Let $\lambda$ be an integer partition, denote the number of Standard Young Tableaux of shape $\lambda$ by $f_{\lambda}$. This number is computed by the formula $$f_{\lambda}=\frac{n!}{\prod_{u\in\lambda}h_u}$$ where $h_u$ is a hook length. It is also well-known that $$\sum_{\lambda\vdash n}f_{\lambda}^2=n!\tag1$$ Recall the notation for the content of a cell $u=(i,j)$ in a partition is $c_u=j-i$.

Question. Is this true? $$\sum_{\lambda\vdash n}f_{\lambda}^2\prod_{u\in\lambda}(t+c_u)=n!\,t^n.$$

NOTE. An affirmative answer would imply (1): divide both sides by $t^n$ and take the limit $t\rightarrow\infty$.

  • 2
    For any cell $u$ of $\lambda$, let $h_{\lambda, u}$ denote the hook length of $u$ in $\lambda$. Using the formula quoted in mathoverflow.net/a/263658 as Corollary 7.21.4, we can rewrite $\prod\limits_{u \in \lambda} \left(t+c_u\right) = s_{\lambda}\left(1^t\right) \prod\limits_{u \in \lambda} h_{\lambda, u} = s_{\lambda}\left(1^t\right) \dfrac{n!}{f_{\lambda}}$ (by the hook-length formula). Thus, the left hand side of your question simplifies to $n! \cdot \sum\limits_{\lambda \vdash n} f_{\lambda} s_{\lambda}\left(1^t\right)$. The rest ... – darij grinberg Oct 11 at 20:49
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    ... is an easy application of RSK: The symmetric function $\sum\limits_{\lambda \vdash n} f_{\lambda} s_{\lambda}$ is the generating function of all $n$-letter words over the alphabet $\left\{x_1,x_2,x_3,\ldots\right\}$ (with the $f_{\lambda}$ counting all possible Q-tableaux and the $s_{\lambda}$ being the generating function of all possible P-tableaux) and thus equals $\left(x_1+x_2+x_3+\cdots\right)^n$ (a fact that also follows by applying the Pieri rule $n$ times); now substitute $1^t$ for the variables and you're done. – darij grinberg Oct 11 at 20:51
up vote 4 down vote accepted

Notice that $f_{\lambda}$ is the number of standard Young tableaux of shape $\lambda$, whereas $f_{\lambda}\frac{\prod_{u\in \lambda} (t+c_u)}{n!}$ is the number of semistandard Young tableaux of shape $\lambda$ and content $t$. It was mentioned in a previous question that the identity $$\sum_{|\lambda|=n}\left|\text{SSYT}(\lambda)\right|\left|\text{SYT}(\lambda)\right|=t^n.$$ can either be proved bijectively from RSK (as Darij explains in the comments) or by appealing to Schur-Weyl duality (as David explains in the link), which gives an isomorphism of $S_n\times GL(V)$ representations $$\sum_{|\lambda|=n} Sp_{\lambda} \boxtimes S_{\lambda}(V) \cong V^{\otimes n}$$ where $V$ is a $t$ dimensional vector space, $S_{\lambda}$ is the Schur functor, and $Sp_{\lambda}$ the corresponding Specht module. Our identity follows by taking the dimension of both sides.

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