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For a cell $\square$ in the Young diagram of a partition $\lambda$, let $h_{\square}$ and $c_{\square}$ denote the hook length and content of $\square$, respectively.

R. Stanley remarked following Theorem 2.2, on page 6 of Some Combinatorial Properties of Hook Lengths, Contents, and Parts of Partitions that: $$\sum_{n\geq0}x^n\sum_{\lambda\vdash n}\prod_{\square\in\lambda}\frac{(t+c_{\square})(v+c_{\square})}{h_{\square}^2}=(1-x)^{-tv}.$$ Something caught my attention:

QUESTION. What is the conceptual or combinatorial reason that the right-hand side of $$\sum_{n\geq0}x^n\sum_{\lambda\vdash n}\prod_{\square\in\lambda}\frac{a+\pmb{q}c_{\square}+c_{\square}^2}{h_{\square}^2}=(1-x)^{-a}$$ is independent of $\pmb{q}$?

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    $\begingroup$ Doesn’t transposition cancel out all the terms involving only the first power of the content? $\endgroup$ Mar 13, 2021 at 22:29
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    $\begingroup$ If $f$ is a polynomial, then $f(x)+f(-x)$ has no odd-power terms (including $x$ term). That doesn't mean $f$ has no $x$ term. Does it? $\endgroup$ Mar 13, 2021 at 23:24
  • $\begingroup$ Sorry, you're right, transposition shows that the expression has to be invariant under $q \mapsto -q$, but not that it is independent of $q$ altogether. $\endgroup$ Mar 13, 2021 at 23:29
  • $\begingroup$ The RHS is the exponential generating function for permutations by number of cycles. By equation (4) in the paper, $t$ counts cycles of $\sigma$ while $v$ counts cycles of $\sigma^{-1}$, but those are the same so you just get powers of $tv$. $\endgroup$
    – lambda
    Mar 14, 2021 at 2:26
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    $\begingroup$ @T.Amdeberhan I guess it's up to you whether you think it answers the question or not, but I left it as a comment and not an answer because I assumed not. It's a combinatorial/conceptual explanation of why the first formula depends only on $tv$, but doesn't readily re-express itself in terms of $q$ and $a$. $\endgroup$
    – lambda
    Mar 14, 2021 at 17:42

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