Is this sequence of polynomials known? $ P_{n,m}(x) = \sum_{j=0}^{\min(n,m)}j! \binom{n}{j}\binom{m}{j}x^{n+m-2j} $ For natural $n,m$ let $P_{n,m}(x)$ be defined by:
$$
\exp(x(s+t) +st) = \sum_{n,m\geq 0}^{\infty}\frac{s^n t^m}{n! m!}P_{n,m}(x)
$$
Alternatively one can write:
$$
P_{n,m}(x) = \sum_{j=0}^{\min(n,m)}j! \binom{n}{j}\binom{m}{j}x^{n+m-2j}
$$
Is this polynomial known by some name? (I need its asymptotics for large $t$) Note that this is superficially similar to the linearization formula fo Hermite polynomials:
$$
H_n(x)H_m(x)=\sum_{j=0}^{\min(n,m)}j! \binom{n}{j}\binom{m}{j}H_{n+m-2j}(x)
$$
but in our case the polynomial doesn't factorize as a product.
 A: Elaborating on the other answer. First note that if $m<n$, then
$$
P_{n,m} (x) = \sum\limits_{j = 0}^m j!\binom{n}{j}\binom{m}{j}x^{n + m - 2j},
$$
whereas if $m \ge n$, then
\begin{align*}
P_{n,m} (x) &= \sum\limits_{j = 0}^n {j!\binom{n}{j}\binom{m}{j}x^{n + m - 2j} }  = \sum\limits_{j = 0}^n {j!\binom{n}{j}\binom{m}{j}x^{n + m - 2j} }  + \underbrace {\sum\limits_{j = n + 1}^m {j!\binom{n}{j}\binom{m}{j}x^{n + m - 2j} } }_0 \\ & = \sum\limits_{j = 0}^m j!\binom{n}{j}\binom{m}{j}x^{n + m - 2j}.
\end{align*}
Thus,
$$
P_{n,m} (x) = \sum\limits_{j = 0}^m j!\binom{n}{j}\binom{m}{j}x^{n + m - 2j}
$$
for all $n,m\geq 0$. Note also that $P_{n,m}(x)=P_{m,n}(x)$. By $(13.2.10)$, it is easy to see that
$$
P_{n,m} (x) = ( - 1)^m x^{n - m} U( - m,1 - m + n, - x^2 ),
$$
where $U$ is one of Kummer's functions. By $(13.6.19)$ and $(13.6.20)$ we obtain representations in terms of the Laguerre and Charlier polynomials:
$$
P_{n,m} (x)= m!x^{n - m} L_m^{(n-m)}(-x^2 ) = x^{n + m} C_m (n; - x^2 ) .
$$
A: $$ P_{n,m}(x) = \sum_{j=0}^{m}j! \binom{n}{j}\binom{m}{j}x^{n+m-2j} $$
As [er mathematica
$$P_{m,n}(x)=x^{m+n} (-x^2)^{-m} \text{HypergeometricU}[-m,1-m+n,-x^2]$$
