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.

  • $\begingroup$ Did you look at Guo Niu Han's papers? He has hooks and contents, too :-) $\endgroup$ Oct 21, 2016 at 5:29
  • $\begingroup$ @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. $\endgroup$ Oct 21, 2016 at 11:07
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    $\begingroup$ 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. $\endgroup$
    – Wouter M.
    Oct 27, 2016 at 11:42

1 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.

  • $\begingroup$ This is a well-documented result which has been reproved, although I don't see how it can be used for the claim above. After all, N-O formula has no involvement of "contents" but "hooks". See my recent question in mathoverflow.net/questions/251068/… and references inside $\endgroup$ Oct 21, 2016 at 2:01
  • $\begingroup$ @T.Amdeberhan You are right... every time you see a result like this you have to start over. At least, part (a) of your conjecture implies part (b) $\endgroup$ Oct 21, 2016 at 2:11
  • $\begingroup$ Indeed, I opted to state (b) due to its simplicity. $\endgroup$ Oct 21, 2016 at 2:19
  • $\begingroup$ 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 $\endgroup$ Oct 21, 2016 at 3:36
  • $\begingroup$ Updated link: arxiv.org/abs/0805.1398 for The Nekrasov-Okounkov hook length formula: refinement, elementary proof, extension and applications, by Guo-Niu Han $\endgroup$
    – David Roberts
    Sep 24, 2021 at 0:51

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