New Hermite polynomial identity? I am trying to prove an identity involving Hermite polynomials using other identities from Wikipedia, but I can't find the way. I have checked the identity in Mathematica for many values of $n$ and it holds for all values of $n$ I have tried. The identity is
$$ \frac{1}{n!}  \Big( \text{He}_n(x) \Big)^2 = \sum_{k=0}^n {n\choose k} \frac{1}{k!} \, \text{He}_{2k} (x) \, ,$$
where $\text{He}_n(x)$ is the probabilists' Hermite polynomial
$$\text{He}_n(x) = (-1)^n e^{\frac{x^2}{2}} \frac{d^n}{dx^n} e^{-\frac{x^2}{2}} \, .$$
Any ideas?
 A: This proof uses operator methods.  From the wiki page we know
$$(1)\quad \operatorname{He}_n(x) =\exp{\big(-\frac{1}{2}\frac{d^2}{dx^2}\big)} \  x^n $$
and also from Mehler's formula
$$(2) \quad \frac{\big(\operatorname{He}_n(x)\big)^2}{n!} = [u^n] \frac{1}{\sqrt{(1-u^2)}} \exp{\big(\frac{u}{1+u} x^2\big)}$$
where $[u^n]$ is the "coefficient of" operator.  We need one more lemma.  (Someone else has done this before me, but I don't have a reference.)
$$ (3) \quad \exp{\big(a \frac{d^2}{dx^2}\big)} \exp{\big(b\ x^2\big)} = 
\frac{1}{\sqrt{1-4\ a\ b}}\exp{\big(\frac{b\ x^2}{1-4a\ b}\big)} $$
Begin proof of lemma: Use well-known Gaussian formula
$$ \exp{(a\ t^2)} = \frac{1}{2\sqrt{a\pi}} \int_{-\infty}^\infty du \exp{(t\ u)} \exp{\big(\frac{-u^2}{4a}\big)}$$
Substitute $t=\frac{d}{dx}.$ Use operational form of Taylor series $\exp{(u d/dx)}f(x) = f(x+u).$  Then the left-hand side of (3) becomes
$$ \frac{1}{2\sqrt{a\pi}} \int_{-\infty}^\infty du \exp{\big(\frac{-u^2}{4a}\big)} \exp{(b(x+u)^2)} $$
Use the penultimate equation again, and algebra.
End proof of lemma:
By eqs. (1) & (2) the OP's formula is equivalent to showing
$$\exp{\big(-\frac{1}{2}\frac{d^2}{dx^2}\big)} \sum_{k=0}^n \binom{n}{k} \frac{x^{2k}}{k!} = [u^n] \frac{1}{\sqrt{(1-u^2)}} \exp{\big(\frac{u}{1+u} x^2\big)} $$
which is equivalent to showing
$$ L_n(-x^2) = \exp{\big(\frac{1}{2}\frac{d^2}{dx^2}\big)}
 [u^n] \frac{1}{\sqrt{(1-u^2)}} \exp{\big(\frac{u}{1+u} x^2\big)}  $$
where I've used the well-known polynomial expression for Laguerre polynomials.  Use eq (3) on the right-hand side of the previous equation.  With algebra it is seen that this is equivalent to
$$ L_n(-x^2) =  [u^n] \frac{1}{1-u} \exp{\big(\frac{u\ x^2}{1-u} \big) } $$
This is a well-known formula; see Gradshteyn & Ryzhik 8.975.1
