This problem is a follow up on [my other MO question](https://mathoverflow.net/questions/269174/hooks-in-a-staircase-partition-part-i).


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? 
> 
> 
> 



Hooks in a rectangle: Part II
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. 


