[28]/bracerightbig , each of them giving rise to m/2 elements of Fix NCm(G27)(φp) sinceiranges from 1 to m/2. Here we have used the short notation of CHEVIEreferring to the internal ordering of the roots of G27inCHEVIE. In total, we obtain 1 + 5m 2=5m+2 2elements in Fix NCm(G27)(φp), which agrees with the limit in (6.39b). In the case that p= 15m/2, we infer from (5.1) that φp/parenleftbig (w0;w1,...,w m)/parenrightbig = (∗;c8wm 2+1c−8,c8wm 2+2c−8,...,c8wmc−8,c7w1c−7,...,c7wm 2c−7/parenrightbig . Supposing that ( w0;w1,...,w m) is fixed by φp, we obtain the system of equations wi=c8wm 2+ic−8, i= 1,2,...,m 2, (6.43a) wi=c7wi−m 2c−7, i=m 2+1,m 2+2,...,m. (6.43b) There are two distinct possibilities for choosing the wi’s, 1≤i≤m: either all the wi’s are equal to ε, or there is an iwith 1≤i≤m 2such that ℓT(wi) =ℓT(wi+m 2) = 1. Writingt1,t2forwi,wi+m 2, respectively, the equations (6.43) reduce to t1=c8t2c−8, (6.44a) t2=c7t1c−7. (6.44b) One of these equations is in fact superfluous: if we substitute (6.44 b) in (6.44a), then we obtaint1=c15t1c−15which is automatically satisfied due to Lemma 29 with d= 2. Since (w0;w1,...,w m)∈NCm(G27), we must have t1t2≤Tc. Combining this with (6.44), we infer that t1(c7t1c−7)≤Tc. (6.45) Using that c5t1c−5=t1, due to Lemma 29 with d= 6, we see that this equation is equivalent with (6.42). Therefore, we are facing exactly the same e numeration problem here as forp= 5m/2, and, consequently, the number of solutions to (6.45) is the same , namely5m+2 2, as required.78 C. KRATTENTHALER AND T. W. M ¨ULLER Finally, we turn to (6.39d). By Remark 3, the only choices for h2andm2to be considered are h2= 6 andm2= 3,h2= 6 andm2= 2,h2=m2= 3, respectively h2=m2= 2. These correspond to the choices p= 5m/3, 5m/2, 10m/3, respectively 15m/2, out of which only p= 5m/3 andp= 10m/3 have not yet been discussed and belong tothecurrent case. If p= 5m/3, thecorresponding actionof φpis givenby (6.6), so that we have to solve for t1withℓT(t1) = 1 in the equation (6.9). A computation with the help of CHEVIEfinds no solution. If p= 10m/3, the corresponding action of φpis given by (6.11), so that we have to solve for t1withℓT(t1) in the equation (6.14). Using that c5t1c−5=t1, due to Lemma 29 with d= 6, we see that this equation is equivalent with the one in (6.9). Hence, in both cases, the left-hand side of (3.3) is equal to 1, as required. CaseG28=F4.The degrees are 2 ,6,8,12, and hence we have Catm(F4;q) =[12m+12]q[12m+8]q[12m+6]q[12m+2]q [12]q[8]q[6]q[2]q. Letζbe a 12m-th root of unity. The following cases on the right-hand side of (3.3) occur: lim q→ζCatm(F4;q) =m+1,ifζ=ζ12, (6.46a) lim q→ζCatm(F4;q) =3m+2 2,ifζ=ζ8,2|m, (6.46b) lim q→ζCatm(F4;q) = (m+1)(2m+1),ifζ=ζ6,ζ3, (6.46c) lim q→ζCatm(F4;q) =(m+1)(3m+2) 2,ifζ=ζ4, (6.46d) lim q→ζCatm(F4;q) = Catm(F4),ifζ=−1 orζ= 1, (6.46e) lim q→ζCatm(F4;q) = 1,otherwise. (6.46f) We must now prove that the left-handside of (3.3) in each case agre es with the values exhibited in (6.46). The only cases not covered by Lemmas 27 and 28 a re the ones in (6.46b) and (6.46f). By Lemma 26, we are free to choose p= 3m/2. In particular, m must be divisible by 2. From (5.1), we infer φp/parenleftbig (w0;w1,...,w m)/parenrightbig = (∗;c2wm 2+1c−2,c2wm 2+2c−2,...,c2wmc−2,cw1c−1,...,cw m 2c−1/parenrightbig . Supposing that ( w0;w1,...,w m) is fixed by φp, we obtain the system of equations wi=c2wm 2+ic−2, i= 1,2,...,m 2, (6.47a) wi=cwi−m 2c−1, i=m 2+1,m 2+2,...,m. (6.47b) There are four distinct possibilities for choosing the wi’s, 1≤i≤m: (i) all thewi’s are equal to ε(andw0=c), (ii) there is an iwith 1≤i≤m 2such that ℓT(wi) =ℓT(wi+m 2) = 2, (6.48a) and all other wj’s are equal to ε,CYCLIC SIEVING FOR GENERALISED NON-CROSSING PARTITIONS 79 (iii) there is an iwith 1≤i≤m 2such that ℓT(wi) =ℓT(wi+m 2) = 1, (6.48b) and the other wj’s, 1≤j≤m, are equal to ε, (iv) there are i1andi2with 1≤i1|S2(W)|,/producttext i∈S1(W)(mh+di)/producttext i∈S2(W)di/producttext i/∈S1(W)(1−ζdi−h) /producttext i/∈S2(W)(1−ζdi),if|S1(W)|=|S2(W)|. (8.4) Since, by Theorem 25, Catm(W;q) is a polynomial in q, the case |S1(W)|<|S2(W)| cannot occur. We claim that, for the case where |S1(W)|=|S2(W)|, the factors in the quotient of products/producttext i/∈S1(W)(1−ζdi−h)/producttext i/∈S2(W)(1−ζdi) cancel pairwise. If we assume the correctness of the claim, it is obv ious that we get the same result if we replace ζbyζk, where gcd( k,(m+1)h/p) = 1, hence establishing (8.2). In order to see that our claim is indeed valid, we proceed in a case-by- case fash- ion, making appeal to the classification of irreducible well-generated complex reflection groups, which werecalled inSection2. Firstofall, since dn=h, thesetS1(W)isalways non-empty as it contains the element n. Hence, if we want to have |S1(W)|=|S2(W)|, the setS2(W) must be non-empty as well. In other words, the integer ( m+ 1)h/p must divide at least one of the degrees d1,d2,...,d n. In particular, this implies that, for each fixed reflection group Wof exceptional type, only a finite number of values of (m+1)h/phas to be checked. Writing Mfor (m+1)h/p, what needs to be checked is whether the multisets (that is, multiplicities of elements must be taken into account) {(di−h) modM:i /∈S1(W)}and{dimodM:i /∈S2(W)} are the same. Since, for a fixed irreducible well-generated complex r eflection group, thereisonlyafinitenumber ofpossibilities for M, thisamountstoaroutineverification. /square Lemma 35. Letpbe a divisor of (m+ 1)h. Ifpis divisible by m+ 1, then(7.2)is true. Proof.According to (8.1), the action of ψponNCm(W) is described by ψp/parenleftbig (w0;w1,...,w m)/parenrightbig = (cp/(m+1)w0c−p/(m+1);cp/(m+1)w1c−p/(m+1),...,cp/(m+1)wmc−p/(m+1)/parenrightbig . Hence, if (w0;w1,...,w m) is fixed