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.