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) identified an error in the above correspondence. So, the answer to my question is negative, unfortunately.

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.

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 $\lambda$.

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) identified an error in the above correspondence. So, the answer to my question is negative, unfortunately.

Let $\lambda=(\lambda_1,\lambda_2,\dots,\lambda_{\ell(\lambda)})$ 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 $(1,2,3,2,0,0,0)$. Resorting (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{align*} &1 \\ &2 \\ &2, 1 \\ & 2, 3 \\ &2, 2, 3 \\ &2, 2, 6, 1 \\ &2, 2, 4, 3, 4. \end{align*} 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 $\lambda$.

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 $\lambda$.