by ψp, then each individual wimust be fixed under conjugation by cp/(m+1). Using the notation W′= Cent W(cp/(m+1)), the previous observation means that wi∈ W′,i= 1,2,...,m. By the theorem of Springer cited in the proof of Lemma 27 and by (5.5), the tuples ( w0;w1,...,w m) fixed byψpare in fact identical with the elements ofCYCLIC SIEVING FOR GENERALISED NON-CROSSING PARTITIONS 13 9 NCm(W′), which implies that |FixNCm(W)(ψp)|=|NCm(W′)|. (8.5) Application of Theorem 1 with Wreplaced by W′and of the “limit rule” (5.4) then yields that |NCm(W′)|=/productdisplay 1≤i≤n (m+1)h p|dimh+di di= Catm(W;q)/vextendsingle/vextendsingle q=e2πip/(m+1)h. (8.6) Combining (8.5) and (8.6), we obtain (7.2). This finishes the proof of t he lemma. /square Lemma 36. Equation (7.2)holds for all divisors pofm+1. Proof.We have Catm(W;q)/vextendsingle/vextendsingle q=e2πip/(m+1)h=/braceleftBigg 0 ifpnthen FixNCm(W)(ψp) =∅. Proof.Let us suppose that ( w0;w1,...,w m)∈FixNCm(W)(ψp) and that there exists a j≥1 such that wj/ne}ationslash=ε. By (8.7), it then follows for such a jthat alsowk/ne}ationslash=εfor all k≡j−lm1b(modm+ 1), where, as before, bis defined as the unique integer with h1=am2+band 0≤b n, a contradiction. This leaves as only possibility ( w0;w1,...,w m) = (c;ε,...,ε). However, this is clearly not an element of Fix NCm(W)(ψp) unlesspis divisible by m+1. This is impossible since p m+1=m1h1 m1m2=h1 m2 is not an integer by our hypotheses. /square Remark 4.(1) If we put ourselves in the situation of the assumptions of Lemma 37, then we may conclude that equation (7.2) only needs to be checked f or pairs (m2,h2) subject to the following restrictions: m2≥2,gcd(h1,m2) = 1,andh2divides all degrees of W. (8.8) Indeed, Lemmas 35 and 37 together imply that equation (7.2) is alway s satisfied except ifm2≥2,h2divides all degrees of W, and gcd(h1,m2) = 1. (2) Still putting ourselves in the situation of Lemma 37, if m2> nandm2h2does not divide any of the degrees of W, then equation (7.2) is satisfied. Indeed, Lemma 38 says that in this case the left-hand side of (7.2) equals 0, while it is obv ious that in this case the right-hand side of (7.2) equals 0 as well. (3)It shouldbeobserved that thisleaves afinitenumber of choices form2to consider, whence a finite number of choices for ( m1,m2,h1,h2). Altogether, there remains a finite number of choices for p=h1m1to be checked.CYCLIC SIEVING FOR GENERALISED NON-CROSSING PARTITIONS 14 1 Lemma 39. LetWbe an irreducible well-generated complex reflection group o f rankn with the property that di|hfori= 1,2,...,n. Then Theorem 33is true for this group W. Proof.By Lemma 34, we may restrict ourselves to divisors pof (m+1)h. Supposethat e2πip/(m+1)hisadi-throotofunityforsome i. Inotherwords, ( m+1)h/p dividesdi. Sincediis a divisor of hby assumption, the integer ( m+1)h/palso divides h. But this is equivalent to saying that m+ 1 divides p, and equation (7.2) holds by Lemma 35. Now assume that ( m+1)h/pdoes not divide any of the di’s. In this case, it follows from (8.4) and the fact that we have S1(W)⊇ {n}andS2(W) =∅that the right- hand side of (7.2) equals 0. Inspection of the classification of all irre ducible well- generated complex reflection groups, which we recalled in Section 2, reveals that all groups satisfying the hypotheses of the lemma have rank n≤2. Except for the groups contained in the infinite series G(d,1,n) andG(e,e,n) for which Theorem 2 has been established in [19], these are the groups G5,G6,G9,G10,G14,G17,G18,G21. We now discuss these groups case by case, keeping the notation of Lemma 37. In order to simplify the argument, we note that Lemma 38 implies that equation (7 .2) holds if m2>2, so that in the following arguments we always may assume that m2= 2. CaseG5. The degrees are 6 ,12, and therefore Remark 4.(1) implies that equa- tion (7.2) is always satisfied. CaseG6. The degrees are 4 ,12, and therefore, according to Remark 4.(1), we need only consider the case where h2= 4 andm2= 2, that is, p= 3(m+1)/2. Then (8.7) becomes ψp/parenleftbig (w0;w1,...,w m)/parenrightbig = (c2wm+1 2c−2;c2wm+3 2c−2,...,c2wmc−2,cw0c−1,...,cw m−1 2c−1/parenrightbig . (8.9) If (w0;w1,...,w m) is fixed by ψp, there must exist an iwith 0≤i≤m−1 2such that ℓT(wi) = 1,wicwic−1=c, and allwj,j/ne}ationslash=i,m+1 2+i, equalε. However, with the help of CHEVIE, one verifies that there is no such solution to this equation. Hence, the left-hand side of (7.2) is equal to 0, as required. CaseG9. The degrees are 8 ,24, and therefore, according to Remark 4.(1), we need only consider the case where h2= 8 andm2= 2, that is, p= 3(m+1)/2. This is the samepas forG6. Again,CHEVIEfinds no solution. Hence, the left-hand side of (7.2) is equal to 0, as required. CaseG10. The degrees are 12 ,24, and therefore, according to Remark 4.(1), we need only consider the case where h2= 12 andm2= 2, that is, p= 3(m+1)/2. This is the samepas forG6. Again,CHEVIEfinds no solution. Hence, the left-hand side of (7.2) is equal to 0, as required. CaseG14. The degrees are 6 ,24, and therefore Remark 4.(1) implies that equa- tion (7.2) is always satisfied. CaseG17. The degrees are 20 ,60, and therefore, according to Remark 4.(1), we need only consider the cases where h2= 20 andm2= 2, respectively that h2= 4 and m2= 2. In the first case, p= 3(m+1)/2, which is the same pas forG6. Again,CHEVIE142 C. KRATTENTHALER AND T. W. M ¨ULLER finds no solution. In the second case, p= 15(m+1)/2. Then (8.7) becomes ψp/parenleftbig (w0;w1,...,w m)/parenrightbig = (c8wm+1 2c−8;c8wm+3 2c−8,...,c8wmc−8,c7w0c−7,...,c7wm−1 2c−7/parenrightbig .