partitions into odd parts vs hooks and symplectic contents

Question

Given a partition $\lambda=(\lambda_1\geq\lambda_2\geq\dots)$, denote the conjugate partition by $\lambda'=(\lambda_1'\geq\lambda_2'\geq\dots)$. For example, if $\lambda=(4,2,2)$ then $\lambda'=(3,3,1,1)$.

The hook length of a cell $(i,j)$ in the Young diagram of $\lambda$ is given by $h(i,j)=\lambda_i+\lambda_j'-i-j+1$. Define the symplectic content of cell $(i,j)$ of $\lambda$ as $$c_{sp}(i,j)=\begin{cases} \lambda_i+\lambda_j-i-j+2 \qquad \text{if $i>j$} \\ i+j-\lambda_i'-\lambda_j' \qquad \qquad \text{if $i\leq j$}.\end{cases}$$ I propose the following claim for which I have no proof. Any ideas?

Claim. Let $\lambda\vdash n$ signify $\lambda$ is a partition of $n$. Then, there is a generating function $$\sum_{n\geq0}x^n\sum_{\lambda\vdash n}\prod_{\square\in\lambda}\frac{c_{sp}^2(\square)}{h^2(\square)}= \prod_{j\geq1}\frac1{1-x^{4j-2}}.$$

If proven, this brings in an interesting consequence. Proof?

Corollary. Let $P_{odd}(n)$ denote the number of partitions of $n$ into odd parts. Then, $$P_{odd}(n)=\sum_{\lambda\vdash 2n}\left(\prod_{\square\in\lambda}\frac{c_{sp}(\square)}{h(\square)}\right)^2$$ where $\lambda\vdash 2n$ runs through all ordinary (unrestricted) partitions of $2n$.

Caveat. This is not an efficient way to compute such partitions.

Answer 1

Buried in the paper by Nekrasov and Okounkov - Seiberg-Witten Theory and Random Partitions - is the following hook length formula:

$$ \eta(q)^{\,\mu^2 -1} = \prod_{n \geq 0} (1 - q^n)^{\mu^2 - 1} = \sum_\mathbf{k} q^{\vert\mathbf{k}\vert}\prod_{\square \in \mathbf{k}} \big( 1 - \frac{\mu^2}{h(\square)^2}\big) $$

This is in Section 6. It is a sum over all partitions $\mathbf{k}$ of all sizes. The "free fermion" or "CFT" methods in that chapter are a bit mysterious and may leave one desiring bijective or probabilistic proof.

This style of result is very general and I could imagine a closed form for your equation.