A formula for this generating function that is similar to the $qt$-Catalan numbers

Question

I came up with the following conjecture: $$ \sum_{n \ge 0} z^n \sum_{\mu \vdash n} \frac{ t^{\sum l}q^{\sum a}}{\prod (q^a - t^{l+1})(t^l - q^{a+1})} = \exp\left(\sum_{n \ge 1} \frac{z^n}{n(q^n-1)(t^n-1)}\right) $$ where each unlabeled sum and product is over the cells of the diagram of the partition, and $l$ and $a$ are the arm and leg lengths.

The coefficients on the left hand side are similar to the $qt$-Catalan numbers but are missing some factors.

I verified it up to order 10 computationally (see here).

With a little work, this conjecture would prove a special case of the conjecture in this question.

I hoping to get some ideas or tools about how to deal with things that look like this.

Answer 1

This identity can be proved using the results in Garsia and Haiman's paper "A Remarkable q,t-Catalan Sequence and q-Lagrange Inversion". In particular theorem 3.10(e) gives $$\sum_{\mu \vdash n} \frac{ t^{\sum l}q^{\sum a}}{\prod (q^a - t^{l+1})(t^l - q^{a+1})}=\sum_{\mu \vdash n} \frac{ t^{\sum a}q^{\sum a}}{\prod (1 - q^h)(1 - t^h)}$$ where $h=a+l+1$ denotes the usual hook length. Writing this expression in terms of principal specializations of Schur functions together with an application of the Cauchy identity gives $$\sum_{n \ge 0} z^n \sum_{\mu \vdash n} \frac{ t^{\sum l}q^{\sum a}}{\prod (q^a - t^{l+1})(t^l - q^{a+1})}=\sum_{\mu}s_{\mu}(1,q,q^2,\dots)s_{\mu}(z,zt,zt^2,\dots)=\prod_{i,j\geq 0}\frac{1}{1-q^it^jz}$$ and it remains to notice that $$\prod_{i,j\geq 0}\frac{1}{1-q^it^jz}=\exp\left(\sum_{n\geq 1}\frac{1}{n}\sum_{i,j\geq 0} q^{ni}t^{nj}z^n\right)=\exp\left(\sum_{n \ge 1} \frac{z^n}{n(q^n-1)(t^n-1)}\right)$$ as desired.