Stuck in solving the Hermite DE? As a part of solving the Schrödinger equation1 for quantum Harmonic oscillator:
\begin{equation}\tag{1}
 \left(-\frac{d^2}{d\tilde{x}^2} + \tilde{x}^2\right) \tilde{\psi}(\tilde{x}) = \tilde{E} \tilde{\psi}(\tilde{x}) \iff \frac{d^2 \tilde{\psi}(\tilde{x}) }{d\tilde{x}^2} = (\tilde{x}^2 - \tilde{E}) \tilde{\psi}(\tilde{x}) 
\end{equation}
I'm looking for a solution of the form:
\begin{equation}
 \tilde{\psi}(\tilde{x}) = \tilde{x}^{k} e^{-\frac{\tilde{x}^2}{2}} = h(\tilde{x})e^{-\frac{\tilde{x}^2}{2}}
\end{equation}
which, when substituted in (1) leads to the Hermite differential equation:
\begin{equation}
 \frac{d^{2}h}{d\tilde{x}^2} - 2\tilde{x}\frac{dh}{d\tilde{x}} + (\tilde{E} - 1)h = 0
\end{equation}
I'm trying to construct a power series solution $h = \sum_{k = 0}^{\infty} a_{k} \tilde{x}^{k}$. Substituting back in the equation:
\begin{equation}
 \sum_{k = 2}^{\infty} k (k - 1) a_{k} \tilde{x}^{k-2} - 2\tilde{x}\sum_{k = 1}^{\infty} k a_{k} \tilde{x}^{k-1} +  (\tilde{E} - 1)\sum_{k = 0}^{\infty} a_{k} \tilde{x}^{k} = 0
\end{equation}
After a shift on the first sum and peeling of the initial term of the first and the last sum:
\begin{equation}
 2a_{0} + \sum_{k = 1}^{\infty} (k + 2)(k + 1) a_{k+2} \tilde{x}^{k} - 2\sum_{k = 1}^{\infty} k a_{k} \tilde{x}^{k} + a_{0}(\tilde{E} - 1) + (\tilde{E} - 1)\sum_{k = 1}^{\infty} a_{k} \tilde{x}^{k} = 0
\end{equation}
Powers and indexes match, so:
\begin{equation}
 2a_{0} + a_{0}(\tilde{E} - 1) + \sum_{k = 1}^{\infty}[(k + 2)(k + 1) a_{k+2} - (2k - \tilde{E} + 1) a_{k} ]\tilde{x}^{k} = 0
\end{equation}
Finally, all the coefficients must be equal to zero, so:  
\begin{equation}
a_{k+2} = \frac{(2k - \tilde{E} + 1)}{(k + 2)(k + 1)} a_{k}
\end{equation}
First (smaller) problem.
From the above, $\tilde{E} = 2k + 1$. But from the peeled of terms $a_{0}(\tilde{E} + 1) = 0$. What should I do with the second relation? 
Second (main) problem.
I expect that by solving the recurrence relation I should somehow deduce that the solution, $h(\tilde{x})$, are the Hermite polynomials, however, I have no idea how to do that. Could you please suggest how to do that? 

1.The unit-free time-independent 1D.
 A: First of all I can only answer your first minor question for sure. Nevertheless I can provide  argumentation for your major problem but it does not involve an explicit solution for the recurrence relation.

You are right about arguing to peal out the initial terms of the first and the third but you made a slight mistake concerning the first sum. Hence your sums look like
$$\sum_{k=0}^{\infty}(k+1)(k+2)a_{k+2}\tilde{x}^k-2\sum_{k=1}^{\infty}ka_k\tilde{x}^k+(\tilde{E}-1)\sum_{k=0}^{\infty}a_k\tilde{x}^k=0$$
the initial term of the first sum is given by $2a_{\color{red}{2}}$ since we are dealing with the subscript $a_{k+2}$ instead of $a_k$ in connection with the first sum. So your condition would be
$$a_2=\frac{(1-\tilde{E})}{2}a_0$$ 
which is simply the same as you would get by writing down recurrence relation in general.

I was only able to deduce the recurrence relation back to two different terms concerning odd and even indices. For them I got
$$\begin{align}
a_{2k}&=\frac{(4k-3-\tilde{E})(4k-7-\tilde{E})\cdots(1-\tilde{E})}{(2k)!}a_0\\
a_{2k+1}&=\frac{(4k-1-\tilde{E})(4k-5-\tilde{E})\cdots(3-\tilde{E})}{(2k+1)!}a_1
\end{align}$$
but similiar to you I have no idea how to show that this relaions equal the Hermite Polynomials. However I guess you can dodge solving these relations by just looking at the DE which define the Hermite Polynomials. The general (physicists') Hermite DE is given by
$$u''-2xu'+2nu=0$$
and the solutions are the Hermite Polynomials of the kind $H_n(x)$. Setting $n=\frac{(\tilde{E}-1)}2$ (and $u=h(\tilde{x})$) leads to your DE
$$h''(\tilde{x})-2\tilde{x}h'(\tilde{x})+(\tilde{E}-1)h(\tilde{x})=0$$
and therefore the solutions of this equation are Hermite Polynomials of $\frac{(\tilde{E}-1)}2$th-order. WolframAlpha agrees on this by providing the solution
$$h(x)=c_1 H_{\frac{\tilde{E}-1}2}(x)+c_2~_1F_1\left(\frac{1-\tilde{E}}4;\frac12;x^2\right)$$
where $_1F_1$ is the confluent hypergeometric function and the defined function is closely related to the Hermite Polynomials.

While working on the solution defined by $_1F_1$ I figured out a way how to deduce this function back to the recurrence relation given for $2k$. Consider the Confluent Hypergeometric Function $_1F_1(a;b;x)$ defined as power series
$$_1F_1(a;b;x)=\sum_{k=0}^{\infty}\frac{(a)_k}{(b)_k}\frac{x^k}{k!}$$
where $(a)_k, (b)_k$ denotes the Pochhammer Symbol. Plugging in $a=\frac{1-\tilde{E}}{4}, b=\frac12$ and $x=x^2$ yields to
$$\begin{align}
_1F_1\left(\frac{1-\tilde{E}}{4};\frac12;x^2\right)=\sum_{k=0}^{\infty}\frac{\left(\frac{1-\tilde{E}}4\right)_k}{\left(\frac12\right)_k}\frac{x^{2k}}{k!}
\end{align}$$
The occuring fraction can be simplified by using the definition of the Pochhammer Symbol and the double factorial
$$\begin{align}
\frac{\left(\frac{1-\tilde{E}}4\right)_k}{\left(\frac12\right)_k k!}=\frac{\frac{(1-\tilde{E})\cdots(4k-7-\tilde{E})(4k-3-\tilde{E})}{4^k}}{\frac{(2k-1)!!}{2^k}k!}&=\frac{(1-\tilde{E})\cdots(4k-7-\tilde{E})(4k-3-\tilde{E})}{2^k~k!~(2k-1)!!}\\
&=\frac{(1-\tilde{E})\cdots(4k-7-\tilde{E})(4k-3-\tilde{E})}{(2k)!}=a_{2k}
\end{align}$$
Hence this way can be gone backwards aswell I guess this is a possibility to show the connection of the power series solution of the DE to the Hermite Polynomials. Howsoever I am still not sure about the odd coefficients.
