Help in solving a recurrence relation

Question

While trying to solve an enumerative combinatorics problem, I reduced the problem to solving the following recurrence relation for $n\geq 2$:

$$x_{n+2}=x_{n+1}+n.x_{n}$$

for all $n\geq 2$, given that $x_2 = x_3=1$. I wanted to know if it is possible to find an explicit solution to this recurrence relation. If not, can we extract any useful information from the equation?

I've tried using the methods used for solving linear recurrence relations with constant coefficients, but they don't seem to work here. Any help/source to read up from will be appreciated.

Answer 1

This is essentially OEIS A000932: their $a_n$ is your $x_{n+2}$. It appears that no nice closed form is known, and even the exponential generating function is rather ugly. The nicest result listed is an asymptotic expression for the ratio of consecutive terms.

Answer 2

Hint:

Let $x(n)=\int_a^be^{ns}K(s)~ds$ ,

Then $\int_a^be^{(n+2)s}K(s)~ds-\int_a^be^{(n+1)s}K(s)~ds-n\int_a^be^{ns}K(s)~ds=0$

$\int_a^be^{2s}e^{ns}K(s)~ds-\int_a^be^se^{ns}K(s)~ds-\int_a^bK(s)~d(e^{ns})=0$

$\int_a^be^{2s}e^{ns}K(s)~ds-\int_a^be^se^{ns}K(s)~ds-[e^{ns}K(s)]_a^b+\int_a^be^{ns}~d(K(s))=0$

$\int_a^be^{2s}e^{ns}K(s)~ds-\int_a^be^se^{ns}K(s)~ds-[e^{ns}K(s)]_a^b+\int_a^be^{ns}K'(s)~ds=0$

$\int_a^be^{2s}e^{ns}K(s)~ds-\int_a^be^se^{ns}K(s)~ds-[e^{ns}K(s)]_a^b+\int_a^b(K'(s)+(e^{2s}-e^s)K(s))e^{ns}~ds=0$

$\therefore K'(s)+(e^{2s}-e^s)K(s)=0$

$\dfrac{K'(s)}{K(s)}=e^s-e^{2s}$