Agreement between two sets of data on partitions

Question

Let $\lambda=(\lambda_1,\lambda_2,\dotsc,\lambda_{\ell(\lambda)})$ be an integer partition of positive numbers where $\ell(\lambda)$ is the length of the partition. One may associate a Ferrer diagram or Young diagram $Y$ to $\lambda$. I read this concept Ferrer's matrix on OEIS but nowhere else.

First, compute $m=\max\{\lambda_1,\ell(\lambda)\}$ and then construct an $m\times m$ matrix by inserting a $1$ in each box of $Y$ while inserting $0$ elsewhere. For example, if $\lambda=(4,3,1)\vdash 8$ then $m=4$ and the corresponding matrix becomes $\begin{pmatrix} 1&1&1&1 \\ 1&1&1&0 \\ 1&0&0&0 \\ 0&0&0&0 \end{pmatrix}$. Now, add elements of all anti-diagonals to get $(1,2,3,2,0,0,0)$. Re-sorting (and ignoring $0$'s) gives $\mu=(3,2,2,1)\vdash 8$.

If one does apply the procedure to all (ordered) partitions of $n=4$, i.e. $\{4, 31, 22, 211, 1111\}$, the resulting partitions form the multi-set $\{1111, 211, 211, 211, 1111\}$. The new statistic (frequencies) reads $2, 3$. Here is a short table from OEIS: $$\begin{array}l 1 \\ 2 \\ 2, 1 \\ 2, 3 \\ 2, 2, 3 \\ 2, 2, 6, 1 \\ 2, 2, 4, 3, 4. \end{array}$$ On a different route, given any partition $\lambda$ of $n$, compute its hook-lengths and add them to generate some statistic. As an example, take $n=4$. Associate the set hook-lengths $\{(4, 3, 2, 1), (4, 1, 2, 1), (3, 2, 2, 1), (4, 2, 1, 1), (4, 3, 2, 1)\}$. The resulting multi-set of sums is $\{10, 8, 8, 8, 10\}$ with frequencies $2, 3$.

QUESTION. Is there a bijection to explain the total agreement between the two statistics (see above table)? I would also be nice if there is a generating function here.

REMARK. The sum of hooks of $\lambda$ equals sum of squares of parts of $\lambda$.

UPDATE. Peter Taylor (comments below 1 2) identified an error in the above correspondence. So, the answer to my question is negative, unfortunately.

Answer 1

First, compute $m=\max\{\lambda_1,\ell(\lambda)\}$ and then construct an $m\times m$ matrix by inserting a $1$ in each box of $Y$ while inserting $0$ elsewhere. For example, if $\lambda=(4,3,1)\vdash 8$ then $m=4$ and the corresponding matrix becomes $\begin{pmatrix} 1&1&1&1 \\ 1&1&1&0 \\ 1&0&0&0 \\ 0&0&0&0 \end{pmatrix}$. Now, add elements of all anti-diagonals to get $(1,2,3,2,0,0,0)$. Re-sorting (and ignoring $0$'s) gives $\mu=(3,2,2,1)\vdash 8$.

The sorting is a bit of a red herring. Denote the anti-diagonal frequency vector, $(1,2,3,2,0,0,0)$ in the example, as $(ad_1, ad_2, ad_3, \ldots)$. It's easy to see that $ad_k > ad_{k-1}$ requires $\forall 1 \le i \le k: ad_i = i$. So if $A = \max(\mu)$ the unique "unsorting" to a legal anti-diagonal frequency vector which produces it is $(1, 2, 3, \ldots, A, ad_{A+1}, ad_{A+2}, \ldots)$ with $A \ge ad_{A+1} \ge ad_{A+2} \ge \cdots$

Let $H$ denote the hook length sum. Suppose we build the Ferrers diagram starting from the unique partition of $1$ and adding a cell at a time in such a way that we always have a valid Ferrers diagram. When we add a cell at $(r, c)$ we increase $ad_{r+c-1}$ by $1$ and $H$ by $(r-1)+(c-1)+1 = r+c-1$, so $H - \sum_k k \cdot ad_k$ is invariant. The initial partition has $H=1$, $ad_1 = 1$, so $$H = \sum_k k \cdot ad_k$$

Thus for every $\mu$ there is a unique hook length sum $H_{\mu}$ which all partitions having that anti-diagonal frequency partition share, but there's nothing to prevent different $\mu$ sharing their hook length sum, and e.g. anti-diagonal frequency vectors $(1,2,2,2,2) \to 1 + 4 + 6 + 8 + 10 = 29$ and $(1, 2, 3, 1, 1, 1) \to 1 + 4 + 9 + 4 + 5 + 6 = 29$ collide.