I have a power series $F_n(x) = \sum_k \binom{n}{2k} x^k$, which has a closed form of $F_n = \frac12 \left((1 + \sqrt{x})^n + (1 - \sqrt{x})^n\right)$.
$$\begin{align} (1 + \sqrt{x})^n + (1 - \sqrt{x})^n & = \sum_i \binom{n}{i}\left(\sqrt{x}\right)^i + \sum_i \binom{n}{i}\left(-\sqrt{x}\right)^i\\ & = \sum_i \binom{n}{i}\left(\sqrt{x}\right)^i(1^i + (-1)^i)\\ \text{We can discard all terms for odd $i$, and set } k = \frac{i}2\\ & = 2\sum_k \binom{n}{2k}\left(\sqrt{x}\right)^{2k}\\ & = 2\sum_k \binom{n}{2k}x^k\\ \end{align}$$
This closed form is inconvenient for me since it involves square roots of $x$. It seems possible to me that there could be a simple closed form involving imaginary numbers and only integral powers of x, but I haven't been able to crack it.
To make things clear, I'm looking for an equivalent expression that does not involve any fractional powers of $x$, or any iterative notation like $\sum$ or $\prod$. I'd also accept a convincing argument that there isn't such a beast.
If you change you generating function you can have easier generating functions for a very similar sequence.
By example, if you had $F_n(x)=\sum_{k\ge 0}\binom{n}{2k}x^{2k}$ instead of just $x^k$ then you can see that
$$F_n(x)=\frac{g(x)+g(-x)}{2},\quad g(x)=\sum_{k\ge 0}\binom{n}{k}x^k=(1+x)^n\\ [x^k]F_n(x)\leftrightarrow\left\{\binom{n}{0},0,\binom{n}{2},0,\binom{n}{4},0,...\right\}$$
I think that from the original opsgf you cant get easier forms that what you get with $\sqrt x$. Maybe the easier or useful you can get is taking some finite number of the Maclaurin expansion, i.e. the summation, and take care of the level of error you can handle, i.e. taking members to some $\mathcal O(x^k)$ as remainder.
Then, from these finite number of addend you can make a closed expression, a finite polynomial, with no roots involved.
My anterior comment is on the way to simply write $F_n(x)=F_n(y^2)$... but it was a joke more than a serious try :p. It depends, after all, where and for what you are using this opsgf.