Let's have Hermite polynomials: $$ e^{2tx - t^{2}} = \sum_{n = 0}^{\infty}H_{n}(x)\frac{t^{n}}{n!}. \qquad (1) $$ I need to write it in terms of confluent hypergeometric Kummer function for index $n = 2k$. The result is $$ H_{2k}(x) = (-1)^{k}\frac{(2k)!}{k!}F_{1}\left(-n, \frac{1}{2}, x^{2}\right). $$ I tried to prove it by the following way. First, I expanded polynomials in Taylor series near zero: $$ H_{n}(x) = \sum_{m}H^{(m)}_{n}(0)\frac{x^{n}}{n!}, \quad H_{n}^{(m)}(0) = \frac{dH_{n}}{dx} \quad at \quad x = 0. \qquad (2) $$ Then I used $(1)$: $$ \partial_{x}^{m}G(x, t)_{x = 0} = (2x)^{m}e^{-t^{2}} = \sum_{k = 0}^{\infty}\frac{(-1)^{k}2^{m}t^{2k + m}}{k!} = \sum_{n = 0}^{\infty}H_{n}^{m}(0)\frac{t^{n}}{n!} \Rightarrow |n = 2k + m| \Rightarrow $$ $$ H_{n}^{m}(0) = \frac{n!(-1)^{\frac{n - m}{2}}2^{m}}{\left( \frac{n - m}{2}\right)!}, \quad m \leqslant n , $$ and $$ H_{n}^{m}(0) , \quad m \geqslant n. $$ Then by returning to $(2)$ I got $$ H_{n}(x) = \sum_{m = 0}^{n}\frac{n!(-1)^{\frac{n - m}{2}}2^{m}}{\left( \frac{n - m}{2}\right)!} \frac{t^{m}}{m!} = n! (-1)^{\frac{n}{2}}\sum_{m = 0}^{n}\frac{(-1)^{-\frac{ m}{2}}2^{m}}{\left( \frac{n - m}{2}\right)!} \frac{t^{m}}{m!}. $$ When $n = 2l$ sum $2k + m = 2l$ will be true only if $m = 2f$, and $$ H_{2l}(x) = (2l)!(-1)^{l}\sum_{f = 0}^{l}\frac{2^{f}(-1)^{f}}{(l - f)!}\frac{x^{2f}}{f!}. $$ But I can't reduce it to hypergeometric function, because I have problem with $2^{f}$.
Where is the mistake?
Edit. It seems that I found it. I represented $(2f)!$ as $2^{f}f!$, which isn't correct.
The second edit. The question is closed. I represented $(2f)!$ as $(2f - 1)!! 2^{f}f!$, which then lead me to the answer.