Operation 7 will be unintelligible, unless it be remembered that if we were calculating for n = 1 instead of n = 4, Operation 6 would have completed the computation of B1 itself, in which case the engine instead of continuing its processes, would have to put B1 on V21; and then either to stop altogether, or to begin Operations 1, 2…7 all over again for value of n(=2), in order to enter on the computation of B3; (having however taken care, previous to this recommencement, to make the number on V3 equal to two, by the addition of unity to the former n=1 on that column).