Taylor coefficients of $e^{-\pi n^2e^{2x}}$. I am trying to write down a closed form expression for the Taylor series of the function $e^{-\pi n^2 e^{2x}}$ expanded at $x=0$, where $n$ is a positive integer.
Differentiating explicitly one finds that
$$e^{-\pi n^2 e^{2x}} = e^{-\pi n^2}(1-2\pi n^2x+2\pi n^2 (\pi n^2-1)x^2-\frac{4}{3}\pi n^2 (\pi^2 n^4-3\pi n^2+1)x^3+O(x^4))$$
From here one can conclude that the general term $\frac{c_m x^n}{m!}$, in the Taylor series, satisfies $c_m \neq 0$, since $\pi$ is trancendental and $c_m$ contains at worst polynomial factors.
I have also tried the usual composition of Taylor series, but i cannot obtain any more information on $c_m$.
Any suggestions will be greatly appreciated!
 A: Expanding $e^{te^x}$ on the outer exponential first, and then on the inner exponential, and finally switching order of summations gives
$$\begin{align}
e^{t e^x}
& = \sum_{m=0}^{\infty} \frac{1}{m!} \left(t e^x \right)^m \\
& = \sum_{m=0}^{\infty} \frac{1}{m!} t^m e^{mx} \\
& = \sum_{m=0}^{\infty} \frac{1}{m!} t^m \left( \sum_{k=0}^{\infty} \frac{1}{k!} (mx)^k \right) \\
& = \sum_{k=0}^{\infty} \frac{x^k}{k!} \left( \sum_{m=0}^{\infty} \frac{t^m}{m!} m^k \right) \\
\end{align}$$
The inner parenthesis can now be written as
$$\sum_{m=0}^{\infty} \frac{t^m}{m!} m^k = \left(t\frac{d}{dt}\right)^k e^t = p_k(t) e^t,$$
where $p_k(t)$ are polynomials satisfying $p_0(t) = 1$ and $p_{k+1}(t) = t \left( p_k'(t) + p_k(t) \right).$ I don't know if there is an explicit formula for the coefficients of these polynomials.
Thus,
$$e^{t e^x} = \sum_{k=0}^{\infty} \frac{x^k}{k!} p_k(t) e^t = e^t \sum_{k=0}^{\infty} \frac{x^k}{k!} p_k(t)$$
and
$$
e^{-\pi n^2 e^{2x}} 
= e^{-\pi n^2} \sum_{k=0}^{\infty} \frac{(2x)^k}{k!} p_k(-\pi n^2) 
= \sum_{k=0}^{\infty} \frac{2^k e^{-\pi n^2} p_k(-\pi n^2) }{k!} x^k.
$$
i.e. the coefficient of $x^k$ is
$$c_k = \frac{2^k e^{-\pi n^2} p_k(-\pi n^2)}{k!}.$$
