I recently proved that the Taylor Series of $\exp(\exp(x))$ is given by $$\exp(\exp(x))=\sum_{n=0}^\infty \frac{eB_n x^n}{n!}$$ where $B_n$ are the Bell Numbers.

However, I can't figure out a Taylor series for the function $$\exp(\exp(\exp(x))) = \text{ ?}$$ Does anyone know how to find this Taylor series, perhaps in terms of the Bell or Stirling numbers, or any other well-known sequences? The first couple terms are $$\frac{e^e}{0!}+\frac{e^{e+1}}{1!}x+\frac{e^{e+2}+2e^{e+1}}{2!}x^2 + \frac{e^{e+3}+6e^{e+2}+5e^{e+1}}{3!}x^3+\cdots$$

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    $\begingroup$ Use Faa di Bruno. $\endgroup$
    – user517969
    Jan 3, 2018 at 15:10
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    $\begingroup$ This oeis.org/A039810 appears to be the number triangle in the numerators of your expansion. $\endgroup$ Jan 3, 2018 at 15:13
  • $\begingroup$ You might find go.helms-net.de/math/tetdocs/TTetrationExactEntries_short.htm helpful. In the table for $T^3$ you find $\lambda$ and $b$. Using this for your case you can replace $\lambda=1$ and $b=e$. Possibly the matrix-multiplication-scheme makes it easier for you to extend the list of coefficients and even for higher iterates mechanically. $\endgroup$ Dec 29, 2018 at 21:38
  • $\begingroup$ A generalization of the coefficients in your series for higher iterates and even polynomial expression in which fractional $h$ for iteration-height can be inserted can be found in an older article go.helms-net.de/math/binomial_new/04_5_SummingBellStirling.pdf with the help of (matrix) logarithm of the Stirlingnumbers 2nd kind. $\endgroup$ Dec 29, 2018 at 21:51

1 Answer 1


It will be cleaner to instead compute the Taylor series of

$$\exp(t (\exp(\exp(x) - 1) - 1) )$$

where $t$ is a parameter, then set $t = e$ and multiply by $e^e$. This is because, like $\exp(\exp(x) - 1)$, whose coefficients are the Bell numbers, the coefficients of this Taylor series have a combinatorial interpretation. More precisely, the yoga of exponential generating functions can be used to show that they count the following:

The coefficient of $t^k \frac{x^n}{n!}$ is the number of ways to partition a set of size $n$ into $k$ nonempty subsets, then to further partition each of those subsets into some number of nonempty subsets. One might call these "2-level" set partitions.

Similarly $\exp(\exp(\exp(\exp(x)))$ is related to "3-level" set partitions, with a bunch of random $e$s around because you haven't subtracted $1$, and so forth.


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