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Let $\lambda$ denote the hook of size $d$ for a fixed integer $d>0$. For $d=2$ there are two kinds of hooks for $d=3$ there are 3 different kind and so on. $c(\Box)$ denote the content of the $\lambda$.

I want to know if following results appear in literature if anyone can give me a reference or proof $$ \sum_{\lambda\in \text{different hook of size d}} \frac{1}{|\lambda|!} (-1)^{ht(\lambda)-1} \, \dim \lambda \, \prod_{\Box \in \lambda} \frac{1}{1-c(\Box)h}$$ that the above equation equal to \begin{equation} \frac{(2d-2)!}{ (d-1)!}h^{d-1}\prod_{i=1}^{d-1} \frac{1}{(1+ih)(1-ih)} \end{equation}

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  • $\begingroup$ The hooks of size $d$ are indexed by the integers $1, 2, \ldots, d$, where integer $i$ corresponds to the hook whose first row has length $i$. This latter hook has height $d-i+1$ and dimension $\dbinom{d-1}{i-1}$. The contents of its cells are $i-d, i-d+1, \ldots, i-1$. So the thing boils down to an identity for binomial coefficients. $\endgroup$
    – darij grinberg
    Aug 5, 2017 at 8:43
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    $\begingroup$ Okay. So after some rewriting and cancellation, your equality becomes $\sum\limits_{i=1}^d \left(-1\right)^{d-i} \dbinom{d-1}{i-1} \prod\limits_{u=1}^d \dfrac{1}{1-h\left(i-u\right)} = d \left(2d-2\right)! \prod\limits_{i=1}^{d-1} \dfrac{h^{d-1}}{\left(1+ih\right)\left(1-ih\right)}$. That said, I suspect the right hand side has too many $h$s. $\endgroup$
    – darij grinberg
    Aug 5, 2017 at 8:50
  • $\begingroup$ I put the equation in maple and checked it $h$'s power are all good. The combinatorial coefficient is $\frac{(2d-2)!}{(d-1)!}$ $\endgroup$
    – GGT
    Aug 5, 2017 at 11:00
  • $\begingroup$ Also I just updated the sign issue. $\endgroup$
    – GGT
    Aug 5, 2017 at 11:02
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    $\begingroup$ You can prove the identity by checking that both sides have the same poles, that these poles are simple, and that they have the same residues. There should be a nicer proof. $\endgroup$
    – Richard Stanley
    Aug 5, 2017 at 16:47

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