Agreement between two sets of data on partitions

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.

• Are you saying you know that for all values of $n$ these two statistics have the same frequencies on partitions of $n$? Commented Apr 9, 2022 at 21:04
• @SamHopkins: I was able to compute the latter data and it agrees with the one shown. Is that what you ask? Commented Apr 9, 2022 at 21:16
• it might make sense to ask findstat, but I can probably only do so on monday. Commented Apr 9, 2022 at 22:05
• The dynamics of the map $\lambda \mapsto \mu$ might still be interesting to explore further. Commented Apr 10, 2022 at 18:04
• @SamHopkins, it turns out to be idempotent. Commented Apr 12, 2022 at 8:33

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.
• The number of distinct $\mu$ among partitions of $n$ is A000009: see Jon Perry's comment of Sep 21 2005. Commented Apr 12, 2022 at 11:48