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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.

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  • $\begingroup$ Not an answer, but I am curious if there is a (conjectural?) symmetric function version of your claimed equality, so that we obtain the right hand side by a plethystic substitution into power sums, and the left hand side reduces to what you have there. $\endgroup$ Commented Nov 12, 2020 at 0:01
  • $\begingroup$ You mean replace the z^n in the rhs with p_n, and then put some symmetric functions in the left hand side somehow, so that where you expand the symmetric functions in single variable z you recover the original? $\endgroup$ Commented Nov 12, 2020 at 18:37
  • $\begingroup$ I tried a bit to interpret the right hand side as a suitable specialization of the cycle index series for the species of permutations (i.e., the Frobenius character for the conjugation action of the symmetric group) but I didn't succeed. I would have thought that writing $\frac{1}{n (1-q^n) (1-t^n)}$ as $\frac{1}{(1-q)(1-t)}(n-1)! [n-1]_q! [n-1]_t! \frac{1}{n! [n]_q! [n]_t!}$ would be a good idea, but it seems that it isn't. $\endgroup$ Commented Nov 12, 2020 at 19:11

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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.

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  • $\begingroup$ I had started to read some Garsia and Haiman papers, but I hadn't quite figured out which part I needed. I'll take a look at this! $\endgroup$ Commented Nov 12, 2020 at 21:50
  • $\begingroup$ To get the principal specializations of Schur functions, we use the Stanley Hook Content theorem? $\endgroup$ Commented Nov 12, 2020 at 23:02
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    $\begingroup$ @Drew You can also find it in section I.3 of Macdonald's book on symmetric functions and Hall polynomials (example 2). $\endgroup$ Commented Nov 12, 2020 at 23:28

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