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Mar 14, 2021 at 19:26 comment added lambda @InesInstitoris I'm not entirely sure why you're asking me specifically, but I don't see an obvious reason you'd get such a characterization.
Mar 14, 2021 at 17:56 comment added Jeanne Scott @lambda Sorry to respond to your question with a question, but does this invariance (i.e. the independence with respect to $q$) lead to a characterization of the content statistic ?
Mar 14, 2021 at 17:42 comment added lambda @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 $a$.
Mar 14, 2021 at 16:01 comment added T. Amdeberhan @lambda: It is not clear if you're answering the question or just remarking.
Mar 14, 2021 at 2:26 comment added lambda 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 $tv$.
Mar 13, 2021 at 23:29 comment added Sam Hopkins 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.
Mar 13, 2021 at 23:24 comment added T. Amdeberhan 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?
Mar 13, 2021 at 22:29 comment added Sam Hopkins Doesn’t transposition cancel out all the terms involving only the first power of the content?
Mar 13, 2021 at 22:27 history edited T. Amdeberhan CC BY-SA 4.0
edited title
Mar 13, 2021 at 22:00 history edited LSpice CC BY-SA 4.0
Name of Stanley's paper
Mar 13, 2021 at 21:54 history asked T. Amdeberhan CC BY-SA 4.0