Suppose you have a formula $ \sum_{n\geq 0}f(n)\frac{x^n}{n!}=\exp\left(x+\frac{x^2}{2}\right). $
There is a recurrence for $f(n)$ found by differentiation, $ \sum_{n\geq 1}f(n)\frac{x^{n-1}}{(n-1)!}=(1+x)e^{x+x^2/2}=(1+x)\sum_{n\geq 0}f(n)\frac{x^n}{n!}. $ So equating coefficients gives $f(n+1)=f(n)+nf(n-1)$ for $n\geq 1$.
Moreover, and explicit formula is found by noting $ \sum_{n\geq 0}f(n)\frac{x^n}{n!}=e^xe^{x^2/2}=\left(\sum_{n\geq 0}\frac{x^n}{n!}\right)\left(\sum_{n\geq 0}\frac{x^{2n}}{2^nn!}\right) $ and so $ f(n)=\sum_{i\geq 0\atop\text{$i$ even}}\binom{n}{i}\frac{i!}{2^{i/2}(i/2)!}=\sum_{j\geq 0}\binom{n}{2j}\frac{(2j)!}{2^jj!}. $
Based on this, what is a nice expression for $\sum_{i=0}^n(-1)^{n-i}\binom{n}{i}f(i)$? (I am grateful for Sasha's current answer, but is it possible to derive such an expression more combinatorially without reference to random variables and moments? If not, that is no problem. I'm just glad to see and answer.)
Thanks!
(I should note that this is motivated by a passage in Richard Stanley's Enumerative Combinatorics following example 1.1.15.)