Is there a closed form for $\sum \frac{1}{k!}\binom{m}{k}\binom{n}{k}x^k$? Let $F = \sum_{k=0}^\infty \frac{1}{k!}a_kx^k$ be a formal power series and define $F_n = \sum_{k=0}^\infty \frac{1}{k!}\binom{n}{k}a_kx^k$. Is it possible to define $F_n$ starting from $F$ by some elementary operations? I don't know what kind of operations I mean, probably sum, multiplication, powering, composition, differentation, integration...
In particular, is there a closed form for $\sum_{k=0}^\infty \frac{1}{k!}\binom{m}{k}\binom{n}{k}x^k$ ?
Sorry if this question lacks of details, but I don't know anything about generating functions, combinatorics and so on. I hope I can apply the answer to another field.
 A: We can use the coefficient of operator $[z^k]$ to denote the coefficient of $z^k$ of a series. This way we can write for instance
\begin{align*}
[z^k](1+z)^n=\binom{n}{k}
\end{align*}

With $F(x)=\sum_{k=0}^\infty a_k\frac{x^k}{k!}$ we have
\begin{align*}
\color{blue}{F_n(x)}&=\sum_{k=0}^{\infty}\binom{n}{k}a_k\frac{x^k}{k!}\\
&=\sum_{k=0}^{\infty}[z^k](1+z)^n a_k\frac{x^k}{k!}\tag{1}\\
&=[z^0](1+z)^n\sum_{k=0}^{\infty}z^{-k}a_k\frac{x^k}{k!}\tag{2}\\
&\,\,\color{blue}{=[z^0](1+z)^nF\left(\frac{x}{z}\right)}
\end{align*}

Comment:

*

*In (1) we use the coefficient of operator $[z^k]$.


*In (2) we use the rule$[z^{p-q}]A(z)=[z^p]z^qA(z)$.
A: Long Comment:
There is a simple pattern for $S_{m,n}=\sum_{k=0}^\infty \frac{1}{k!}\binom{m}{k}\binom{n}{k}x^k$
$$n=1:\;\;\;\;S_{m,1}=1+m$$
$$n=2:\;\;\;\;S_{m,2}=1+2mx+\frac{(m-1)m}{(2!)^2}x^2$$
$$n=3:\;\;\;\;S_{m,3}=1+3mx+\frac{3(m-1)m}{(2!)^2}x^2+\frac{(m-2)(m-1)m}{(3!)^2}x^3$$
$$n=4:\;\;\;\;S_{m,4}=1+4mx+\frac{6(m-1)m}{(2!)^2}x^2+\frac{4(m-2)(m-1)m}{(3!)^2}x^3+\frac{(m-3)(m-2)(m-1)m}{(4!)^2}x^4$$
The relations are symmetric between $m$ and $n$.
