finite sum - closed form solution?

Question

To answer a question in another post, the solution included a finite sum $\sum_{k=0}^L \frac {x^{-k}}{k!(n-2k)!}$, where $L=\lfloor \frac{n}{2}\rfloor$. Can this sum be expressed more concisely as a function of $x$ and $n$?

Answer 1

Well, $$ \sum_{k=0}^{\lfloor n/2\rfloor}\frac{1}{x^k k!}\cdot\frac{1}{(n-2k)!} $$ is the coefficient of $z^{n}$ in the product between $$ \sum_{k\geq 0}\frac{z^{2k}}{x^k k!}=e^{z^2/x}\qquad\text{and}\qquad \sum_{k\geq 0}\frac{z^k}{k!}=e^z, $$ hence by the residue theorem it can be written as $$ \frac{1}{2\pi i}\oint_{\|z\|=1}\exp\left(\frac{z^2}{x}+z\right)\,\frac{dz}{z^{n+1}}=\frac{1}{2\pi i x^n}\oint_{\|z\|=1}\exp\left(x(z^2+z)\right)\,\frac{dz}{z^{n+1}} $$ and approximated via the Laplace method.

Answer 2

The good answer is yes.

The bad answer is that the result involves the Tricomi's confluent hypergeometric function $U(a,b,z)$.

If $n=2m$ the result would be $$\frac{4^m (-x)^{-m}}{(2 m)!}\, U\left(-m,\frac{1}{2},-\frac{x}{4}\right)$$

If $n=2m+1$ the result would be $$\frac{4^m (-x)^{-m}}{(2 m+1)!}\, U\left(-m,\frac{3}{2},-\frac{x}{4}\right)$$