(8.10) By Lemma 29, every element of NC(W) is fixed under conjugation by c3, and, thus, on elements fixed by ψp, the above action of ψpreduces to the one in (8.9). This action was already discussed in the first case. Hence, in both cases, the le ft-hand side of (7.2) is equal to 0, as required. CaseG18. The degrees are 30 ,60, and therefore Remark 4.(1) implies that equa- tion (7.2) is always satisfied. CaseG21. The degrees are 12 ,60, and therefore, according to Remark 4.(1), we need only consider the cases where h2= 5 andm2= 2, respectively that h2= 15 and m2= 2. In the first case, p= 5(m+1)/2, so that (8.7) becomes ψp/parenleftbig (w0;w1,...,w m)/parenrightbig = (c3wm+1 2c−3;c3wm+3 2c−3,...,c3wmc−3,c2w0c−2,...,c2wm−1 2c−2/parenrightbig .(8.11) If (w0;w1,...,w m) is fixed by ψp, there must exist an iwith 0≤i≤m−1 2such that ℓT(wi) = 1 and wic2wic−2=c. However, with the help of CHEVIE, one verifies that there is no such solution to this equation. In the second case, p= 15(m+1)/2. Then (8.7) becomes the action in (8.10). By Lemma 29, every element of NC(W) is fixed under conjugation by c5, and, thus, on elements fixed by ψp, the action of ψpin (8.10) reduces to the one in the first case. Hence, in both cases, the left -hand side of (7.2) is equal to 0, as required. This completes the proof of the lemma. /square 9.Case-by-case verification of Theorem 33 We now perform a case-by-case verification of Theorem 33. It sho uld be observed that theactionof ψ(givenin (7.1)) isexactly thesame astheactionof φ(given in(3.1)) withmreplaced by m+1on the components w1,w2,...,w m+1, that is, if we disregard the 0-th component of the elements of the generalised non-cross ing partitions involved. The only difference which arises is that, while the ( m+ 1)-tuples ( w0;w1,...,w m) in (7.1) must satisfy w0w1···wm=c, forw1,w2,...,w m+1in (3.1) we only must have w1w2···wm+1≤Tc. The condition for ( w0;w1,...,w m) of being in Fix NCm(W)(ψp) is therefore exactly the same as the condition on w1,w2,...,w m+1for the element (ε;w1,...,w m,wm+1)beinginFix NCm+1(W)(φp). Consequently, wemayusethecounting results from Section 6, except that we have to restrict our atten tion to those elements (w0;w1,...,w m,wm+1)∈NCm+1(W) for which w1w2···wm+1=c, or, equivalently, w0=ε. As before, we write ζdfor a primitive d-th root of unity. CaseG4.The degrees are 4 ,6, and hence we have Catm(G4;q) =[6m+6]q[6m+4]q [6]q[4]q.CYCLIC SIEVING FOR GENERALISED NON-CROSSING PARTITIONS 14 3 Letζbe a 6(m+ 1)-th root of unity. As before, in what follows we abbreviate the assertion that “ ζis a primitive d-th root of unity” as “ ζ=ζd.” The following cases on the right-hand side of (7.2) occur: lim q→ζCatm(G4;q) =m+1,ifζ=ζ6,ζ3, (9.1a) lim q→ζCatm(G4;q) =3m+3 2,ifζ=ζ4,2|(m+1), (9.1b) lim q→ζCatm(G4;q) = Catm(G4),ifζ=−1 orζ= 1, (9.1c) lim q→ζCatm(G4;q) = 0,otherwise. (9.1d) We must now prove that the left-hand side of (7.2) in each case agre es with the values exhibited in (9.1). The only cases not covered by Lemmas 35 an d 36 are the ones in (9.1b) and (9.1d). On the other hand, the only case left to co nsider according to Remark 4 is the case where h2=m2= 2, that is the case (9.1b) where p= 3(m+1)/2. In particular, m+ 1 must be divisible by 2. The action of ψpis the same as the one in (8.9). Hence, the counting problem is the same as there, except t hat the underlying group now is G4. With the help of CHEVIE, one finds that each of the 3 (complex) reflections in G4which are less than the (chosen) Coxeter element is a valid choice for wi, and each of these choices gives rise to ( m+1)/2 elements in Fix NCm(G4)(ψp) since the indexiranges from 0 to ( m−1)/2. Hence, in total, we obtain 3m+1 2=3m+3 2elements in Fix NCm(G4)(ψp), which agrees with the limit in (9.1b). CaseG8.The degrees are 8 ,12, and hence we have Catm(G8;q) =[12m+12]q[12m+8]q [12]q[8]q. Letζbe a 12(m+ 1)-th root of unity. The following cases on the right-hand side of (7.2) occur: lim q→ζCatm(G8;q) =m+1,ifζ=ζ12,ζ6,ζ3, (9.2a) lim q→ζCatm(G8;q) =3m+3 2,ifζ=ζ8,2|(m+1), (9.2b) lim q→ζCatm(G8;q) = Catm(G8),ifζ=ζ4,−1,1, (9.2c) lim q→ζCatm(G8;q) = 0,otherwise. (9.2d) We must now prove that the left-hand side of (7.2) in each case agre es with the values exhibited in (9.2). The only cases not covered by Lemmas 35 an d 36 are the ones in (9.2b) and (9.2d). On the other hand, the only case left to co nsider according to Remark 4 is the case where h2= 4 andm2= 2, that is the case (9.2b) where p= 3(m+ 1)/2. In particular, m+1 must be divisible by 2. The action of ψpis the same as the one in (8.9). Hence, the counting problem is the same as t here, except that the underlying group now is G8. With the help of CHEVIE, one finds that each of the 3 (complex) reflections in G8which are less than the (chosen) Coxeter element is a valid choice for wi, and each of these choices gives rise to ( m+ 1)/2 elements in FixNCm(G8)(ψp) since the index iranges from 0 to ( m−1)/2.144 C. KRATTENTHALER AND T. W. M ¨ULLER Hence, in total, we obtain 3m+1 2=3m+3 2elements in Fix NCm(G8)(ψp), which agrees with the limit in (9.2b). CaseG16.The degrees are 20 ,30, and hence we have Catm(G16;q) =[30m+30]q[30m+20]q [30]q[20]q. Letζbe a 30(m+ 1)-th root of unity. The following cases on the right-hand side of (7.2) occur: lim q→ζCatm(G16;q) =m+1,ifζ=ζ30,ζ15,ζ6,ζ3, (9.3a) lim q→ζCatm(G16;q) =3m+3 2,ifζ=ζ20,ζ4,2|(m+1), (9.3b) lim q→ζCatm(G16;q) = Catm(G16),ifζ=ζ10,ζ5,−1,1, (9.3c) lim q→ζCatm(G16;q) = 0,otherwise. (9.3d) We must now prove that the left-hand side of (7.2) in each case agre es with the values exhibited in (9.3). The only cases not covered by Lemmas 35 an d 36 are the ones in (9.3b) and (9.3d). On the other hand, the only cases left to c onsider according to Remark 4 are the cases where h2= 10 andm2= 2, respectively h2=m2= 2. Both cases belong to (9.3b). In the first case, we have p= 3(m+1)/2, while