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After like 100 hours searching info to unravel a mystery about expressions on $e$ like $\frac{e^{6} A_{5}\left(e^{-1}\right)}{e^{6}\left(1-e^{-1}\right)^{6}} \approx 5 !$ talked by me in here, user Jair Taylor redirected me to this generating functions $\frac{1}{1-e^{x-1}}$ as they generate exactly the coefficients I'm interested in.

So I'm asking for help hopping to get a true final combinatorial (or even probabilistic) interpretation of this coefficients (so I can stop searching for esoteric papers and maybe focus on my other subjects).

Anyway sadly although $\frac{1}{1-e^{x+1}}$ is related to $Seq(Set^*(z+1))$ a very natural looking construction from Flajolet and Sedgewick's classic book, this has problems. In fact it's a very exact variation of the surjection generating function $$\frac{1}{2-e^z}$$ first the surjection evaded the set with no elements (it was $Seq(Seq(Z)-1)$). My expression on the other hand even has an extra empty element $Z+1$ entering the set construction, or $Seq(Set(Z+\varepsilon)$.

I have also imagined maybe an $e$ "cost" factor (rewriting $\frac{1}{1-e^{x+1}} = \frac{1}{1-e*e^{x}}$) but with the problem of having the empty "urn" going this way doesn't seem promising enough.

For a maybe less contrived approach this seems to be related to seemingly niche problem called Simon Newcomb's problem: A counting problem about compositions with $k$ falls, in generic multisets.

The expression that pertains me the most is $$\sum_{d=0}^{\infty} \frac{A_{d}(t) x^{d}}{(1-t)^{d+1} d !}=\frac{1}{1-t e^{x}}$$ where replacing $t$ follows the patterns I see in wolfram-alpha for the two functions in the title, and going with $t=1$ gives an identity in wikipedia, but as both expressions don't seem to have any justification, it's just a formula for now.

Trying to see the connection with the general problem and reading some of the papers the only thing I found was a curious $q-$nomial identity $e_q(x^{-1}) = (x ; q)_{\infty}=\prod_{n=0}^{\infty}\left(1-x q^{n}\right)$

that in the limit would give a crazy equality suggested in that post $$e^{-x}= \prod_{n=0}^{\infty}\left(1-x\right)$$

The $q$ version property appears in the Wikipedia page for the so called $q$-exponential and the quantum dilogarithm but along "polytopes","matrix generating functions" or other scary names in papers of this Newcomb problem, suggest me I'm maybe not ready to solve this problem.

Alejandro Quinche
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    Not sure how helpful this is, but $\frac{-x}{1-e^x}$ is an exponential generating function for the Bernoulli numbers which resembles your generating function. Maybe since they're more well studied there is a clear analogous relation between that GF and a combinatorial derivation you can use. – Merosity Jun 22 '21 at 17:08
  • Sorry, but I don't think this formula can be allright: $$e^{-x}= \prod_{n=0}^{\infty}\left(1-x\right)$$ – Han de Bruijn Jun 22 '21 at 18:49
  • I'm not sure if this helps, but $e^x$ has the Maclaurin series $e^x=\sum\limits_{n=0}^\infty\frac{x^n}{n!}$ which implies $\frac{1}{1-e^{x+1}}=\frac{1}{1-\sum\limits_{n=0}^\infty\frac{(x+1)^n}{n!}}$ and $\frac{1}{1-e^{x-1}}=\frac{1}{1-\sum\limits_{n=0}^\infty\frac{(x-1)^n}{n!}}$. – Steven Clark Jun 22 '21 at 21:57
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    As was point out in the comment by @Merosity: $\frac{1}{1-e^{x+1}}=-\sum_{n=0}^{\infty } \frac{B_n}{n!}(x+1)^{n-1}$ and $\frac{1}{1-e^{x-1}}=-\sum_{n=0}^{\infty } \frac{B_n}{n!} (x-1)^{n-1}$. – Steven Clark Jun 22 '21 at 22:24
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    Check this paper, the former series in the title is related to the Genocchi numbers: https://cs.uwaterloo.ca/journals/JIS/VOL7/Domaratzki/doma23.pdf – Phicar Jun 23 '21 at 15:21
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    It's a little unclear to me what you're asking exactly. What would a combinatorial interpretation for $\frac{1}{1-e^{x-1}}$ mean given that the coefficients are not integers? On the other hand if you're looking for a combinatorial interpretation for $\frac{1}{1-t e^{x}}$, that's a well-known thing we can discuss. – Jair Taylor Jun 24 '21 at 20:49
  • Hi again Jair, I would like the combinatorial interpretation of that series at least, I guess i naively tough I could extrapolate an interpretation from it even when applied to a trascendental. – Alejandro Quinche Jun 26 '21 at 00:13

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If $f(z)=\displaystyle\frac1{1-t\,e^z}$, then $(1-t)^{n+1}n![z^n]f(z)$ is $t$ times the $n$th Eulerian polynomial (in $t$).

So, if $t$ is a constant, $n![z^n]f(z)$ is $t/(1-t)^{n+1}$ times a weighted count of $n$-permutations, where a permutation with $k$ ascents has weight $t^k$.

David Bevan
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