Function such that $f^{(n)}(0)=\frac{(n!)^2}{(2n+1)!}$ I was trying to solve another problem and come up with the problem if there is a function with closed form such that $$f^{(n)}(0)=\frac{(n!)^2}{(2n+1)!};(n\ge1).$$
I tried to check the condition for compositions of some elementary functions but could not find such function.
Any hints and suggestions would be appreciated.
Thanks!
 A: Notice that if we set:
$$ f(x) = \sum_{n=0}^{+\infty}\frac{n!}{(2n+1)!}x^n $$
and 
$$g(x)=x\cdot f(x^2)=\sum_{n=0}^{+\infty}\frac{n!}{(2n+1)!}x^{2n+1}$$
then
$$ g'(x) = 1+\frac{x}{2}g(x)\tag{1}$$
that is a quite easy ODE to solve. By solving it we get:
$$ g(x) = \int_{0}^{x}\exp\left(\frac{x^2-y^2}{4}\right)\,dx=\sqrt{\pi}e^{\frac{x^2}{4}}\operatorname{Erf}\left(\frac{x}{2}\right),\tag{2} $$
from which:
$$ f(x) = \sqrt{\frac{\pi}{x}} e^{\frac{x}{4}}\operatorname{Erf}\left(\frac{\sqrt{x}}{2}\right).\tag{3}$$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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Given $\ds{\fermi^{\rm\pars{n}}\pars{0}={\pars{n!}^{2} \over \pars{2n + 1}!}\,,
\quad n\ \geq\ 1}$:

\begin{align}\color{#66f}{\large\fermi\pars{x}}&
=\sum_{n\ =\ 0}^{\infty}{\fermi^{\rm\pars{n}}\pars{0} \over n!}\,x^{n}
=\fermi\pars{0}
+\sum_{n\ =\ 1}^{\infty}{1 \over n!}\,{\pars{n!}^{2} \over \pars{2n + 1}!}\,x^{n}
\\[5mm]&=\fermi\pars{0}
+\sum_{n\ =\ 1}^{\infty}{1 \over n!}\,
{\Gamma\pars{n + 1}\Gamma\pars{n + 1} \over \Gamma\pars{2n + 2}}\,x^{n}
=\fermi\pars{0} +\sum_{n\ =\ 1}^{\infty}{x^{n} \over n!}\,
\int_{0}^{1}t^{n}\pars{1 - t}^{n}\,\dd t
\\[5mm]&=\fermi\pars{0} + \int_{0}^{1}
\sum_{n\ =\ 1}^{\infty}{\bracks{xt\pars{1 - t}}^{n} \over n!}\,\dd t
=\fermi\pars{0} + \int_{0}^{1}\braces{\exp\pars{xt\bracks{1 - t}} - 1}\,\dd t
\\[5mm]&=\fermi\pars{0} - 1
+\int_{0}^{1}\exp\pars{xt\bracks{1 - t}}\,\dd t
\\[5mm]&=\color{#66f}{\large\fermi\pars{0} - 1
+\root{\pi}x^{-1/2}\expo{x/4}\,{\rm erf}\pars{\root{x} \over 2}}
\end{align}