in the second case we have p= 15(m+ 1)/2. In particular, m+ 1 must be divisible by 2. In the first case, the action of ψpis the same as the one in (8.9). Hence, the counting problem is the same as there, except that the underlying group now is G16. With the help of CHEVIE, one finds that each of the 3 (complex) reflections in G16which are less than the (chosen) Coxeter element is a valid choice for wi, and each of these choices gives rise to (m+ 1)/2 elements in Fix NCm(G16)(ψp) since the index iranges from 0 to ( m−1)/2. On the other hand, if p= 15(m+1)/2, then the action of ψpis the same as the one in (8.10). By Lemma 29, every element of NC(G16) is fixed under conjugation by c3, and, thus, on elements fixed by ψp, the action of ψpreduces to the one in the first case. Hence, in total, we obtain 3m+1 2=3m+3 2elements in Fix NCm(G16)(ψp), which agrees with the limit in (9.3b). CaseG20.The degrees are 12 ,30, and hence we have Catm(G20;q) =[30m+30]q[30m+12]q [30]q[12]q. Letζbe a 30(m+ 1)-th root of unity. The following cases on the right-hand side of (7.2) occur: lim q→ζCatm(G20;q) =m+1,ifζ=ζ30,ζ15,ζ10,ζ5, (9.4a) lim q→ζCatm(G20;q) =5m+5 2,ifζ=ζ12,ζ4,2|(m+1), (9.4b) lim q→ζCatm(G20;q) = Catm(G20),ifζ=ζ6,ζ3,−1,1, (9.4c) lim q→ζCatm(G20;q) = 0,otherwise. (9.4d)CYCLIC SIEVING FOR GENERALISED NON-CROSSING PARTITIONS 14 5 We must now prove that the left-hand side of (7.2) in each case agre es with the values exhibited in (9.4). The only cases not covered by Lemmas 35 an d 36 are the ones in (9.4b) and (9.4d). On the other hand, the only cases left to c onsider according to Remark 4 are the cases where h2= 6 andm2= 2, respectively h2=m2= 2. Both cases belong to (9.4b). In the first case, we have p= 5(m+1)/2, while in the second case we have p= 15(m+1)/2. In particular, m+1 must be divisible by 2. In the first case, the action of ψpis the same as the one in (8.11). Hence, the counting problem is the same as there, except that the underlying group now is G20. With the help of CHEVIE, one finds that each of the 5 (complex) reflections in G20which are less than the (chosen) Coxeter element is a valid choice for wi, and each of these choices gives rise to (m+ 1)/2 elements in Fix NCm(G20)(ψp) since the index iranges from 0 to ( m−1)/2. On the other hand, if p= 15(m+1)/2, then the action of ψpis the same as the one in (8.10). By Lemma 29, every element of NC(G20) is fixed under conjugation by c5, and, thus, on elements fixed by ψp, the action of ψpreduces to the one in the first case. Hence, in total, we obtain 5m+1 2=5m+5 2elements in Fix NCm(G20)(ψp), which agrees with the limit in (9.4b). CaseG23=H3.The degrees are 2 ,6,10, and hence we have Catm(H3;q) =[10m+10]q[10m+6]q[10m+2]q [10]q[6]q[2]q. Letζbe a 10(m+ 1)-th root of unity. The following cases on the right-hand side of (7.2) occur: lim q→ζCatm(H3;q) =m+1,ifζ=ζ10,ζ5, (9.5a) lim q→ζCatm(H3;q) =5m+5 3,ifζ=ζ6,ζ3,3|(m+1), (9.5b) lim q→ζCatm(H3;q) = Catm(H3),ifζ=−1 orζ= 1, (9.5c) lim q→ζCatm(H3;q) = 0,otherwise. (9.5d) We must now prove that the left-hand side of (7.2) in each case agre es with the values exhibited in (9.5). The only cases not covered by Lemmas 35 an d 36 are the ones in (9.5b) and (9.5d). On the other hand, the only cases left to c onsider according to Remark 4 are the cases where h2= 1 andm2= 3,h2= 2 andm2= 3, and h2=m2= 2. These correspond to the choices p= 10(m+ 1)/3,p= 5(m+ 1)/3, respectively p= 5(m+1)/2. The first two cases belong to (9.5b), while p= 5(m+1)/2 belongs to (9.5d). In the case that p= 5(m+1)/3, the action of ψpis given by ψp/parenleftbig (w0;w1,...,w m)/parenrightbig = (c2wm+1 3c−2;c2wm+4 3c−2,...,c2wmc−2,cw0c−1,...,cw m−2 3c−1/parenrightbig . Hence, for an iwith 0≤i≤m−2 3, we must find an element wi=t1, wheret1satisfies (6.9), andall other wj,j /∈/braceleftbig i,i+m+1 3,i+2(m+1) 3/bracerightbig , are set equal to ε. We have found five solutions to the counting problem (6.9) in (6.10). Each of them gives r ise to (m+1)/3 elements in Fix NCm(H3)(ψp) since the index iranges from 0 to ( m−2)/3. On the other146 C. KRATTENTHALER AND T. W. M ¨ULLER hand, ifp= 10(m+1)/3, then the action of ψpis given by ψp/parenleftbig (w0;w1,...,w m)/parenrightbig = (c4w2m+2 3c−4;c4w2m+5 3c−4,...,c4wmc−4,c3w0c−3,...,c3w2m−1 3c−3/parenrightbig . By Lemma 29, every element of NC(H3) is fixed under conjugation by c5, and, thus, on elements fixed by ψp, the action of ψpreduces to the one in the first case. Hence, in total, we obtain 5m+1 3=5m+5 3elements in Fix NCm(H3)(ψp), which agrees with the limit in (9.5b). Ifp= 5(m+ 1)/2, then the action of ψpis the same as the one in (8.11). The computation at the end of Case H3in Section 6 did not find any solutions, which is in agreement with (9.5d). CaseG24.The degrees are 4 ,6,14, and hence we have Catm(G24;q) =[14m+14]q[14m+6]q[14m+4]q [14]q[6]q[4]q. Letζbe a 14(m+ 1)-th root of unity. The following cases on the right-hand side of (7.2) occur: lim q→ζCatm(G24;q) =m+1,ifζ=ζ14,ζ7, (9.6a) lim q→ζCatm(G24;q) =7m+7 3,ifζ=ζ6,ζ3,3|(m+1), (9.6b) lim q→ζCatm(G24;q) = Catm(G24),ifζ=−1 orζ= 1, (9.6c) lim q→ζCatm(G24;q) = 0,otherwise. (9.6d) We must now prove that the left-hand side of (7.2) in each case agre es with the values exhibited in (9.6). The only cases not covered by Lemmas 35 an d 36 are the ones in (9.6b) and (9.6d). On the other hand, the only cases left to c onsider according to Remark 4 are the cases where h2= 1 andm2= 3,h2= 2 andm2= 3, and h2=m2= 2. These correspond to the choices p= 14(m+ 1)/3,p= 7(m+ 1)/3, respectively p= 7(m+1)/2. The first two cases belong to (9.6b), while p= 7(m+1)/2 belongs to (9.6d). In the case that p= 14(m+1)/3 orp= 7(m+1)/3, we have found seven solutions to the counting problem (6.18) in (6.19), and each