The question accounts my curiosity only, and may not be as deep as I think.
One of recent talks at our local seminar was devoted to the proof of the classical formula $$ F(q)=\sum_{n=0}^\infty r_4(n)q^n =\biggl(\sum_{n\in\mathbb Z}q^{n^2}\biggr)^4 =1+8\sum_{n=1}^\infty\frac{q^n}{(1+(-1)^nq^n)^2} $$ for the generating function of the number $r_4(n)$ of representations of $n$ as a sum of four squares of integers. The key ingredient is the fact that the function $F(e^{2\pi i\tau})$ is a modular form of weight 2 and level 4. As an illustrative example, the speaker gave $r_4(100)=744$ which, of course, can be also given in a more "general" form $r_4(2^\ell\cdot25)=744$ for $\ell=1,2,\dots$. This number 744 is not quite random: it appears as the constant term in the modular invariant $$ J(q)=\frac1q+\sum_{m=0}^\infty c_mq^m =\frac1q+744+196884q+21493760q^2+\cdots, $$ a (weak) modular form of weight 0 viewed as a function of $\tau$ where $q=e^{2\pi i\tau}$. The next coefficient $c_1=196884$ is not divisible by 8, hence cannot be given as $r_4(n)$ for some $n$, but then the problem of finding solutions to the equation $c_m=r_4(n)$ becomes messy. (In fact, I cannot find any other with $m\ge2$.) My particular problem is as follows.
Question. Are there (in)finitely many $m$ such that $c_m=r_4(n)$ for some $n$?
Well, it seems to be quite natural to ask even the following.
General question. Suppose $A(q)=\sum_{n\gg-\infty}a_nq^n$ and $B(q)=\sum_{m\gg-\infty}b_mq^m$ are two modular forms of different weight with, say, nonnegative coefficients and $b_{m+1}>b_m$ holding for $m\ge m_0$. Is there a way to decide whether the equation $a_n=b_m$ holds for infinitely many $m\ge m_0$?
