So there's an elementary (but in my opinion quite interesting!) result which is that the Laurent series expansion of


$$\frac{1}{1-e^x} = -\frac{1}{x} + \frac{1}{2} - \frac{1}{12}x - \frac{1}{720}x^3 \cdots$$


Now the reason that is interesting is because each of those coefficients are equal to $\frac{1}{n!}B\_n$ where $B\_n$ is $n$th Bernoulli number and so more deeply one has the result


$$ \Gamma(\alpha + 1) \frac{d^\alpha}{dx^\alpha} \left[ \frac{1}{1-e^x} - 1 \right] = (-1)^{\alpha-1} \alpha \zeta(1-\alpha) $$


If you choose your fractional derivative carefully. ^^ (there might be an off by 1 error there, I need to double check)


So naturally I was inspired to go hunting for other elementary functions and to look at their coefficients/fractional derivatives in the hopes of finding something cool.


So I decided to look at $\frac{1}{1-e^{e^x-1}}$ the laurent series expansion of this is


$$\frac{1}{1-e^{e^x-1}} = -\frac{1}{x} + 1 - \frac{x}{6} - \frac{x^2}{24} - \frac{x^3}{90} - \frac{x^4}{720} + \frac{59}{60480}x^5 + \frac{17}{20160}x^6 + \frac{169}{453600} x^7 + \frac{47}{483840}x^8 - \frac{181}{119750400}x^9 \cdots$$


This was generated using this python code:



```
import sympy as sym
import math

x = sym.symbols('x')

w = 1/(1 - sym.exp(sym.exp(x)-1))
result = w.series(x, 0, 10).removeO()
print(result)

```

So I went over to OEIS and tried looking for those denominators and found nothing, I also looked at the denominators scaled by n! and similarly came up empty handed.


So this has me rather surprised since this is a pretty simple series to write down. My first question then that I want to ask is: is there some recurrence or functional equation that these coefficients/some scaled version of these coefficients obey?


If no recurrence then at least some kind of interpretation of their combinatorial significance would be cool.



Is there a recurrence for the coefficients of the Laurent series expansion of $\frac{1}{1-e^{e^x - 1}}$?
$e^{e^x-1}$ is the exponential generating function for [Bell numbers](https://en.wikipedia.org/wiki/Bell_number) ${\cal B}\_n$:
$$e^{e^x-1} = \sum\_{n\geq 0} {\cal B}\_n \frac{x^n}{n!}.$$
Then
$$g(x) := \frac{e^{e^x-1}-1}{x} = \sum\_{n\geq 0} {\cal B}\_{n+1} \frac{x^n}{(n+1)!}.$$


Correspondingly, the coefficient of $x^n$ in $g(x)^{-1}$ can be computed via [Faà di Bruno's formula](https://en.wikipedia.org/wiki/Fa%C3%A0_di_Bruno%27s_formula) as
\begin{split}
\frac1{n!}\left.\left(\frac{d}{dx}\right)^n g(x)^{-1}\right|\_{x=0} &= \frac1{n!}\sum\_{k=1}^n (-1)^k k! B\_{n,k}( \frac{{\cal B}\_2}2, \frac{{\cal B}\_3}3, \dots) \\
&=\sum\_{k=1}^n (-1)^k \hat B\_{n,k}( \frac{{\cal B}\_2}{2!}, \frac{{\cal B}\_3}{3!}, \dots),
\end{split}
where $B\_{n,k}$ and $\hat B\_{n,k}$ are the exponential and ordinary [Bell polynomials](https://en.wikipedia.org/wiki/Bell_polynomials), respectively. Up to a sign this gives the coefficient of $x^{n-1}$ in $\frac1{1-e^{e^x-1}} = -\frac{g(x)^{-1}}x$.


PS. [This Sage code](https://sagecell.sagemath.org/?z=eJxNj80KgzAQhO-C77A3d7X-pEehb1IqahMJMRtJ46GI715_KDiHhfkY2Jm3VDDI0PROKoWW6jiCTQwPsJk4jZdh9gw5fGaLgLmgl4EUVNsH53U7oqHNdnIcm8mNX3Z2h3wzhCkemGfbSY86u1N5XFDOgwbN4FseJHJuMkF0crNzFEXBGygvf5jiKI72iD0iVVGI6l958poDqmSxa_0My3XUmtAPp-1DxA==&lang=sage&interacts=eJyLjgUAARUAuQ==) computes the coefficients using the derived formula.