of them gives rise to ( m+1)/3 elements in Fix NCm(G24)(ψp) (in the style as discussed in Case H3). Hence, in total, we obtain 7m+1 3=7m+7 3elements in Fix NCm(G24)(ψp), which agrees with the limit in (9.6b). Ifp= 7(m+ 1)/2, the relevant counting problem is (6.25). However, no element (w0;w1,...,w m)∈FixNCm(G24)(ψp) can be produced in this way since the counting problem imposes the restriction that ℓT(w0) +ℓT(w1) +···+ℓT(wm) be even, which contradicts the fact that ℓT(c) =n= 3. This is in agreement with the limit in (9.6d). CaseG25.The degrees are 6 ,9,12, and hence we have Catm(G25;q) =[12m+12]q[12m+9]q[12m+6]q [12]q[9]q[6]q.CYCLIC SIEVING FOR GENERALISED NON-CROSSING PARTITIONS 14 7 Letζbe a 12(m+ 1)-th root of unity. The following cases on the right-hand side of (7.2) occur: lim q→ζCatm(G25;q) =m+1,ifζ=ζ12,ζ4, (9.7a) lim q→ζCatm(G25;q) =4m+4 3,ifζ=ζ9,3|(m+1), (9.7b) lim q→ζCatm(G25;q) = (m+1)(2m+1),ifζ=ζ6,−1 (9.7c) lim q→ζCatm(G25;q) = Catm(G25),ifζ=ζ3,1, (9.7d) lim q→ζCatm(G25;q) = 0,otherwise. (9.7e) We must now prove that the left-hand side of (7.2) in each case agre es with the values exhibited in (9.7). The only cases not covered by Lemmas 35 an d 36 are the ones in (9.7b) and (9.7e). On the other hand, the only case left to co nsider according to Remark4isthecasewhere h2=m2= 3. Thiscorrespondstothechoice p= 4(m+1)/3, which belongs to (9.7b). We have found four solutions to the countin g problem (6.30) in (6.31), and each of them gives rise to ( m+1)/3 elements in Fix NCm(G25)(ψp) (in the style as discussed in Case H3). Hence, in total, we obtain 4m+1 3=4m+4 3elements in FixNCm(G25)(ψp), which agrees with the limit in (9.7b). CaseG26.The degrees are 6 ,12,18, and hence we have Catm(G26;q) =[18m+18]q[18m+12]q[18m+6]q [18]q[12]q[6]q. Letζbe a 14(m+ 1)-th root of unity. The following cases on the right-hand side of (7.2) occur: lim q→ζCatm(G26;q) =m+1,ifζ=ζ18,ζ9, (9.8a) lim q→ζCatm(G26;q) = Catm(G26),ifζ=ζ6,ζ3,−1,1, (9.8b) lim q→ζCatm(G26;q) = 0,otherwise. (9.8c) We must now prove that the left-handside of (7.2) in each case agre es with the values exhibited in (9.8). The only case not covered by Lemmas 35 and 36 is th e one in (9.8c). On the other hand, the only cases left to consider according to Rem ark 4 are the cases whereh2= 6 andm2= 2, respectively h2=m2= 2. These correspond to the choices p= 3(m+1)/2,respectively p= 9(m+1)/2,bothofwhichbelongto(9.8c). Therelevant counting problem is (6.35). However, no element ( w0;w1,...,w m)∈FixNCm(G26)(ψp) can be produced in this way since the counting problem imposes the re striction that ℓT(w0)+ℓT(w1)+···+ℓT(wm) be even, which is absurd. This is in agreement with the limit in (9.8c). CaseG27.The degrees are 6 ,12,30, and hence we have Catm(G27;q) =[30m+30]q[30m+12]q[30m+6]q [30]q[12]q[6]q.148 C. KRATTENTHALER AND T. W. M ¨ULLER Letζbe a 14(m+ 1)-th root of unity. The following cases on the right-hand side of (7.2) occur: lim q→ζCatm(G27;q) =m+1,ifζ=ζ30,ζ15,ζ10,ζ5, (9.9a) lim q→ζCatm(G27;q) = Catm(G27),ifζ=ζ6,ζ3,−1,1, (9.9b) lim q→ζCatm(G27;q) = 0,otherwise. (9.9c) We must now prove that the left-handside of (7.2) in each case agre es with the values exhibited in (9.9). The only case not covered by Lemmas 35 and 36 is th e one in (9.9c). On the other hand, the only cases left to consider according to Rem ark 4 are the cases whereh2= 6 andm2= 3,h2=m2= 3,h2= 6 andm2= 2, respectively h2=m2= 2. These correspond to the choices p= 5(m+1)/3, 10(m+1)/3, 5(m+1)/2, respectively 15(m+1)/2, all of which belong to (9.9c). Ifp= 5(m+ 1)/3 orp= 10(m+ 1)/3, the computation with the help of CHEVIE at the end of Case G27in Section 6 did not find any solutions for the corresponding counting problem. This is in agreement with the limit in (9.9c). In the case that 5( m+1)/2 or 15(m+1)/2, the relevant counting problem is (6.42). However, no element ( w0;w1,...,w m)∈FixNCm(G27)(ψp) can be produced in this way since the counting problem imposes the restriction that ℓT(w0)+ℓT(w1)+···+ℓT(wm) be even, which is absurd. This is again in agreement with the limit in (9.9c) . CaseG28=F4.The degrees are 2 ,6,8,12, and hence we have Catm(F4;q) =[12m+12]q[12m+8]q[12m+6]q[12m+2]q [12]q[8]q[6]q[2]q. Letζbe a 12(m+ 1)-th root of unity. The following cases on the right-hand side of (7.2) occur: lim q→ζCatm(F4;q) =m+1,ifζ=ζ12, (9.10a) lim q→ζCatm(F4;q) =3m+3 2,ifζ=ζ8,2|(m+1), (9.10b) lim q→ζCatm(F4;q) = (m+1)(2m+1),ifζ=ζ6,ζ3, (9.10c) lim q→ζCatm(F4;q) =(m+1)(3m+2) 2,ifζ=ζ4, (9.10d) lim q→ζCatm(F4;q) = Catm(F4),ifζ=−1 orζ= 1, (9.10e) lim q→ζCatm(F4;q) = 0,otherwise. (9.10f) We must now prove that the left-handside of (7.2) in each case agre es with the values exhibited in (9.10). The only cases not covered by Lemmas 35 and 36 a re the ones in (9.10b) and (9.10f). On the other hand, according to Remark 4, th e are no choices for h2andm2left to be considered. CaseG29.The degrees are 4 ,8,12,20, and hence we have Catm(G29;q) =[20m+20]q[20m+12]q[20m+8]q[20m+4]q [20]q[12]q[8]q[4]q.CYCLIC SIEVING FOR GENERALISED NON-CROSSING PARTITIONS 14 9 Letζbe a 20(m+ 1)-th root of unity. The following cases on the right-hand side of (7.2) occur: lim q→ζCatm(G29;q) =m+1,ifζ=ζ20,ζ10,ζ5, (9.11a) lim q→ζCatm(G29;q) = Catm(G29),ifζ=ζ4,−1,1, (9.11b) lim q→ζCatm(G29;q) = 0,otherwise. (9.11c) We must now prove that the left-hand side of (7.2) in each case agre es with the values exhibited in (9.11). The only case not covered by Lemmas 35 an d 36 is the one in(9.11c). Ontheother hand, theonly cases left to consider accor ding to Remark4, the only choices for h2andm2to be considered are h2= 1 andm2= 3,h2= 2 andm2= 3, h2= 4 andm2= 3,h2= 4 andm2= 2, respectively h2=m2= 4. These correspond to the choices p= 20(m+ 1)/3,p= 10(m+ 1)/3,p= 5(m+ 1)/3,p= 