For a cell $\square$ in the Young diagram of a partition $\lambda$, let $h\_{\square}$ and $c\_{\square}$ denote the [hook length and content](http://www-math.mit.edu/~rstan/transparencies/hooks.pdf) of $\square$, respectively.


R Stanley [proved the following](http://www-math.mit.edu/~rstan/papers/hooks.pdf) in Theorem 2.2:
$$\sum\_{n\geq0}x^n\sum\_{\lambda\vdash n}\prod\_{\square\in\lambda}\frac{t+c\_{\square}^2}{h\_{\square}^2}=(1-x)^{-t}.$$
In the same spirit, I ask:



> 
> **Question.** if $\mathcal{O}(\pi)$ and $\mathcal{E}(\pi)$ stand for the number of odd and even cycles in $\pi$, respectively, then is this averaging formula true?
>  $$\frac1{n!}\sum\_{\pi\in\mathfrak{S}\_n}t^{\mathcal{O}(\pi)+2\mathcal{E}(\pi)}=\sum\_{\lambda\vdash n}\prod\_{\square\in\lambda}\frac{t+c\_{\square}}{h\_{\square}}.$$
> 
> 
> 



hooks and contents: Part I
Specialize the Cycle Index Formula to get


$$\prod\_{n=1}^\infty \exp \bigl( \frac{a\_i}{i}z^i \bigr) = \sum\_{n=0}^\infty \frac{z^n}{n!} \sum\_{\pi \in \mathfrak{S}\_n} a\_1^{\mathrm{cyc}\_1(\pi)} a\_2^{\mathrm{cyc}\_2(\pi)} \ldots, $$


by setting $a\_i = t^2$ if $i$ is even and $a\_i = t$ if $i$ is odd. We get


$$\sum\_{n=1}^\infty \frac{z^n}{n!} \sum\_{\pi \in \mathfrak{S}\_n} t^{\mathcal{O}(\pi)+2\mathcal{E}(\pi)} = \frac{1}{(1-z^2)^{t^2/2}} \frac{(1+z)^{t/2}}{(1-z)^{t/2}}. $$


It is well known (see Corollary 7.21.4 in Stanley, *Enumerative Combinatorics II*) that 
$$\prod\_{\Box \in \lambda} \frac{d+c\_\Box}{h\_\Box} = s\_\lambda(1^d)$$ 
is the number $|\mathrm{SSYT}\_d(\lambda)|$ of semistandard Young tableaux of shape $\lambda$ with entries from $\{1,\ldots,d\}$, or equivalently, the dimension of the irreducible polynomial representation of $\mathrm{GL}\_d(\mathbb{C})$ labelled by the partition $\lambda$. 


Therefore it is equivalent to show that if $d \in \mathbb{N}$ then


$$ \frac{1}{(1-z^2)^{d^2/2}} \frac{(1+z)^{d/2}}{(1-z)^{d/2}} = \sum\_{n=0}^\infty z^n \sum\_{\lambda \in \mathrm{Par}(n) } |\mathrm{SSYT}\_d(\lambda)| \tag{$\star$} $$


Now, as suggested by Gjerji Zaimi in a comment on my first answer, we use Littlewood's Identity. It states that


$$ \sum s\_\lambda(x\_1,\ldots,x\_d) = \prod\_{1 \le i < j \le d} \frac{1}{1-x\_ix\_j} \prod\_{1 \le k \le d} \frac{1}{1-x\_k} $$


where the sum is over all partitions $\lambda$. (For an online source, see (6) in [this paper by Betea and Wheeler](https://arxiv.org/pdf/1402.0229.pdf), and multiply by $\prod\_{k=1}^d \frac{1}{1-x\_kz} = \sum\_{n=0}^\infty h\_n(x\_1,\ldots,x\_d)z^n$ using Young's rule.) Specialize Littlewood's Identity by setting $x\_i = z$ for all $i \in \{1,\ldots, d\}$ and sum over $n$ to get


$$ \sum\_{n=0}^\infty \sum\_{\lambda \in \mathrm{Par}(n)} s\_\lambda(1^d) z^n = \frac{1}{(1-z^2)^{(d^2-d)/2}(1-z)^d}. $$ 


The right-hand side agrees with the left-hand side of ($\star$), so we are done.


