Ordinary Power series

Question

How do I find an exponential generating function for $$\{p(n)/n!\}_{1}^{\infty}$$ where $$ p \in \mathbb{Z}^{+}[x] $$I see a method to find an opsgf for the same sequence by applying $x \frac{d}{dx}$ operator to $e^x$ but I am not able to figure out a good way for egf. Help would be appreciated. Thank You.

Answer 1

With the modified Bessel function $I_0$ we have

$$ \begin{array}{ll} \displaystyle\sum_{n=0}^\infty p(n)\frac{x^n}{n!^2} & \displaystyle = \sum_{n=0}^\infty \Big(\sum_{k=0}^d a_kn^k\Big)\frac{x^n}{n!^2} \\[5pt] & \displaystyle = \Big[\sum_{k=0}^d a_k\Big(x\frac{d}{dx}\Big)^k\Big]\sum_{n=0}^{\infty} \frac{x^n}{n!^2} \\[5pt] & = \displaystyle p\Big(x\tfrac{d}{dx}\Big) I_0(2\sqrt{x}). \end{array} $$