Discretized normal distribution The normal distribution $e^{-x^2/2}$ satisfies the ODE y'=-xy on the real line.
Is there a non trivial, non-negative solution to the discretized version of this ODE $u(n+1)-u(n-1)=-2nu(n)$.
 A: First, any solution $u$ satisfies $u(-n)=u(n)$ for all $n$ — because both $u(n)$ and $u(-n)$ satisfy the same recurrence, $u(0)=u(-0)$ and $u(1)=u(-1)\impliedby u(1)-u(-1)=-2\cdot 0\cdot u(0).$
Next, the function
$$U(x)=\sum_{n=0}^{\infty}u(n)\frac{x^n}{n!},\qquad |x|<\frac12,$$
satisfies $(1+2x)U''(x)+2U'(x)-U(x)=0$ which is solved by
$$U(x)=C_1 I_0(\sqrt{1+2x})+C_2 K_0(\sqrt{1+2x})$$
where $I_0$ and $K_0$ are modified Bessel functions (and $C_1, C_2$ are constants).
This gives general solutions of the recurrence in question. Further, as we have
$$I_0(\sqrt{1+2x})=\sum_{k=0}^{\infty}\frac{1}{k!^2}\Big(\frac{1+2x}{4}\Big)^k\\=\sum_{k=0}^{\infty}\frac{2^{-2k}}{k!^2}\sum_{n=0}^{k}\frac{k!(2x)^n}{n!(k-n)!}\\=\sum_{n=0}^{\infty}\frac{(2x)^n}{n!}\sum_{k=n}^{\infty}\frac{2^{-2k}}{k!(k-n)!}\\=\sum_{n=0}^{\infty}\frac{1}{n!}\Big(\frac{x}{2}\Big)^n\sum_{k=0}^{\infty}\frac{2^{-2k}}{k!(n+k)!},$$
the choice $C_1=1, C_2=0$ gives a nonnegative solution:
$$u(n)=\sum_{k=0}^{\infty}\frac{2^{-|n|-2k}}{k!(|n|+k)!}.$$
This is just $I_n(1)$; actually, the general solution can be seen as
$$u(n)=C_1 I_n(1) + C_2 (-1)^n K_n(1).$$
Note that $I_n(1)$ decays (and $K_n(1)$ grows) when $n\to\infty$.
A: Edit : in fact, the sequence one gets through the given second order recurrence relationship is, whatever $u_0$ and $u_1$, most often oscillating and rapidly divergent.
I propose to take instead the following discretization with a non-symmetrical derivative (with a step $\Delta x=1$) :
$$-n u_n=u_n-u_{n-1} \ \ \ \text{with} \ u_0=1\tag{1}$$
giving the first order (instead of second order) recurrence relationship  :
$$u_n=\tfrac{1}{1+n}u_{n-1}\tag{2}$$
One gets in this way a decreasing sequence tending to $0$ from above,  already "in the same spirit" as Laplace-Gauss. 
But a nice surprise comes, when one takes a much smaller step, for example $\Delta x = 0.01$, instead of step $\Delta x = 1$, i.e. by replacing (2) by :
$$u_n=\tfrac{1}{1+n/100}u_{n-1}\tag{3}$$
When one plots $u_n$ as a function of variable $n$, one obtains the following "curve", very close to the (right part of) Laplace-Gauss curve : 

Fig. 1 : The 21 first terms of recurrent sequence defined by (3) $u_0=1, u_1=\tfrac{100}{101}, \cdots$ represented as a curve.
Remark : One should make a study of the "most adequate" $\Delta x$. We will not do it because we are already far enough from the initial question of the OP.  

Previous "answer", merely a comment now.
For sequences, It is often advisable to take a look at the OEIS Encyclopedia of sequences. But in this case, I have only found results for the cousin sequence A001053 : https://oeis.org/A001053.
$a(n+1) = n*a(n) + a(n-1)$ with $a(0)=1, a(1)=0,$
mentionning solutions with modified Bessels, etc. (as some of those found by @Metamorphy and explained in his very interesting answer).
