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 $\square=(i,j)$ 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 had an [MO question here](https://mathoverflow.net/questions/252688/partitions-into-odd-parts-vs-hooks-and-symplectic-contents) without any solution. So, I decided to rephrase the problem in case it helps.


Define $\mathcal{syP}\_0(n)$ to be the set of all partitions $\lambda\vdash n$ which has no zero symplectic content for any $\square\in\lambda$. Here are a few examples for $n=2, 4, 6, 8, 10$ and $12$:
\begin{align\*} &11 \\ &211 \\ &222, 3111 \\ &3221, 41111 \\ &3322, 42211, 511111 \\
&3333, 43221, 522111, 6111111.
\end{align\*}  

On the other hand, if ${}\_2\mathcal{P}\_4(n)$ denotes the set of all partitions $\lambda\vdash n$ in which parts are congruent to $2$ mod $4$, then here are a few examples for $n=2, 4, 6, 8, 10$ and $12$:
\begin{align\*} &2 \\ &22 \\&222, 6 \\ &2222, 62 \\ &22222, 622, (10) \\ &222222, 6222, 66, (10,2). 
\end{align\*}
Also, if ${}\_0\mathcal{P}\_2(n)$ denotes the set of all partitions $\lambda\vdash n$ in which parts are distinct and congruent to $0$ mod $2$, then here are a few examples for $n=2, 4, 6, 8, 10$ and $12$:
\begin{align\*} &2 \\ &4 \\ &42, 6 \\ &62, 8 \\ &64, 82, (10) \\ &642, 84, (10,2), (12).
 \end{align\*}


We already know (due to Euler) that $\#{}\_2\mathcal{P}\_4(n)=\#{}\_0\mathcal{P}\_2(n)$. So, I would like to inquire that:



> 
> **QUESTION 1.** Is there a bijection proving either $\#\mathcal{syP}\_0(n)=\#{}\_2\mathcal{P}\_4(n)$ or $\#\mathcal{syP}\_0(n)=\#{}\_0\mathcal{P}\_2(n)$?
> 
> 
> 



> 
> **QUESTION 2.** Is it true that $\lambda\in\mathcal{syP}\_0(n)$ iff $\vert c\_{sp}(\square)\vert=h(\square)$?
> 
> 
> 



In search of a combinatorial proof on particular set of partitions
**Proposition.** A partition $\lambda \in \mathcal{syP}\_0$ iff it is empty, or both of the following hold:


* $\lambda'\_1 - \lambda\_1 = 1$,
* the partition $\mu$ obtained by removing first row and column of $\lambda$ is also in $\mathcal{syP}\_0$.


Informally, all partitions in $\mathcal{syP}\_0$ (and only those) are obtained by starting from the empty partition, and repeatedly adding "L-shapes" to the bottom left corner, each L-shape one cell wider than it is taller, given that each intermediate step yields a proper (column-monotonic) partition. For example, $\lambda = 43221$ looks like this:



```
A
AB
ABBB
AAAAA

```

and $\lambda = 3333$ looks like this



```
ABCC
ABBB
AAAA

```

To see this, reverse the sequence $\lambda\_0 = \lambda, \lambda\_1 = \mu, \ldots, \lambda\_k = \varnothing$, where each subsequent partition is obtained by "shedding" the first row and column of the one before it. As explained below in the proof, if at any point $\lambda'\_{i, 1} - \lambda\_{i, 1} \neq 1$, then we present a cell with $c\_{sp}(\square) = 0$, otherwise the claim is established by induction.


*Proof.* Observe that symplectic content of cells of $\mu$ is carried over to corresponding cells of $\lambda$, as such it has to be/stays non-zero.


If $\lambda'\_1 - \lambda\_1 > 1$, then $c\_{sp}(i, j) = 0$ for $(i, j) = (\lambda\_1 + 2, 1)$. Indeed $i > j$ and $\lambda\_i = 1$, and $c\_{sp}(i, j) = 1 + \lambda\_1 - (\lambda\_1 + 2) - 1 + 2 = 0$.


If $\lambda'\_1 - \lambda\_1 < 1$, then similarly $c\_{sp}(1, \lambda'\_1) = 1+\lambda'-\lambda'-1=0$.


Finally, if $\lambda'\_1 - \lambda\_1 = 1$, then for any $x \in \{1, \ldots, \lambda\_1\}$ we have $c\_{sp}(1, x) = 1 + x - \lambda'\_1 - \lambda'\_x \leq 1 + \lambda\_1 - \lambda'\_1 - \lambda'\_x = -\lambda'\_x < 0$. Similarly one obtains that $c\_{sp}(x + 1, 1)=\lambda\_{x+1}+\lambda\_1-1-(x+1)+2=\lambda\_{x+1}+\lambda\_1-x\geq \lambda\_{x+1} > 0$. $\square$


The bijection to partitions into even distinct parts is now obvious: use the parts as descending even sizes (hook-lengths $h(i,i)$ of the diagonals) of L-shapes, which answers **Q1**.


**Q2** is now also easy to answer in the positive. Both $h(\square)$ and $c\_{sp}(\square)$ are preserved after adding/removing the first row and column. Direct substitution yields $h(1, x) = -c\_{sp}(1, x)$, $h(x + 1, 1) = c\_{sp}(x + 1, 1)$ iff we are allowed to substitute $\lambda'\_1 = \lambda\_1 + 1$.


