I'm trying to solve the Hermite's differential equation
\begin{equation*} y''(x)-2xy'(x)+2ny(x) = 0 \end{equation*}
I look for power series solutions of the form
\begin{equation} y(x) = \displaystyle\sum_{\nu=0}^\infty a_\nu x^\nu \end{equation}
Then I found the recurrence,
\begin{equation} a_{\nu+2} = \frac{2(\nu-n)}{(\nu+2)(\nu+1)} a_\nu \label{eq:anHer} \end{equation} for $\nu = 0,...,n$.
So i am looking for even and odd powers of x solutions. For the even solutions I have
\begin{equation} a_{2k} = \frac{4\left(k-1-\frac{n}{2}\right)}{2k(2k-1)}a_{2(k-1)} \end{equation} for $k = 1,2,...$
In terms of $a_0$ I have, \begin{align*} &a_2=\frac{4(-\frac{n}{2})}{2\cdot1}a_0\\ &a_4 =\frac{4\left(1-\frac{n}{2}\right)}{4\cdot3}a_2 = \frac{4^2\left(1-\frac{n}{2}\right)\left(-\frac{n}{2}\right)}{4\cdot3\cdot2\cdot1}a_0 \\ & a_6 = \frac{4\left(2-\frac{n}{2}\right)}{6\cdot5}a_4 =\frac{4^3\left(2-\frac{n}{2}\right)\left(1-\frac{n}{2}\right)\left(-\frac{n}{2}\right)}{6\cdot5\cdot4\cdot3\cdot2\cdot1}a_0 \\ \vdots\\ &a_{2k} = \frac{2^{2k}(-\frac{n}{2})_k}{(2k)!}a_0 \end{align*} where $(x)_\mu$ is the Pochhammer symbol.
Then the even solutions are \begin{equation} y_{even}(x) = \displaystyle\sum_{k = 0}^\infty a_{2k}x^{2k} = a_0\left(1+ \displaystyle\sum_{k=1}^\infty \frac{2^{2k}(-\frac{n}{2})_k}{(2k)!}x^{2k}\right) \end{equation} in https://mathworld.wolfram.com/HermitePolynomial.html the even solutions are given by \begin{equation} y_{2n}(x) = c_0 \left(1+\displaystyle\sum_{k=1}^n\frac{2^{2k}(-n)_k}{(2k)!}x^{2k}\right) \end{equation}
I don't see why I have $(-\frac{n}{2})_k$ instead of $(-n)_k$. Since the series I've obtained terminates for $k = 1+\frac{n}{2}$ maybe I can truncate the series and write my series in this form but I'm not sure. Or maybe is just because the normalization of $c_0$ given in the previous link, which is $c_0= (-1)^n2^n(2n-1)!!$, while my $a_0$ is a general normalization constant without a particular expression so is a ''more general'' solution.