Closed form for Fibonacci-like recurrence with $\frac{2}{n}$ coefficient

Question

I came across the following recurrence relation, and was wondering if there might be some way to get a closed form:

$F\left(0\right)=0$

$F\left(1\right)=1$

$F\left(n\right)=F\left(n-1\right)+\left(\frac{2}{n}\right)F\left(n-2\right)$

Answer 1

The DE associated with the recurrence relation $$ n a_n = n a_{n-1} + 2 a_{n-2} \tag{1} $$ is given by $$ x f' = xf+x^2 f'+2x^2 f \tag{2}$$ that is separable. By solving it we get that our coefficients depend on the coefficient of the Taylor series of $$g(x)=\frac{e^{-2x}}{(1-x)^3}\tag{3}$$ in a neighbourhood of zero, that can be computed by convolution. By stars and bars and the fact that $g(x)$ has a triple pole at $x=1$ we have that the sequence $\{a_n\}$ behaves like a second-degree polynomial in $n$ plus a small perturbation, so it is way different from the Fibonacci sequence.