I have found this limit in https://oeis.org/A019609 and I was wondering how to prove it (if it is actually correct): $$\lim_{n\to\infty} \frac{4n}{a^2_n}=\pi e$$ where $$a_1=0,a_2=1, a_{n+2}=a_{n+1}+\frac{a_n}{2n}.$$

By computer evaluation, it is correct for $2$ digits after decimal point at about $n\approx 24100$, so if it is correct, it converges really slow.

I've attempted to prove this by first considering generating function $f(x)=\sum_{n \geq 1}a_nx^n$ and then trying to get asymptotics of its coefficients. By using recurrence, we get $f(x)/x^2-1=f(x)/x+\sum \frac{a_n}{2n}x^n$, and after differentiation we get differential equation which solves to $$f(x)=\frac{e^{-x/2}x^2}{(1-x)^{3/2}}.$$ Now I think this is a step away from getting asymptotics of $a_n$, but I don't know how. Can anybody show how to finish this? Or maybe there is another way?

Also, I don't think it is useful, but here is at least closed form obtained from the $f(x)$ using binomial series and exponential function series: $$ a_n=\sum_{i=0}^{n-2}\frac{(-1)^n}{2^i i!}\binom{-3/2}{n-i-2}. $$

Closest to this question seems to be Mirror algorithm for computing $\pi$ and $e$ - does it hint on some connection between them?, where there are two sequences approaching $\pi$ and $e$ and solutions seem to use same approach using generating functions, so this seems to be on the right track.