Evaluating the sum: $S_n=\sum_{k=1}^\infty\dfrac{k^n}{(k!)^2}$ The sum:
$$S_2=\sum_{k=1}^\infty\dfrac{k^2}{(k!)^2}$$ is equal to:
$S_2=I_0(2)$ where $I_0(2)$ is the modifyed Bessel function of the first kind $I_0(x)$ calculated for $x=2$. $$S_3=\sum_{k=1}^\infty\dfrac{k^3}{(k!)^2}$$
is equal to: $S_3=I_0(2)+I_1(2)$. For $n\gt3$ the result is a more complicated form involving Hypergeometric functions.
My question is: is it possible to find a closed formula for $$S_n=\sum_{k=1}^\infty\dfrac{k^n}{(k!)^2}?$$ Thanks in advance for any answer or hint.
 A: By using Stirling numbers of the second kind we have:
$$ k^n = \sum_{j=0}^{n}{n \brace j}(k)_j $$
where $(k)_j$ is the falling Pochhammer symbol: $(k)_j = k(k-1)\cdot\ldots\cdot(k-j+1).$
Since:
$$\sum_{k=1}^{+\infty}\frac{(k)_j}{(k!)^2}=I_{-j}(2)$$
it follows that:

$$\sum_{k=1}^{+\infty}\frac{k^n}{(k!)^2}=\sum_{j=0}^{n}{n \brace j}I_{-j}(2)=\sum_{j=0}^{n}{n \brace j}I_{j}(2).$$

Have a look at this OEIS entry, too.
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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\begin{align}
\color{#f00}{S}_{n + 2} & = \sum_{k = 1}^{\infty}{k^{n + 2} \over \pars{k!}^{2}} =
\sum_{k = 1}^{\infty}{k^{n} \over \bracks{\pars{k - 1}!}^{2}} =
\sum_{k = 0}^{\infty}{\pars{k + 1}^{n} \over \pars{k!}^{2}} = 
1 + \sum_{k = 1}^{\infty}{1 \over \pars{k!}^{2}}
\sum_{\ell = 0}^{n}{n \choose \ell}k^{\ell}
\\[3mm] & = 
1 + \sum_{\ell = 0}^{n}{n \choose \ell}
\sum_{k = 1}^{\infty}{k^{\ell} \over \pars{k}!^{2}}
\end{align}

\begin{equation}
\imp\quad S_{n + 2} =
1 + \sum_{\ell = 0}^{n}{n \choose \ell}S_{\ell}
\,;\qquad n \geq 0\,,\quad
\left\lbrace\begin{array}{rcl}
\ds{S_{0}} & \ds{=} & \ds{\,\mathrm{I}_{0}\pars{2} - 1}
\\[1mm]
\ds{S_{1}} & \ds{=} & \ds{\,\mathrm{I}_{1}\pars{2}}
\end{array}\right.\tag{1}
\end{equation}

$$
\mbox{A few of them,}\quad
\left\lbrace\begin{array}{rcrcr}
\ds{S_{2}} & \ds{=} &
\ds{\mathrm{I}_{0}\pars{2}}&&
\\[1mm]
\ds{S_{3}} & \ds{=} &
\ds{\mathrm{I}_{0}\pars{2}} & \ds{+} & \ds{\mathrm{I}_{1}\pars{2}}
\\[1mm]
\ds{S_{4}} & \ds{=} &
\ds{2\,\mathrm{I}_{0}\pars{2}} & \ds{+} & \ds{2\,\mathrm{I}_{1}\pars{2}}
\\[1mm]
\ds{S_{5}} & \ds{=} &
\ds{5\,\mathrm{I}_{0}\pars{2}} & \ds{+} & \ds{4\,\mathrm{I}_{1}\pars{2}}
\\[1mm]
\ds{S_{6}} & \ds{=} &
\ds{13\,\mathrm{I}_{0}\pars{2}} & \ds{+} & \ds{10\,\mathrm{I}_{1}\pars{2}}
\\[1mm]
\ds{S_{7}} & \ds{=} &
\ds{36\,\mathrm{I}_{0}\pars{2}} & \ds{+} & \ds{29\,\mathrm{I}_{1}\pars{2}}
\\[1mm]
\ds{S_{8}} & \ds{=} &
\ds{109\,\mathrm{I}_{0}\pars{2}} & \ds{+} & \ds{90\,\mathrm{I}_{1}\pars{2}}
\\[1mm]
\ds{S_{9}} & \ds{=} &
\ds{359\,\mathrm{I}_{0}\pars{2}} & \ds{+} & \ds{295\,\mathrm{I}_{1}\pars{2}}
\\[1mm]
\ds{S_{10}} & \ds{=} &
\ds{1266\,\mathrm{I}_{0}\pars{2}} & \ds{+} & \ds{1030\,\mathrm{I}_{1}\pars{2}}
\end{array}\right.
$$
