Is there a recurrence for the coefficients of the Laurent series expansion of $\frac{1}{1-e^{e^x - 1}}$?

Question

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.

Answer 1

$e^{e^x-1}$ is the exponential generating function for Bell numbers ${\cal B}_n$: $$e^{e^x-1} = \sum_{n\geq 0} {\cal B}_n \frac{x^n}{n!}.$$ 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 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, 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 computes the coefficients using the derived formula.

Answer 2

You can use FriCAS to find a differential equation for $\frac{x}{1-e^{e^x-1}}$ without any work. I'll do it from within Sagemath here:

sage: L.<x> = LazyLaurentSeriesRing(QQ)
sage: fricas.guessADE((1/(1-exp(exp(x)-1)))[:60])
[
  [
      n
    [x ]f(x):
                   2    2      ,,                   2  ,   2
          (- x f(x)  + x f(x))f  (x) + (2 x f(x) - x )f (x)

        + 
                      2    2      ,          3              2
          ((x - 2)f(x)  - x f(x))f (x) - f(x)  + (x + 1)f(x)

      = 
        0
    ,
                     2    3
                    x    x       4
   f(x) = - 1 + x - -- - -- + O(x )]
                     6   24
  ]
```