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Given a partition $\lambda\vdash n$ of $n$, look at its Young diagram $Y_{\lambda}$. Let $a(\lambda)$ be the number of squares (of all sizes) in $Y_{\lambda}$. For example, if $n=4$ then $a(4)=4, a(3,1)=4, a(2,2)=5, a(2,1,1)=4$, and $a(1,1,1,1)=4$. So, we have a total of $4+4+5+4+4=21$ squares.

Together with Richard Stanley (unpublished, Section 9), we proved that $$\sum_{n=0}^{\infty}F_n(q)\, x^n=\sum_{k=0}^{\infty}\frac{x^{k^2}q^{\frac{k(k+1)(2k+1)}6}} {\prod_{j=1}^k\left(1-x^j\,q^{\frac{j(j+1)}2}\right)^2}, \qquad \text{where} \qquad F_n(q)=\sum_{\lambda\vdash n}q^{a(\lambda)}.$$

QUESTION. Now, instead consider $b(\lambda)$ to be the number of rectangles in $Y_{\lambda}$. Can you provide a generating function for $G_n(q)=\sum_{\lambda\vdash n}q^{b(\lambda)}$?

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  • $\begingroup$ We have $b(\lambda) = \sum_{(i,j) \in \lambda} i \cdot j$. $\endgroup$
    – Sam Hopkins
    Commented Jul 28, 2021 at 16:35
  • $\begingroup$ (Whereas by comparison $a(\lambda) = \sum_{(i,j)\in \lambda} \min(i,j)$.) $\endgroup$
    – Sam Hopkins
    Commented Jul 28, 2021 at 16:50
  • $\begingroup$ Both are correct. Also that $\frac{d}{dq}F_n(q)$ and $\frac{d}{dq}G_n(q)$ evaluated at $q=1$ equal the total number of squares (resp. rectangles) in all partitions of $n$. $\endgroup$
    – T. Amdeberhan
    Commented Jul 28, 2021 at 16:59
  • $\begingroup$ It also suggests that looking at $H_n(q)=\sum_{\lambda\vdash n} q^{c(\lambda)}$ where $c(\lambda) = \sum_{(i,j)\in \lambda}i+j$ might be interesting... $\endgroup$
    – Sam Hopkins
    Commented Jul 28, 2021 at 17:01

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