5(m+ 1)/2, respectively p= 5(m+1)/4, all of which belong to (9.11c). In the case that p= 20(m+1)/3,p= 10(m+1)/3, orp= 5(m+1)/3, the relevant counting problem is (6.55). However, no element ( w0;w1,...,w m)∈FixNCm(G27)(ψp) can be produced in this way since the counting problem imposes the re striction that ℓT(w0)+ℓT(w1)+···+ℓT(wm) be divisible by 3, which is absurd. This is in agreement with the limit in (9.11c). In the case that p= 5(m+1)/2, the relevant counting problem is (6.64), for which we did not find any solutions. This is again in agreement with the limit in (9.1 1c). In the case that p= 5(m+1)/4, the computation at the end of Case G29in Section 6 did not find any solutions, which is as well in agreement with the limit in (9.1 1c). CaseG30=H4.The degrees are 2 ,12,20,30, and hence we have Catm(H4;q) =[30m+30]q[30m+20]q[30m+12]q[30m+2]q [30]q[20]q[12]q[2]q. Letζbe a 30(m+ 1)-th root of unity. The following cases on the right-hand side of (7.2) occur: lim q→ζCatm(H4;q) =m+1,ifζ=ζ30,ζ15, (9.12a) lim q→ζCatm(H4;q) =3m+3 2,ifζ=ζ20,2|(m+1), (9.12b) lim q→ζCatm(H4;q) =5m+5 2,ifζ=ζ12,2|(m+1), (9.12c) lim q→ζCatm(H4;q) =(m+1)(3m+2) 2,ifζ=ζ10,ζ5, (9.12d) lim q→ζCatm(H4;q) =(m+1)(5m+2) 2,ifζ=ζ6,ζ3, (9.12e) lim q→ζCatm(H4;q) =(m+1)(15m+1) 4,ifζ=ζ4,2|(m+1), (9.12f) lim q→ζCatm(H4;q) = Catm(H4),ifζ=−1 orζ= 1, (9.12g) lim q→ζCatm(H4;q) = 0,otherwise. (9.12h) We must now prove that the left-hand side of (7.2) in each case agre es with the values exhibited in (9.12). The only cases not covered by Lemmas 35 a nd 36 are150 C. KRATTENTHALER AND T. W. M ¨ULLER the ones in (9.12b), (9.12c), (9.12f), and (9.12h). On the other ha nd, the only cases left to consider according to Remark 4 are the cases where h2= 2 andm2= 4, respectively h2=m2= 2. Thesecorrespondtothechoices p= 15(m+1)/2,respectively p= 15(m+1)/4, out of which the first belongs to (9.12f), while the second belongs to (9.12h). In the case that p= 15(m+1)/2, the action of ψpis the same as the one in (8.10). We have found eight solutions to the counting problem (6.80) in (6.83) , each of them giving rise to ( m+1)/2 elements in Fix NCm(H4)(ψp) since the index i(in (6.80)) ranges from 0 to ( m−1)/2, and we have found 30 solutions to the counting problem (6.82) in (6.84), each of them giving rise to/parenleftbig(m+1)/2 2/parenrightbig elements in Fix NCm(H4)(ψp) since 0 ≤ i1< i2≤(m−1)/2 (in (6.82)). Hence, we obtain 8m+1 2+ 30/parenleftbig(m+1)/2 2/parenrightbig =(m+1)(15m+1) 4 elements in Fix NCm(H4)(ψp), which agrees with the limit in (9.12f). Ifp= 15(m+1)/4, the computation at the end of Case H4in Section 6 did not find any solutions, which is in agreement with the limit in (9.12h). CaseG32.The degrees are 12 ,18,24,30, and hence we have Catm(G32;q) =[30m+30]q[30m+24]q[30m+18]q[30m+12]q [30]q[24]q[18]q[12]q. Letζbe a 30(m+ 1)-th root of unity. The following cases on the right-hand side of (7.2) occur: lim q→ζCatm(G32;q) =m+1,ifζ=ζ30,ζ15,ζ10,ζ5, (9.13a) lim q→ζCatm(G32;q) =5m+5 4,ifζ=ζ24,ζ8,4|(m+1), (9.13b) lim q→ζCatm(G32;q) =(5m+5)(5m+3) 8,ifζ=ζ12,ζ4,2|(m+1), (9.13c) lim q→ζCatm(G32;q) = Catm(G32),ifζ=ζ6,ζ3,−1,1, (9.13d) lim q→ζCatm(G32;q) = 0,otherwise. (9.13e) We must now prove that the left-hand side of (7.2) in each case agre es with the values exhibited in (9.13). The only cases not covered by Lemmas 35 a nd 36 are the ones in (9.13b), (9.13c), and (9.13e). On the other hand, the only c ases left to consider according to Remark 4 are the cases where h2= 2 andm2= 4,h2= 6 andm2= 4, h2=m2= 3,h2= 6 andm2= 3,h2=m2= 2, respectively h2= 6 andm2= 2. These correspond to the choices p= 15(m+1)/4,p= 5(m+1)/4,p= 10(m+ 1)/3, p= 5(m+1)/3,p= 15(m+1)/2, respectively p= 5(m+1)/2, out of which the first two belong to (9.13b), the next two belong to (9.13e), and the last two b elong to (9.13c). In the case that p= 15(m+1)/4 orp= 5(m+1)/4, we have found five solutions to the counting problem (6.88) in (6.89), each of them giving rise to ( m+1)/4 elements in Fix NCm(G32)(ψp). Hence, we obtain 5m+1 4=5m+5 4elements in Fix NCm(G32)(ψp), which agrees with the limit in (9.13b). In the case that p= 10(m+1)/3 orp= 5(m+1)/3, the relevant counting problem is (6.95). However, no element ( w0;w1,...,w m)∈FixNCm(G32)(ψp) can be produced in this way since the counting problem imposes the restriction that ℓT(w0)+ℓT(w1)+ ···+ℓT(wm) be divisible by 3, which is absurd. This is in agreement with the limit in (9.13e).CYCLIC SIEVING FOR GENERALISED NON-CROSSING PARTITIONS 15 1 In the case that p= 15(m+1)/2 orp= 5(m+1)/2, we have found ten solutions to the counting problem (6.102) in (6.105), each of them giving rise to ( m+1)/2 elements in Fix NCm(G32)(ψp), and we have found 25 solutions to the counting problem (6.104) in (6.106), each of them giving rise to/parenleftbig(m+1)/2 2/parenrightbig elements in Fix NCm(G32)(ψp). Hence, we obtain 10m+1 2+ 25/parenleftbig(m+1)/2 2/parenrightbig =(5m+5)(5m+3) 8elements in Fix NCm(G32)(ψp), which agrees with the limit in (9.13c). CaseG33.The degrees are 4 ,6,10,12,18, and hence we have Catm(G33;q) =[18m+18]q[18m+12]q[18m+10]q[18m+6]q[18m+4]q [18]q[12]q[10]q[6]q[4]q. Letζbe a 18(m+ 1)-th root of unity. The following cases on the right-hand side of (7.2) occur: lim q→ζCatm(G33;q) =m+1,ifζ=ζ18,ζ9, (9.14a) lim q→ζCatm(G33;q) =9m+9 5,ifζ=ζ10,ζ5,5|(m+1), (9.14b) lim q→ζCatm(G33;q) =(m+1)(3m+2)(3m+1) 2,ifζ=ζ6,ζ3, (9.14c) lim q→ζCatm(G33;q) = Catm(G33),ifζ=−1 orζ= 1, (9.14d) lim q→ζCatm(G33;q) = 0,otherwise. (9.14e) We must now prove that the left-handside of (7.2) in each case agre es with the values exhibited in (9.14). The only cases not covered by Lemmas 35 and 36 a re the ones in (9.14b) and (9.14e). On the other hand, the only cases left to cons ider according to Remark 4 