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}{1-c(\Box)h}&= \frac{(2d-2)!}{ (d-1)!}h^{d-1}\prod_{i=1}^{d-1} \frac{1}{(1+ih)(1-ih)} \end{align} where $\lambda$ denote the hook of size $d$,$c(\Box)$ denote the content of the hook. In the right-hand side of the equation if I replace $ \prod_{\Box \in \lambda} \frac{1}{1-c(\Box)h}$ by $ \prod_{\Box \in \lambda} (1-c(\Box)h)^2$ that is the following identity $$\sum_{\lambda\in \text{different hook of size d}} \frac{1}{|\lambda|!} (-1)^{ht(\lambda)-1} \, \dim \lambda \, \prod_{\Box \in \lambda}(1-c(\Box)h)^2 $$ 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?

  • $\begingroup$ 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 Jucys-Murphy elements (see hal.archives-ouvertes.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

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