As the title suggests, I am looking for the solutions to the recurrence relation
$a(n+3) = B \frac{a(n)}{n^2}$.
In particular, I am attempting to solve a differential equation using the power series method and came across this formula for the coefficients. If you're curious, the differential equation is something like
$\frac{y'(x)}{x^2} - \frac{y''(x)}{x} - By(x) = 0. $
I was wondering if this were solvable in closed-form. Since I'm not an expert on recurrence relations, I didn't really know where to start (though the internet tells me nonlinear recurrence relations are usually not easy to solve).
Nevertheless, I note that if one defines $b(n) \equiv 1/\Gamma(n)$, one gets the relation $b(n+2) = \frac{b(n)}{n^2}$. This gives me hope it's solvable in terms of (scaled/shifted) gamma functions and exponentials.
Any advice?
Note: Not homework or anything, I'm just playing around with trying to find functions orthogonal to each other given the weight function w(x) = 1/x.
Check again your Gamma function understanding. If you set $$ b_n=Γ(n/3)^2a_n, $$ then $$ b_{n+3}=Γ(n/3+1)^2a_{n+3}=(n/3)^2Γ(n/3)^2\frac{Ba_n}{n^2}=\frac{B}{9}b_n $$ which is a simple geometric sequence for the sub-sequences of every third sequence element.