Recently I have proven the following identity
\begin{align}
\sum_{\lambda\in \text{different hook of size d}} \frac{1}{d!} (1)^{ht(\lambda)1} \, \dim \lambda \, \prod_{\Box \in \lambda} \frac{1}{1c(\Box)h}&= \frac{(2d2)!}{ (d1)!}h^{d1}\prod_{i=1}^{d1} \frac{1}{(1+ih)(1ih)}
\end{align}
where $\lambda$ denote the hook of size $d$,$c(\Box)$ denote the content of the hook.
In the righthand side of the equation if I replace $ \prod_{\Box \in \lambda} \frac{1}{1c(\Box)h}$ by $ \prod_{\Box \in \lambda} (1c(\Box)h)^2$ that is the following identity
$$\sum_{\lambda\in \text{different hook of size d}} \frac{1}{\lambda!} (1)^{ht(\lambda)1} \, \dim \lambda \, \prod_{\Box \in \lambda}(1c(\Box)h)^2 $$
I could not predict a close form like the one I did for my first equation.
Is there exist one?
More interesting it looks close to the following formula which I have seen in a problem book of Enumerative Combinatorics vol 2 that is
$$\sum_{w\in S_d}q^{k(w^2)}=\sum_{\Gamma}dim{\Gamma}\prod_{\Box\in \Gamma}(q+c(\Box))$$
where $\Gamma \vdash d$ is all the partititon of $d$ not only hook,$k(w^2)$ denote the number of cycles of $w^2\in S_n$.
So my identity has some combinatorial explanation like above?
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$\begingroup$ Is it really $ \lambda ! $ in your first expression ? In which case, you could just replace it by $ d! $ since the hooks are all of size $d$. $\endgroup$ – Synia Aug 24 '17 at 20:15

$\begingroup$ have seen the related question, this seems to be the case. As remarked in your previous question, you have a particular polynomials in h. If you want a closed expression, first translate your sum into a sum on integers (the height of your hook) as done by D. Grinberg and then feed Maple/Mathematica with it. You will have immediately the answer. In case, if what comes out is too complicated, you can go to OEIS. $\endgroup$ – Synia Aug 24 '17 at 20:30

$\begingroup$ You are right it should be $d!$, I wrote in crude version. $\endgroup$ – GGT Aug 25 '17 at 6:04

$\begingroup$ We tried that out there is some symmetry notice but we could not get any nice form like previous ones. $\endgroup$ – GGT Aug 25 '17 at 6:06

$\begingroup$ Now, I am less convinced that there is a nice form. This is a convolution in the symmetric group, and the coefficients count some geodesic on it. I would try to see what are these coefficients. On the other hand, you found a nice form for the previous sum, and it involved homogeneous functions in JucysMurphy elements (see hal.archivesouvertes.fr/file/index/docid/512865/filename/…, this is related with unitary Weingarten elements). I am curious : where does your problem come from ? $\endgroup$ – Synia Aug 25 '17 at 9:32