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](http://garsia.math.yorku.ca/MPWP/qtanalogs/catalan/definition.html#:%7E:text=The%20q%2C%20t-Catalan%20numbers%20arise%20naturally%20in%20the,and%20there%20exists%20a%20proof%20in%20the%20article) but are missing some factors.


I verified it up to order 10 computationally [(see here)](https://github.com/uberparagon/sum-over-parts-conjecture/blob/master/conj2.ipynb).


With a little work, this conjecture would prove a special case of the conjecture in [this question](https://mathoverflow.net/questions/375271/a-conjecture-about-sums-over-partitions-arising-from-hilbert-scheme-of-points).


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



A formula for this generating function that is similar to the $qt$-Catalan numbers
This identity can be proved using the results in Garsia and Haiman's paper ["A Remarkable q,t-Catalan Sequence and q-Lagrange Inversion"](https://link.springer.com/article/10.1023/A:1022476211638). 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.


