On Wikipedia it's said that one could express the Hermite polynomials using the $_1F_1$ function and the following formulae are provided:
$$H_{2n}(x)=(-1)^n \frac{(2n)!}{n!}(_1F_1)(-n,1/2;x^2),$$
$$H_{2n+1}(x)=(-1)^n \frac{(2n+1)!}{n!}2(_1F_1)(-n,3/2;x^2),$$
Even though I've made quite an extensive search online I haven't been able to find some detailed proof of the latter.
If you have some idea on how to prove it, or you know some source where an explanation is given I would be thankful.
Thanks in advance!
Best Answer
The equations you have provided are equivalent to the Wikipedia ones, when a slight typo is fixed. $$H_{2n}(x)=(-1)^n \frac{(2n)!}{n!}(_1F_1)(-n,1/2;x^2), \tag{1} $$ $$H_{2n+1}(x)=(-1)^n \frac{(2n+1)!}{n!}2x(_1F_1)(-n,3/2,x^2). \tag{2}$$ Notice the $\,2x\,$ instead of $\,2\,$ in equation $(2)$ which is the fix. I used the Wolfram language code
which evaluates to
They give the same results except the Wikipedia single case sum gives the terms in reverse order.