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For a partition $\lambda=(\lambda_1\geq \lambda_2\geq\ldots\geq \lambda_k)$ of $n$, let the set of standard Young tableau of shape $\lambda$ be denoted by $SYT(\lambda)$ with boxes at $(i,j)$ denoted by $B_{ij}$ taking in distinct values in $\{1,2,\ldots,n\}$. Also let $f_\lambda:=|SYT(\lambda)|$. Finally, let $N_{ij}(k)$ be the set of young tableau of shape $\lambda$ with $B_{ij}=k$. We will exclusively be talking about boxes on the right edge of the Young tableaux (boxes that have no boxes directly south or east of them). We will call these corner boxes. In the diagram below, 8,9 and 10 are corner boxes. On the other hand, 5 and 6 are not corners.

enter image description here

The usual branching rule says that

$$f_\lambda=\sum_{\mu\rightarrow\lambda}f_\mu,$$

where the sum is taken over all partitions $\mu$ of $n-1$ that are contained ($\rightarrow$) in $\lambda$. It is easy to see that when $(i,j)$ is a corner box, $N_{ij}(n)=f_{\lambda- (i,j)}$, the number of SYT of shape $\lambda$ minus the corner box in question. This kind of reasoning can be extended to $N_{ij}(n-1),N(n-2)$. Unfortunately, $N_{ij}(n-k)$ becomes exceeding difficult to compute for $k>2$.

I would like to ask what is known about $N_{ij}(k)$, when $(i,j)$ is a corner box. Specifically, are there any recurrences that these satisfy? Are they at all related to coefficients of certain weighted hook-walk algorithms? Have they been considered in any enumeration problems? Does this question have an answer for certain fixed nontrivial shapes $\lambda$?

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  • $\begingroup$ "boxes on the diagonal of the Young tableaux"? How are these boxes with no boxes directly south or east of them? $\endgroup$ – Amritanshu Prasad Mar 19 '14 at 3:46
  • $\begingroup$ @AmritanshuPrasad: I've added a clarifying picture. $\endgroup$ – Alex R. Mar 19 '14 at 4:37
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    $\begingroup$ I understand what the usual definition of a corner is, but "right edge" is not correct either. 5 and 6 should not be considered corners of the tableau conjugate to the one you have drawn. The parenthetical definition is the usual one. $\endgroup$ – Amritanshu Prasad Mar 19 '14 at 6:40
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    $\begingroup$ There is another easy case: $N(i,j)=ij$. All integers up to $ij$ form then a standard tableau of a rectangular Young tableaux whose removal leaves two ordinary Young tableaux (suitably relabelled) $\mu_1,\mu_2$ of content $k$ and $n-ij-k$. The total number is then ${n-ij\choose k}$ times the product of the number of standard tableaux of the three possible shapes. Using skew tableaux, this approach can be slightly refined for counting $N{i,j}(ij+\alpha)$ with small $\alpha$. $\endgroup$ – Roland Bacher Mar 19 '14 at 8:21
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    $\begingroup$ It is an immediate consequence of the case $k=1$ of Theorem 3.1 of math.mit.edu/~rstan/pubs/pubfiles/48.pdf that the sequence $N_{ij}(1), N_{ij}(2),\dots,N_{ij}(n)$ is log-concave. This is true for any box $(i,j)$, not just a corner box. $\endgroup$ – Richard Stanley Dec 23 '14 at 22:16
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This is a developement of my comment above:

A not very satisfying (from a computational point of view) answer is as follows: Let $\mathcal L=\mathcal L(\lambda,i,j,n)$ be the set of all Young tableaux contained in $\lambda$, with a total number of $n$ boxes and containing the box labelled $(i,j)$. Each element $\mu$ of $\mathcal L$ gives a contribution $\alpha(\mu)$ as follows: Since $n$ is a corner of $\mu$, standard tableaux with box $(i,j)$ labelled $n$ are easy to count (as you observed) by removing the box $(i,j)$ and counting standard tableaux of $\mu\setminus(i,j)$. The difference $\lambda\setminus \mu$ decomposes into two skew tableaux $\mu_1,\mu_2$ which have to be labelled suitably by $n+1,n+2,\dots$ in order to get a standard tableau of $\lambda$. This contributes a binomial factor times the product of the number standard tableaux of the skew tableaux $\mu_1,\mu_2$ (which can be counted by a result of Stanley (On the enumeration of skew Young tableaux, Advances in Applied Mathematics, Volume 30, Issues 1–2, February 2003, Pages 283–-294)). Your number $N_{ij}(n)$ is then given by $\sum_{\mu\in \mathcal L}\alpha(\mu)$.

The number of elements in $\mathcal L$ can be computed as follows: Remove the rectangular Young diagram defined by box $(i,j)$ from $\lambda$ leaving two ordinary Young diagrams $\lambda_1,\lambda_2$ at the right and below the rectangular tableau. The number of elements in $\mathcal L$ is then given by the number of pairs $(\mu_1,\mu_2)$ of tableaux containing a total number of $n-ij$ boxes with $\mu_1$ contained in $\lambda_1$ and $\mu_2$ contained in $\lambda_2$.

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