Context:
I've derived these two series that both converge to the same value:
\begin{align} f:&=\sum_{n=-\infty}^{\infty} \mathrm{e}^{-\pi n^2} \sum_{k=0}^{\infty}\frac{(16)^{-k}\,H_{4k}\,(n\,\sqrt{2\pi})}{(2k+1)!} =\frac{\pi}{12}+\frac74 \\ \\ g:&=\sum_{n=-\infty}^{\infty} \mathrm{e}^{\pi n^2} \left(\sqrt{2}\,\mathrm{E}_{\frac32}\left(\pi\, n^2\right)-\mathrm{E}_{\frac32}\left(2\pi\, n^2\right) \right)=\frac{\pi}{12}+\frac74 \end{align}
where $\mathrm{E}_a(b)$ is the generalized Exponential Integral .
It is nice to see how the exponentials, with either a negative or a positive power, both get 'tamed' towards the same value.
Question:
Does a closed form expression also exist for the infinite sum of Hermite polynomials?
$$\sum_{k=0}^{\infty}\frac{(16)^{-k}\,H_{4k}\,(n\,\sqrt{2\pi})}{(2k+1)!}$$
Checked the 15 summation formulae on this page, however they all seem to rely on the requirement that the factorial and the first parameter of $H$ have the same structure. In my case these are different, hence I expect some additional trick is required.
Thanks.