Exponential Generating Function Transformation Say we have an exponential generating function:
$$F(x) = \sum_{n\geq 0} f_n \frac{x^n}{n!}.$$
Is there a simple transformation from $F(x)$ to $G(x)$ where 
$$G(x) = \sum_{n\geq 0} f_n \frac{x^{2n}}{(2n)!}?$$
 A: If $F$ is exponentially bounded and entire, the Laplace transforms of $F(x)$ and $G(x)$ are 
$$ \mathcal L F(s) = \sum_{n=0}^\infty \int_0^\infty f_n \frac{x^n}{n!} e^{-sx}\; dx  = \frac{1}{s} \sum_{n=0}^\infty f_n s^{-n} $$
and
$$ \mathcal L G(s) = \sum_{n=0}^\infty \int_0^\infty f_n \frac{x^{2n}}{(2n)!} e^{-sx}\; dx  = \frac{1}{s} \sum_{n=0}^\infty f_n s^{-2n} = s \mathcal L F(s^2) $$
for $\text{Re}(s)$ sufficiently large.
That is, $G = \mathcal L^{-1} (s \mathcal L F(s^2))$.
A: I should point out that yes in fact there is a transformation for this if you're willing to accept an integral of the original sequence generating function as a solution! I actually proved this result to find integral formulas for a special case in this article (2017). First, we notice that 
$$\sum_{n \geq 0} \frac{f_{2n}}{(2n)!} z^{2n} = \frac{1}{2}(F(z) + F(-z)).$$ 
Next, we can expand the single factorial function via Gauss' multiplication formula for the gamma function as 
$$\frac{1}{n!} = \frac{4^n \Gamma(n+1/2)}{\sqrt{\pi} \cdot (2n)!} = \frac{2^{n+1}}{\sqrt{2\pi} \cdot (2n)!} \int_0^{\infty} t^{2n} e^{-t^2/2} dt.$$ 
Now we need only apply this integral termwise to the transformed generating function above:
$$G(z) = \frac{1}{\sqrt{2\pi}} \int_0^{\infty} \left[F\left(\sqrt{2} tz\right) + F\left(-\sqrt{2} tz\right)\right]  e^{-t^2/2} dt.$$ 
This method is also given in a more concrete example in this Wikipedia article on generating function transformations. 
Enjoy!
A: Let's try an example ... $f_n = n!$
\begin{align}
F(n) &= \sum_{n=0}^\infty \frac{n!\;x^n}{n!}= \sum_{n=0}^\infty x^n = \frac{1}{1-x}
\\
G(x) &= \sum_{n=0}^\infty \frac{n! \;x^{2n}}{(2n)!} = 
1+\frac{x\sqrt {\pi }\;{\rm erf} \left(x/2\right){{e}^{{x}^{2}/4}}}{2}
\end{align}
Is there a simple transformation from $F(x)$ to $G(x)$ in this case?  I doubt it.
