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I am looking for an answer to the following problem.

Let $S$ be the multiset $\{1^{d_1},2^{d_2},\dots,m^{d_m}\}$ $A_{S,k}$ is the number of permutations of $S$ with $k-1$ descents and no descent at the end.

  1. Let $A_S(t)=\sum\limits_k A_{S,k}t^k$ and $d=\sum\limits_{i\leq m} d_i$. I want to show that $$\sum\limits_S \frac{A_S(t)}{(1-t)^{d+1}} \prod\limits_i x_i^{d_i}=(1-t\prod\limits_i(i-x_i)^{-1})^{-1}$$

  2. The Eulerian polynomial $A_d(t)$ is the special case $A_{[d]}$. I want to show that $$\sum\limits_{d=0}^{\infty}\frac{A_d(t)x^d}{(1-t)^{d+1}d!}=\frac{1}{1-te^x}$$

Thanks for your help.

bronko
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  • For 1. I somehow have to use that $A_{\emptyset}(t)=1$ and $\frac{A_S(t)}{(1-t)^{d+1}}=\sum\limits_{n=0}^{\infty}t^n\prod\limits_i\binom{n+d_i-1}{d_i}$ – bronko Dec 13 '12 at 13:38
  • If you are still active in stack exchange I'm very interested in knowing what introduced you to this subject? I was wondering about the coefficients where one evaluates $t$ as $1/e$ or $e$ in this post https://math.stackexchange.com/questions/4174786/combinatorial-interpretation-of-rational-function-on-e – Alejandro Quinche Jun 17 '21 at 17:27
  • Going from 1 to 2 seems to be related to the q-exponential or the "quantum" dilogarithm – Alejandro Quinche Jun 18 '21 at 04:01

0 Answers0