# Wherefore art thou a Borcherds Product?

This question essentially asks how can one recognize (or rule out) that a generating function of combinatorial origin may be given as a Borcherds type product. I'll start with a motivational example: When studying generating functions of $t$-residues of partitions one encounters an infinite product at $t=2$ (for example here). However already for $t=3$ we encounter (a variation of) $$F_3(a,b,c)=\sum_{n,m\in \mathbb Z}a^{m^2}b^{n^2}c^{mn}.$$ I read in a previous question that one shouldn't expect this to have an infinite product formula since if we specialize at $a=b=c=q$ we get a zero at $q=-\exp(\pi/\sqrt{3})$, and this prevents any right hand side of the form $\prod_n (1-q^n)^{p(n)}$ for any polynomial $p(n)$. However the keyword here for the exponents is "polynomial in $n$". If we allow nonpolynomial sequences of exponents we have a beautiful formula due to Borcherds: $$F_3(a,b,c)=\prod_{u+v+w>0}\left(\frac{1-(-1)^{u+v}a^ub^vc^w}{1+(-1)^{u+v}a^ub^vc^w}\right)^{k(uv-w^2)}$$ where $$\sum_{n\geq 0}k(n)q^n=\frac{1}{\sum_{n\geq 0} (-1)^nq^{n^2}}.$$ In geometry these types of infinite products appear when calculating generating functions of the elliptic genus due to the DMVV formula and its extensions. (There are purely combinatorial applications)

Here is an attempt to make this into a specific question: A famous generating function that was originally conjectured to be of the form $\prod_n(1-q^n)^{p(n)}$ for $p(n)=\frac{n-n^2}{2}$ is the case of solid partitions (partitions with a 4 dimensional Ferrers diagram). However this turned out to be false and the exponents actually end up oscillating in sign, therefore excluding the possibility of polynomial exponents. One may hope that these exponents may still be a specific set of coefficients of some quasimodular form.

Question: Is it possible to expect the generating function of solid partitions to have a Borcherds type infinite product expansion or can this be ruled out?