Solution to $u_{n+1}=u_n/n+u_{n-1}/(n-1)$

Question

What is the solution to the following recurrence relation $$u_{n+1}=\frac{u_n}{n}+\frac{u_{n-1}}{n-1}\ \forall n\geq 2$$ where $u_2=u_1=1$?

Answer 1

Let $a_n=(n-1)!u_n$, and multiply the recurrence by $n!$:

$$\begin{align*} a_{n+1}&=n!u_{n+1}\\ &=\frac{n!}nu_n+\frac{n!}{n-1}u_{n-1}\\ &=(n-1)!u_n+n(n-2)!u_{n-1}\\ &=a_n+na_{n-1}\;, \end{align*}$$

with initial values are $a_1=a_2=1$. If you shift the index by setting $b_n=a_{n+1}$, then $b_0=b_1=1$, and the recurrence $a_{n+1}=a_n+na_{n-1}$ becomes

$$b_n=b_{n-1}+nb_{n-2}\;.$$

This is OEIS A000932, which doesn’t seem to have a simple generating function or closed form, so I think it unlikely that there’s a nice closed form for your sequence. The OEIS entry does have a closed form for the modified sequence in terms of several hypergeometric functions; if that’s good enough for your purposes, you can simply divide it by $n!$, since $b_n=a_{n+1}=n!u_{n+1}$.