An interesting calculation of derivative 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.  
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}
 A: $$\eqalign{G(s) &= \sum_{j=0}^\infty \dfrac{a^j (s - 1)^{2j}}{j!} = \sum_{j=0}^\infty \sum_{n=0}^{2j} {2 j \choose n} \dfrac{(-1)^{2j-n} a^j}{j!} s^n\cr & 
= \sum_{n=0}^\infty \sum_{j = \lceil n/2 \rceil}^\infty  {2 j \choose n} \dfrac{(-1)^{n} a^j}{j!} s^n}$$
so your coefficients are
$$p(n) = (-1)^n \sum_{j=\lceil n/2 \rceil}^\infty {2j \choose n} \dfrac{a^j}{j!}$$ 
It appears each $p(n)$ is $\exp(a)$ times a polynomial $q_n(a)$ satisfying
the recurrence
$$-2\,a \left( 2+n \right) q_{{n+1}} \left( a \right) -4\,a \left( n+3
 \right) q_{{n+2}} \left( a \right) - \left( n+4 \right)  \left( -n+2
\,a-2 \right) q_{{n+3}} \left( a \right) + \left( n+5 \right)  \left( 
n+4 \right) q_{{n+4}} \left( a \right) 
 = 0 $$
A: Robert Israel noted that $p(n)$ is a $e^a$ times a polynomial of degree $n$
in $a$ with a zero of order $\lceil n/2 \rceil$ at $a=0$.
We express this polynomial in terms of a
Hermite
polynomial $H_n$ evaluated at an imaginary argument $\alpha := \pm \sqrt{-a}$.
Start from the generating function
$$
\exp(2xt-t^2) = \sum_{n=0}^\infty H_n(x) \frac{t^n}{n!},
$$
and set $(x,t) = (\alpha, \alpha s)$.  Then
$$
2xt-t^2 = \alpha^2 (2s-s^2) = a (s^2-2s) = a(s-1)^2 - a,
$$
whence
$$
e^{a(s-1)^2} = e^a e^{a(s^2-2s)} 
= e^a \sum_{n=0}^\infty H_n(\alpha) \frac{(\alpha s)^n}{n!},
$$
and $p(n) = \alpha^n H_n(\alpha) / n!$.  Known formulas and identities
for the Hermite polynomials can then be used to study the sequence $\{ p_n \}$.
