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Let $\lambda\vdash n$ denote an integer partition of $n$ and $\frak{H}_{\lambda}$ be the multiset of hook lengths of $\lambda$. Further, let $o(\lambda)=\#$ of odd entries and $e(\lambda)=\#$ of even entries in $\frak{H}_{\lambda}$.

Example. If $\lambda=(4,3,1,1)$ then $\frak{H}_{\lambda}=(7,4,3,1,5,2,1,2,1)$. Also, $o(\lambda)=6$ and $e(\lambda)=3$.

QUESTION. Is there a generating function for the following "parity distribution" function? $$F_n(q,t):=\sum_{\lambda\vdash n}q^{o(\lambda)}t^{e(\lambda)}.$$

Examples. Here is a short list of such polynomials: \begin{align*} F_1(q,t)&=q, \qquad \qquad F_2(q,t)=2qt, \qquad F_3(q,t)=q^3+2q^2t, \\ F_4(q,t)&=5q^2t^2, \qquad F_5(q,t)=2q^4t+5q^3t^2. \end{align*}

Remark. Observe that $\sum_{n\geq0}F_n(q,q)\,x^n=\prod_{i\geq1}\frac1{1-q^ix^i}$.

Remark. The coefficient of $q^nt^n$ in $F_{2n}(q,t)$ is the convolution of $\sum_{k=0}^np(k)p(n-k)$ where $p(n)$ are the partitions numbers.

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Yes, the generating function is $$\sum_{n\geq 0} F_{n}(q,t)x^n=\prod_{k\geq 1}\frac{(1-q^{2k}x^{2k})^2}{1-q^kx^k}\cdot \prod _{k\geq 1}\frac{1}{(1-q^kt^kx^{2k})^2}.$$ This follows from the usual bijection between partitions and 2-cores (corresponding to the first product on the right) and 2-quotients (corresponding to the second product) upon realizing that the boxes in a partition with even hook length are in bijection with the boxes of its 2-quotient.

Incidentally these polynomials are also homogenized versions of the Poincare polynomials of G-Hilbert schemes for $G=\mathbb Z/2\mathbb Z$. For example, see Theorem 2 of "On generating series of classes of equivariant Hilbert schemes of fat points", by S.M. Gusein-Zade, I. Luengo, A. Melle Hernandez, arXiv:0905.1779.

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  • $\begingroup$ Do you think this generalizes? That is, what if you look for hooks modulo $3$; a 3-parameter distribution. $\endgroup$ Mar 10, 2021 at 14:16
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    $\begingroup$ @T.Amdeberhan There is a straightforward generalization mod $n$, where the statistic is "number of hooks divisible by $n$". $\endgroup$ Mar 10, 2021 at 17:30
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    $\begingroup$ Nice! It might be informative if you can add it to your answer. $\endgroup$ Mar 10, 2021 at 17:44

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