Given an integer partition $\lambda$ and its Young diagram $Y\_{\lambda}$, let $h\_{\lambda}(i,j)$ stand for the corresponding [hook length](https://en.wikipedia.org/wiki/Hook_length_formula) of the cell $(i,j)\in Y\_{\lambda}$. Write $\lambda\vdash n$ for $\lambda$ a partition of $n$.


Recall the [Gaussian binomials](https://en.wikipedia.org/wiki/Gaussian_binomial_coefficient) denoted by
$$\mathbf{\binom{n}k\_q}.$$



> 
> **QUESTION.** If $[q^k]F(q)=$ the coefficient of $q^k$ in the polynomial $F(q)$, is this true?
>  $$\sum\_{\lambda\vdash n}q^{h\_{\lambda}(1,1)}
> =\sum\_{i,j=0}^nq^j\cdot[q^{n-j}]\mathbf{\binom{j-1}i\_q}.$$
> 
> 
> 


**POSTSCRIPT.** If $p(n)$ is the number of partitions of $n$ then clearly we have
$$p(n)=\sum\_{j=1}^n\sum\_{i=0}^{j-1}\,\,\, [q^{n-j}]\mathbf{\binom{j-1}i\_q}.$$



Reading off top hook-lengths in partitions
Let $j$ be the size of the $(1,1)$-hook, and $i+1$ the number of rows of this hook. To complete this hook to a partition of $n$, we must place to its southeast a partition of $n-j$ with at most $j-i-1$ columns and at most $i$ rows. The number of such partitions is $[q^{n-j}]\boldsymbol{{j-1\choose i}}$, and the proof follows.


