Closed form for $\sum_{n=0}^{\infty}\frac{x^n}{n!}\sum_{m=0}^{n+1}\frac{y^m}{(n+1-m)!}$?

Question

For the following infinite series

$$\sum_{n=0}^{\infty}\frac{x^n}{n!}\sum_{m=0}^{n+1}\frac{y^m}{(n+1-m)!}$$

Is there any closed expression for this infinite sum in terms of $x,y$? My intuitive thought leads me to power series related to gamma function, inverse gamma function, beta function or Bessel functions. Or maybe there is none. I welcome any hint or thought.

Answer 1

There may be a closed form with the Marcum Q function since @Claude Leibovici noted:

$$y\,e^{\frac{1}{y}}\,\sum_{n=0}^{\infty}\frac{ \Gamma \left(n+2,\frac{1}{y}\right)}{n! (n+1)!}(xy)^n = y\,e^{\frac{1}{y}}\,\sum_{n=0}^{\infty}\frac{Q \left(n+2,\frac{1}{y}\right)}{n!}(xy)^n $$

and

$$e^aQ_m(\sqrt{2 a},\sqrt{2 b})=\sum_{n=0}^\infty \frac{\Gamma(m+n,b)}{n!\Gamma(m+n)}a^n= \sum_{n=0}^\infty \frac{Q(m+n,b)}{n!}a^n $$

with gamma regularized $Q(a,z)$. Therefore:

$$\sum_{n=0}^{\infty}\frac{x^n}{n!}\sum_{m=0}^{n+1}\frac{y^m}{(n+1-m)!}=ye^{xy+\frac1y}Q_2\left(\sqrt{2xy},\sqrt\frac2y\right)$$

which is true

Answer 2

Beside $$\sum_{n=0}^{\infty}\frac{x^n}{n!}\sum_{m=0}^{n+1}\frac{y^m}{(n+1-m)!}=y\,e^{\frac{1}{y}}\,\sum_{n=0}^{\infty}\frac{ \Gamma \left(n+2,\frac{1}{y}\right)}{n! (n+1)!}(xy)^n$$ I do not see what we could write.

Even $y=x$ does not lead to anything.