This quest has its impetus in [a paper](https://arxiv.org/abs/1305.6083) by Stanley and Zanello. I became curious about


*What is the sum of all hooks lengths of all partitions that fit
inside the $n$-th staircase partition?*


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



> 
> **Question.** Let $\lambda=(n,n-1,\dots,1)$ be the staircase partition, $h\_{\square}$ the hook-length of a cell $\square$ in the Young diagram of a partition. Then,  
> 
> $$\sum\_{\mu\subset\lambda}\sum\_{\square\in\mu}h\_{\square}=\frac{(n+4)(2n+3)}6\binom{2n+2}{n+1} - (n+2) 2^{2n+1}.$$
> Is this true? If so, any proof?
> 
> 
> 


**Update.** I'm still hoping for a complete solution for this evaluation; it's not common to get such a nice closed from in general. [Part II of this problem](https://mathoverflow.net/questions/269184/hooks-in-a-rectangle-part-ii) has received a fine answer.


**Update.** To help Stanley's mention of a recurrence, denote the RHS quantity by $w\_n:=\frac{(n+4)(2n+3)}6\binom{2n+2}{n+1} - (n+2) 2^{2n+1}$ then we have ("shaloshable")
$$n(n-2)w\_n-2(4n^2-5n-3)w\_{n-1}+8n(2n-1)w\_{n-2}=0.$$



Hooks in a staircase partition: Part I
The following method should give a proof, but I haven't done the
computation. Let $f\_n$ denote the desired sum. I use the result of
Douglas Zare's answer that
 $$ f\_n = \sum\_{\mu\subseteq \delta\_n} \sum\_i \mu\_i^2, $$
where $\delta\_n=(n,n-1,\dots,1)$. Let
 $$ g\_n = \sum\_{\mu\subseteq \delta\_n} \sum\_i \mu\_i. $$
By OEIS A029760 we have
 $$ g\_n = \frac 12 (n+2)^2C\_n-2^{2n+1}, $$
where $C\_n$ is a Catalan number.


We now break up the sum for $f\_n$ into the following parts:


(1) No $\mu\_i=n-i+1$. In other words, $\mu\subseteq\delta\_{n-1}$, so
the contribution from these terms is $f\_{n-1}$.


(2) $\mu\_1=n$. Below the first row is any partition $\nu\subseteq
\delta\_{n-1}$, so the contribution from these terms is
$f\_{n-1}+n^2C\_n$ since there are $C\_n$ choices for $\nu$.


(3) $\mu\_k=n-k+1$ for some $k\geq 2$, and $\mu\_j<n-j+1$ for $j<k$. The
first $k-2$ rows of $\mu$ consist of a $(k-2)\times(n-k+1)$ rectangle
with a partition $\nu\subseteq\delta\_{k-2}$ on the right. Rows $k-1$
and $k$ of $\mu$ have length $n-k+1$. The remaining rows ($k+1$ to
$n$) consist of a partition $\rho\subseteq \delta\_{n-k}$. Each $\nu$
occurs $C\_{n-k+1}$ times (the number of choices for $\rho$), and each
$\rho$ occurs $C\_{k-1}$ times. Hence the contribution to the sum for
$f\_n$ is
 $$ C\_{n-k+1}\sum\_{\nu\subseteq\delta\_{k-2}}\sum\_i
 (\nu\_i+n-k+1)^2+ C\_{k-1}\sum\_{\rho\in\delta\_{n-k}}
 \sum\_i \rho\_i^2+2C\_{n-k+1}C\_{k-1}(n-k+1)^2 $$
 $$ =
C\_{n-k+1}(f\_{k-2}+2(n-k+1)g\_{k-2}+C\_{k-1}(k-2)(n-k+1)^2)+
 C\_{k-1}f\_{n-k}+2C\_{n-k+1}C\_{k-1}(n-k+1)^2. $$


Summing over all possibilities gives a recurrence satisfied by
$f\_n$. It should be routine (i.e., "shaloshable") to check that the
conjectured answer satisfies the recurrence. Perhaps there is some
inaccuracy in the details above, but I think that the general idea
should give a solution.


