Let $\lambda$ be an integer partition, denote the number of [Standard Young Tableaux](https://en.wikipedia.org/wiki/Young_tableau) of shape $\lambda$ by $f\_{\lambda}$. This number is [computed by](https://en.wikipedia.org/wiki/Hook_length_formula) 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?
>  $$\sum\_{\lambda\vdash n}f\_{\lambda}^2\prod\_{u\in\lambda}(t+c\_u)=n!\,t^n.$$
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**NOTE.** An affirmative answer would imply (1): divide both sides by $t^n$ and take the limit $t\rightarrow\infty$.



SYT and contents of a partition
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](https://mathoverflow.net/questions/245076/an-identity-related-to-partitions-into-n-parts-and-schur-polynomials) 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.


