The problem is to find a recurrence relation for the , defined in (4.1). An examination of (3.18) reveals that we need to get in terms of . From (3.14) we see that this is equivalent to finding a relation between , and . Here we have written as a function of q and not r. By using (3.15), it can be shown that is a modular form of weight two on a certain congruence subgroup.
This implies that all three functions , and are modular forms of weight two on a certain congruence subgroup and hence must satisfy an algebraic relation of the formwhere P is a certain rational homogeneous polynomial of degree k, say. The crucial observation is that the left-hand side of (4.4) is a modular form of weight 2k. It is well-known that
the dimension of the space of modular forms (above) of weight 2k is asymptotic to some postive constant times k (see ). It is also well-known that the number of monomials is asymptotically equal to . Hence there will always be a relation for large enough k. Such a relation can be found and proved symbolically. This is really a linear problem. The q-series expansion of each monomial up to a certain power of q can be easily computed and stored as a column in a matrix. Finding homogeneous relations is then equivalent to finding the nullspace of a certain matrix. Such relations can be proved by verifying them to a high enough power of q using the theory of modular forms. See  for more details. We illustrate the case p=2 with a MAPLE session.
> read funcs: > read findhom: > A2:=Aseries(2);> findhom([A2(q),A2(q),A2(q)],2,100);
# of terms, 22
-----RELATIONS-----of order---,2We now check the relations to O(q t)
---RELATION----, 1, ---checks to order---The function Aseries(2) gives the q-series expansion of . Using our function findhom we found that , , seem to satisfy the equation at least to , which is enough for a proof. By solving this equation we see that