$$f\left(x\right)=\sum_{n=0}^{\infty}\left(\frac{x^{2n}\left(n+1\right)^{x}\ }{n!}\right)$$ This is a self made problem I came across while researching something, so I do not have high hopes for it having a closed form.
First of all, Wolfram Does Not give a Closed Form for the Function.
But it does give the function value for values of $x$ as follows:
$$f(-2)=\frac{\text{Ei}(4)-\gamma-\ln4}{4}$$
$$f(-1)=e-1$$
$$f(0)=1$$
$$f(1)=2e$$
$$f(2)=29e^4$$
$$f(3)=1279e^9$$
$$f(4)=113137e^{16}$$
Now, for $f(n)$, $n\in \mathrm N$:
$$f(n)=a_ne^{n^2}$$
The Sequence "$2, 29, 1279, 113137$" led to no results on OEIS.
EDIT:
Using Binomial Expansion:
$$f(x)=\sum_{r=0}^{x}\frac{x!}{\left(x-r\right)!r!}\sum_{n=0}^{\infty}\frac{x^{2n}n^{r}}{n!}$$
Now you could say the Problem Comes Down to Expanding $\frac{n^r}{n!}$:
Example:
$$\frac{n^4}{n!}=\frac{1}{\left(n-4\right)!}+\frac{6}{\left(n-3\right)!}+\frac{7}{\left(n-2\right)!}+\frac{1}{\left(n-1\right)!}$$
But then, although I can do this by hand, how do I write it as a function of $r$.
Second Edit:
Using @SangchulLee's help, I was able to deduce this form for natural numbers:
$$f(n)=\left(\sum_{r=0}^{n}\binom{n}{r}\text{T}_r\left(n^2\right)\right)e^{n^{2}}$$
where, $\text{T}_r(x)$ are Touchard Polynomials.
Then after browsing through I found a wonderful property:
$$\frac{T_{n+1}\left(x\right)}{x}=\sum_{k=0}^{n}\binom{n}{k}T_{k}\left(x\right)$$
This leads to the closed form in terms of Touchard Polynomial:
$$f(n)=\frac{\text{T}_{n+1}\left(n^2\right)}{n^{2}}e^{n^2}$$
$$\color{green}{f(x)=\frac{\left(x+1\right)!}{\pi x^{2}}\int_{0}^{\pi}\left(e^{x^{2}\left(e^{\cos\left(t\right)}\cos\left(\sin\left(t\right)\right)\right)}\cos\left(x^{2}e^{\cos\left(t\right)}\sin\left(\sin t\right)-\left(x+1\right)t\right)\right)dt}$$
Well this seems to be the closed form.
This needs verification though.
