Given an integer partition $\lambda$ and its Young diagram $Y_{\lambda}$, let $h_{\lambda}(i,j)$ stand for the corresponding hook length of the cell $(i,j)\in Y_{\lambda}$. Write $\lambda\vdash n$ for $\lambda$ a partition of $n$.

Recall the Gaussian binomials 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 $$p(n)=\sum_{j=1}^n\sum_{i=0}^{j-1}\,\,\, [q^{n-j}]\mathbf{\binom{j-1}i_q}.$$

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    $\begingroup$ Looking at the generating function at findstat.org/St000459 and then clicking on "search the OEIS" yields that this was observed in the OEIS (oeis.org/A049597) in 2008, but no reference to a proof is given. $\endgroup$ – Christian Stump May 7 '19 at 8:56

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.

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