I was trying to get the probability distribution $p(n)$ from a generating function $G(s)$ like this:
$G(s) = e^{a(s-1)^2}=\sum s^np(n)$
I need first to do Maclaurin expansion of the exponential and then get the $n$th order term for $p(n)$.
My first thinking was it would be simple to calculate the derivatives. But it turns out to be much more difficult and also very interesting to generalize the $n$ order derivative.
I list a table for the powers and coefficients of each derivative order, finding that the powers are odd numbers for odd $n$, even numbers for even $n$, the coefficients are associated to $(n-1)$th order's powers and coefficients. It is easy to see the association but I cannot generalize it.
This calculation is also like $z$ transform but no existing result for it.
Anyone could give a shot and help me out?
UPDATE: to give some feedbacks to the question for other users
Actually the generating function I gave above cannot generate a proper probability sequence since negative values will show up. One way of modifying it is to add one more term in the power:
\begin{equation} e^{a(s-1)^2+b(s-1)}=\sum s^np(n) \end{equation}
with constraint $b>2a$ guaranteeing the positiveness of sequence elements. Using hermite polynomials and changing of variables, one can obtain
\begin{equation} \nonumber p(n)=e^{a-b}\frac{\alpha^n}{n!}H_n(\frac{b+2\alpha^2}{2\alpha}) \end{equation}