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This problem is a follow up on my other MO question.

On the basis of experimental data, I'm prompted to ask:

Question. Let $R(a,b)$ an $a\times b$ rectangular grid, $h_{\square}$ the hook-length of a cell $\square$ in the Young diagram of a partition. Then,
$$\sum_{\mu\subset R(a,b)}\sum_{\square\in\mu}h_{\square}=\binom{a+b-2}{a-1}\binom{a+b+1}3.$$ Is this true? If so, any proof?

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    $\begingroup$ This looks like a problem about lattice paths in a rectangle, not so much about Young diagrams. After all, partitions that fit inside a rectangle are in bijection with lattice paths between two opposite corners of a rectangle. A hook can be subdivided into its arm and leg (and its source). I suspect that the sum of arms and the sum of legs can be computed separately. $\endgroup$ – darij grinberg May 7 '17 at 5:26
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    $\begingroup$ Yes, just replace the rhs by $\binom{a+b-3}{a-1}\binom{a+b}{3}$ $\endgroup$ – Martin Rubey May 7 '17 at 8:27
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As Darij suggests, we may count separately the sum of legs and the sum of arms. Denote the rows sizes $t_1,t_1+t_2,\dots,t_1+\dots+t_a$, where $t_i,1\leqslant i\leqslant a+1$, are non-negative integers and $\sum_{i=1}^{a+1} t_i=b$. Then the sum of arms (let's think that the arm includes the half of the source of the hook and the leg another half) equals $\sum_{i=1}^a \frac12(t_1+\dots+t_i)^2$. By symmetry, we know that the sums of $t_i^2$, $t_it_j$ over all diagrams do not depend on $i$, or respectively on a pair $i<j$. From $\sum t_i=b$ we see that the sum of $t_i$ (with fixed $i$) equals $\binom{a+b}b\cdot \frac{b}{a+1}$. Next, since $t_1^2+t_1t_2+\dots+t_1t_{a+1}=t_1b$, we see that $X+aY=\binom{a+b}b\cdot \frac{b}{a+1}$, where $X=\sum t_1^2$, $Y=\sum t_1t_2$. For finding $X$ we may observe that it is a coefficient of $x^b$ in $(\sum_{i=0}^\infty i^2x^i)(\sum_{i=0}^\infty x^i)^a$. We have $\sum_{i=0}^\infty i^2x^i=(x+x^2)(1-x)^{-3}$, so $X=[x^b](x+x^2)(1-x)^{-a-3}=\binom{a+b+1}{a+2}+\binom{a+b}{a+2}$. Rest is straightforward.

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This is essentially lemma 2.5 in the paper AVERAGE SIZE OF A SELF-CONJUGATE (s, t)-CORE PARTITION.

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  • $\begingroup$ Uhmm... No, it not, Harry. $\endgroup$ – T. Amdeberhan May 15 '17 at 13:58

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