are the cases where h2= 1 andm2= 5,h2= 2 andm2= 5,h2= 2 and m2= 4, respectively h2=m2= 2. These correspond to the choices p= 18(m+1)/5, p= 9(m+ 1)/5,p= 9(m+ 1)/4, respectively p= 9(m+ 1)/2, out of which the first two belong to (9.14b), while the others belong to (9.14e). In the case that p= 18(m+1)/5 orp= 9(m+1)/5, we have found nine solutions to the counting problem (6.115) in (6.116). Hence, we obtain 9m+1 5=9m+9 5elements in FixNCm(G33)(ψp), which agrees with the limit in (9.14b). Ifp= 9(m+1)/4, the computation at the end of Case G33in Section 6 did not find any solutions, which is again in agreement with the limit in (9.13e). In the case that p= 9(m+ 1)/2, the relevant counting problems are (6.122) and (6.124). However, no element ( w0;w1,...,w m)∈FixNCm(G33)(ψp) can be produced in this way since the counting problem imposes the restriction that ℓT(w0)+ℓT(w1)+···+ ℓT(wm) be even, which is absurd. This is in agreement with the limit in (9.14e). CaseG34.The degrees are 6 ,12,18,24,30,42, and hence we have Catm(G34;q) =[42m+42]q[42m+30]q[42m+24]q [42]q[30]q[24]q ×[42m+18]q[42m+12]q[42m+6]q [18]q[12]q[6]q.152 C. KRATTENTHALER AND T. W. M ¨ULLER Letζbe a 42(m+ 1)-th root of unity. The following cases on the right-hand side of (7.2) occur: lim q→ζCatm(G34;q) =m+1,ifζ=ζ42,ζ21,ζ14,ζ7, (9.15a) lim q→ζCatm(G34;q) = Catm(G34),ifζ=ζ6,ζ3,−1,1, (9.15b) lim q→ζCatm(G34;q) = 0,otherwise. (9.15c) We must now prove that the left-hand side of (7.2) in each case agre es with the values exhibited in (9.15). The only case not covered by Lemmas 35 an d 36 is the one in (9.15c). On the other hand, the only cases left to consider accor ding to Remark 4 are the cases where h2= 1 andm2= 5,h2= 2 andm2= 5,h2= 3 andm2= 5,h2= 6 andm2= 5,h2= 2 andm2= 4,h2= 6 andm2= 4,h2=m2= 3,h2= 6 andm2= 3, h2=m2= 2,h2= 6 andm2= 2, respectively h2= 6 andm2= 6. These correspond to the choices p= 42(m+1)/5,p= 21(m+1)/5,p= 14(m+1)/5,p= 7(m+ 1)/5, p= 21(m+1)/4,p= 7(m+1)/4,p= 14(m+1)/3,p= 7(m+1)/3,p= 21(m+1)/2, p= 7(m+1)/2, respectively p= 7(m+1)/6, all of which belong to (9.15c). Inthecasethat p= 42(m+1)/5,p= 21(m+1)/5,p= 14(m+1)/5,orp= 7(m+1)/5, the relevant counting problem is (6.128). However, no element ( w0;w1,...,w m)∈ FixNCm(G34)(ψp) can be produced in this way since the counting problem imposes the restriction that ℓT(w0)+ℓT(w1)+···+ℓT(wm) be divisible by 5, which is absurd. This is in agreement with the limit in (9.15c). In the case that p= 21(m+1)/4 orp= 7(m+1)/4, the relevant counting problem is (6.140). However, no element ( w0;w1,...,w m)∈FixNCm(G34)(ψp) can be produced in this way since the counting problem imposes the restriction that ℓT(w0)+ℓT(w1)+ ···+ℓT(wm) be divisible by 4, which is absurd. This is in agreement with the limit in (9.15c). In the case that p= 14(m+1)/3 orp= 7(m+1)/3, the relevant counting problems are (6.146) and (6.148). However, the computations with the help o fCHEVIEperformed in CaseG34in Section 6 did not find any solutions for (6.146) or (6.148). This is in agreement with the limit in (9.15c). In the case that p= 21(m+1)/2, the relevant counting problems are (6.157), (6.158), and (6.159). However, the computations with the help of CHEVIEperformed in Case G34 in Section 6 found no wiwithℓT(wi) = 3 in (6.157), and hence no solutions for( wi1,wi2) withℓT(wi1)+ℓT(wi2) = 3 in (6.158), and no solutions for ( wi1,wi2,wi3) in (6.159). This is in agreement with the limit in (9.15c). Ifp= 7(m+1)/6, the computation at the end of Case G34in Section 6 did not find any solutions, which is also in agreement with the limit in (9.15c). CaseG35=E6.The degrees are 2 ,5,6,8,9,12, and hence we have Catm(E6;q) =[12m+12]q[12m+9]q[12m+8]q[12m+6]q[12m+5]q[12m+2]q [12]q[9]q[8]q[6]q[5]q[2]q.CYCLIC SIEVING FOR GENERALISED NON-CROSSING PARTITIONS 15 3 Letζbe a 12(m+ 1)-th root of unity. The following cases on the right-hand side of (7.2) occur: lim q→ζCatm(E6;q) =m+1,ifζ=ζ12, (9.16a) lim q→ζCatm(E6;q) =4m+4 3,ifζ=ζ9,3|(m+1), (9.16b) lim q→ζCatm(E6;q) =3m+3 2,ifζ=ζ8,2|(m+1), (9.16c) lim q→ζCatm(E6;q) = (m+1)(2m+1),ifζ=ζ6, (9.16d) lim q→ζCatm(E6;q) =(m+1)(3m+2) 2,ifζ=ζ4, (9.16e) lim q→ζCatm(E6;q) =(m+1)(4m+3)(2m+1) 3,ifζ=ζ3, (9.16f) lim q→ζCatm(E6;q) =(m+1)(3m+2)(2m+1)(6m+1) 2,ifζ=−1, (9.16g) lim q→ζCatm(E6;q) = Catm(E6),ifζ= 1, (9.16h) lim q→ζCatm(E6;q) = 0,otherwise. (9.16i) We must now prove that the left-hand side of (7.2) in each case agre es with the values exhibited in (9.16). The only cases not covered by Lemmas 35 a nd 36 are the ones in (9.16b), (9.16c), and (9.16i). On the other hand, the only c ases left to consider according to Remark 4 are the cases where h2= 1 andm2= 5, respectively h2= 2 and m2= 5. These correspond to the choices p= 12(m+1)/5, respectively p= 6(m+1)/5, both of which belong to (9.16i). In the case that p= 12(m+1)/5, the relevant counting problem is (6.172). However, no element ( w0;w1,...,w m)∈FixNCm(E6)(ψp) can be produced in this way since the counting problemimposes therestriction that ℓT(w0)+ℓT(w1)+···+ℓT(wm)bedivisible by 5, which is absurd. This is in agreement with the limit in (9.16i). Ifp= 6(m+1)/5, the computation at the end of Case E6in Section 6 did not find any solutions, which is also in agreement with the limit in (9.16i). CaseG36=E7.The degrees are 2 ,6,8,10,12,14,18, and hence we have Catm(E7;q) =[18m+18]q[18m+14]q[18m+12]q [18]q[14]q[12]q ×[18m+10]q[18m+8]q[18m+6]q[18m+2]q [10]q[8]q[6]q[2]q.154 C. KRATTENTHALER AND T. W. M ¨ULLER Letζbe a 18(m+ 1)-th root of unity. The following cases on the right-hand side of (7.2) occur: lim q→ζCatm(E7;q) =m+1,ifζ=ζ18,ζ9, (9.17a) lim q→ζCatm(E7;q) =9m+9 7,ifζ=ζ14,ζ7,7|(m+1), (9.17b) lim q→ζCatm(E7;q) =(m+1)(3m+2)(3m+1) 2,ifζ=ζ6,ζ3, (9.17c) lim q→ζCatm(E7;q) = Catm(E7),ifζ=−1 orζ= 