Does this function have a closed form? $$k_a(x)=\sum^\infty_{n=0}{\frac{n^2}{(n+x)!}}$$
$$k_b(x)=\sum^\infty_{n=0}{\frac{(n+x)^2}{n!}}$$
I noticed these functions closely relate to $e$. By looking at them I was able to determain a closed from for $k_b$.
$$k_b(x)=e(x^2+2x+2)$$
Though I still have not found a short form for $k_a$. I do think it exists. Some integer inputs are as follows.
$k_a(0)=2e$
$k_a(1)=e-1$
$k_a(2)=2e-5$
$k_a(3)=5e-\frac{27}{2}$
Can any of you take a stab at it?

This is too long for a comment.
Considering $$k_a(x)=\sum^\infty_{n=0}{\frac{n^2}{(n+x)!}}$$ and using the result from Wolfram Alpha, it seems that the expression could simplify to
$$k_a(x)=\frac{(2+x-x^2)+e\,(2-2x+x^2)\left(\Gamma (x+2)-(x+1)\, \Gamma (x+1,1) \right) }{\Gamma (x+2)}$$ which, as  user404188  noticed,  for integer values of $x$, reduces to $$k_a(x)=(x^2-2x+2)\,e + \mathrm{rational}$$ as shown in the table.
$$\left(
\begin{array}{cc}
 x & k_a(x) \\
 0 & 2 e \\
 1 & -1+e \\
 2 & -5+2 e \\
 3 & -\frac{27}{2}+5 e \\
 4 & -\frac{163}{6}+10 e \\
 5 & -\frac{1109}{24}+17 e \\
 6 & -\frac{2827}{40}+26 e \\
 7 & -\frac{14483}{144}+37 e \\
 8 & -\frac{685007}{5040}+50 e \\
 9 & -\frac{2374691}{13440}+65 e \\
 10 & -\frac{80885629}{362880}+82 e
\end{array}
\right)$$
The expression can rewrite $$k_a(x)=e \left(x^2-2 x+2\right)-\frac{x^2-x-2}{\Gamma (x+2)}-e \left(x^3-x^2+2\right)\frac{ \Gamma (x+1,1)}{\Gamma (x+2)}$$
