# partitions into odd parts vs hooks and symplectic contents

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.

• Did you look at Guo Niu Han's papers? He has hooks and contents, too :-) – Martin Rubey Oct 21 '16 at 5:29
• @MartinRubey: Thanks. I know his papers on the subject, intimately. I've mentioned one of them in the comments below. None applies to our problem shown above. In fact, Han computes with ordinary "contents" while here it is about "symplectic contents". It seems to me this is the first time anyone considers the latter in the context of hooks. – T. Amdeberhan Oct 21 '16 at 11:07
• It is remarkable that the symplectic content equals the hook length (up to sign) if and only if the product of the symplectic contents is non-zero. It's sign being opposite to the sign of the (ordinary) content. – Wouter M. Oct 27 '16 at 11:42

$$\eta(q)^{\,\mu^2 -1} = \prod_{n \geq 0} (1 - q^n)^{\mu^2 - 1} = \sum_\mathbf{k} q^{\vert\mathbf{k}\vert}\prod_{\square \in \mathbf{k}} \big( 1 - \frac{\mu^2}{h(\square)^2}\big)$$
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.
• You may like to compare a proof for the N-O formula, using Macdonald identities and $t$-cores, at front.math.ucdavis.edu/0805.1398 – T. Amdeberhan Oct 21 '16 at 3:36