1, (9.17d) lim q→ζCatm(E7;q) = 0,otherwise. (9.17e) We must now prove that the left-handside of (7.2) in each case agre es with the values exhibited in (9.17). The only cases not covered by Lemmas 35 and 36 a re the ones in (9.17b) and (9.17e). On the other hand, the only cases left to cons ider according to Remark 4 are the cases where h2= 1 andm2= 7,h2= 2 andm2= 7,h2= 1 andm2= 5,h2= 2 andm2= 5,h2= 2 andm2= 4, respectively h2=m2= 2. These correspond to the choices p= 18(m+1)/7,p= 9(m+1)/7,p= 18(m+ 1)/5, p= 9(m+ 1)/5,p= 9(m+ 1)/4, respectively p= 9(m+ 1)/2, out of which the first two belong to (9.16b), and all others belong to (9.16i). In the case that p= 18(m+1)/7 orp= 9(m+1)/7, we have found nine solutions to the counting problem (6.176) in (6.177). Hence, we obtain 9m+1 7=9m+9 7elements in FixNCm(E7)(ψp), which agrees with the limit in (9.17b). In the case that p= 18(m+1)/5 orp= 9(m+1)/5, the relevant counting problem is (6.186). However, no element ( w0;w1,...,w m)∈FixNCm(E7)(ψp) can be produced in this way since the counting problem imposes the restriction that ℓT(w0)+ℓT(w1)+ ···+ℓT(wm) be divisible by 5, which is absurd. This is in agreement with the limit in (9.17e). In the case that p= 9(m+1)/4, the relevant counting problem is (6.192). However, no element ( w0;w1,...,w m)∈FixNCm(E7)(ψp) can be produced in this way since the counting problemimposes therestriction that ℓT(w0)+ℓT(w1)+···+ℓT(wm)bedivisible by 4, which is absurd. This is again in agreement with the limit in (9.17e). In the case that p= 9(m+1)/2, the relevant counting problems are (6.195), (6.196), and (6.197). However, no element ( w0;w1,...,w m)∈FixNCm(E7)(ψp) can be produced in this way since the counting problem imposes the restriction that ℓT(w0)+ℓT(w1)+ ···+ℓT(wm) be even, which is absurd. This is also in agreement with the limit in (9.17e). CaseG37=E8.The degrees are 2 ,8,12,14,18,20,24,30, and hence we have Catm(E8;q) =[30m+30]q[30m+24]q[30m+20]q[30m+18]q [30]q[24]q[20]q[18]q ×[30m+14]q[30m+12]q[30m+8]q[30m+2]q [14]q[12]q[8]q[2]q.CYCLIC SIEVING FOR GENERALISED NON-CROSSING PARTITIONS 15 5 Letζbe a 30(m+ 1)-th root of unity. The following cases on the right-hand side of (7.2) occur: lim q→ζCatm(E8;q) =m+1,ifζ=ζ30,ζ15, (9.18a) lim q→ζCatm(E8;q) =5m+5 4,ifζ=ζ24,4|(m+1), (9.18b) lim q→ζCatm(E8;q) =3m+3 2,ifζ=ζ20,2|(m+1), (9.18c) lim q→ζCatm(E8;q) =(5m+5)(5m+3) 8,ifζ=ζ12,2|(m+1), (9.18d) lim q→ζCatm(E8;q) =(m+1)(3m+2) 2,ifζ=ζ10,ζ5, (9.18e) lim q→ζCatm(E8;q) =(5m+5)(15m+7) 16,ifζ=ζ8,4|(m+1), (9.18f) lim q→ζCatm(E8;q) =(m+1)(5m+4)(5m+3)(5m+2) 24,ifζ=ζ6,ζ3,(9.18g) lim q→ζCatm(E8;q) =(m+1)(5m+3)(15m+7)(15m+1) 64,ifζ=ζ4,2|(m+1), (9.18h) lim q→ζCatm(E8;q) = Catm(E8),ifζ=−1 orζ= 1, (9.18i) lim q→ζCatm(E8;q) = 0,otherwise. (9.18j) We must now prove that the left-handside of (7.2) in each case agre es with the values exhibited in (9.18). The only cases not covered by Lemmas 35 and 36 a re the ones in (9.18b), (9.18c), (9.18d), (9.18f), (9.18h), and (9.18j). On the o ther hand, the only cases left to consider according to Remark 4 are the cases where h2= 2 andm2= 8, h2= 1 andm2= 7,h2= 2 andm2= 7,h2= 2 andm2= 4, respectively h2=m2= 2. These correspond to the choices p= 15(m+1)/8,p= 30(m+1)/7,p= 15(m+1)/7, p= 15(m+1)/4, respectively 15( m+1)/2, out of which the first three belong to (9.18j), the fourth belongs to (9.18f), and the last belongs to (9.18h). Ifp= 15(m+1)/8, the relevant counting problem is (6.242). However, the computa - tion at the end of Case E8in Section 6 did not find any solutions, which is in agreement with the limit in (9.18j). Hence, the left-hand side of (7.2) is equal to 0 , as required. In the case that p= 30(m+1)/7 orp= 15(m+1)/7, the relevant counting problem is (6.213). However, no element ( w0;w1,...,w m)∈FixNCm(E8)(ψp) can be produced in this way since the counting problem imposes the restriction that ℓT(w0)+ℓT(w1)+ ···+ℓT(wm) be divisible by 7, which is absurd. This is also in agreement with the limit in (9.18j). In the case that p= 15(m+ 1)/4, the relevant counting problems are (6.227) and (6.228). We have found 45 solutions wito (6.227) of type A2 1in (6.229), and we have found 20 solutions wito (6.227) of type A2in (6.230), which implied 150 solutions for (wi1,wi2) to (6.228). The first two give rise to to (45 + 20)m+1 4= 65m+1 4elements in FixNCm(E8)(ψp), while the third give rise to 150/parenleftbig(m+1)/4 2/parenrightbig elements in Fix NCm(E8)(ψp). Hence, we obtain 65m+1 4+ 150/parenleftbig(m+1)/4 2/parenrightbig =(5m+5)(15m+7) 16elements in Fix NCm(E8)(ψp), which agrees with the limit in (9.18f).156 C. KRATTENTHALER AND T. W. M ¨ULLER In the case that p= 15(m+1)/2, the relevant counting problems are (6.233), (6.234), (6.235), and(6.236). Wehave found15 solutions wito(6.233) of type A2 1∗A2in(6.237), we have found 45 solutions wito (6.233) of type A1∗A3in (6.238), we have found 5 solutionswito (6.233) of type A2 2in (6.239), we have found 18 solutions wito (6.233) of typeA4in (6.240), we have found 5 solutions wito (6.233) of type D4in (6.241), each giving rise to ( m+ 1)/2 elements in Fix NCm(E8)(ψp). Using the notation from there, these imply 2 n3,1+n2,2= 2·660+1195 = 2515 solutions for ( wi1,wi2) to (6.234) withℓT(wi1) +ℓT(wi2) = 4, each giving rise to/parenleftbig(m+1)/2 2/parenrightbig elements in Fix NCm(E8)(ψp). They also imply 3 n2,1,1= 3·2850 = 8550 solutions for ( wi1,wi2,wi3) to (6.235) with ℓT(wi1)+ℓT(wi2)+ℓT(wi3) = 4, each giving rise to/parenleftbig(m+1)/2 3/parenrightbig elements in Fix NCm(E8)(ψp). Finally, they implyaswell n1,1,1,1= 6750solutionsfor( wi1,wi2,wi3,wi4)to(6.236), each giving rise to/parenleftbig(m+1)/2 4/parenrightbig elements in Fix NCm(E8)(ψp). 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