Find all $n \in N$ such that $11 | a_n$, with the following sequence defined $\{a_n\}_{n \in \Bbb N}$ with $a_1=1, a_2=3$
and for $n \ge 1$:
$a_{n+2} = (n+3)a_{n+1}-(n+2)a_n$
as the title states, find all $n \in \Bbb N$ such that $11 | a_n$
 A: Edit: I originally computed the table from the incorrect value $a_2=2$. Fortunately, finial caught this, and I’ve recomputed it with the correct initial value, modifying the rest of the hint accordingly.
HINT: If you calculate the first few values of $a_n\bmod 11$, you get this:
$$\begin{array}{r|c}
n:&1&2&3&4&5&6&7&8&9&10&11&12&13\\ \hline
a_n:&1&3&9&0&10&4&6&0&1&0&0&0&0
\end{array}$$
That string of $0$’s should suggest a conjecture: $a_n\equiv 0\pmod{11}$ for $n\ge 10$. Since the $a_n$’s are defined by a recurrence, induction is the natural choice for a method of proof.
A: HINT: If you let $b_n=a_n-a_{n-1}$ you find that $$b_{n+2}=(n+2)b_{n+1}$$
We have $b_2=2$ (the first defined value) and it is then easy to see (and prove) a simple formula for $b_n$.
You still have a little work to do, and the idea of working modulo 11 may be quicker, but this is another way of seeing what is happening.
A: By calculation, we can get that $a_{n}=1!+2!+\cdots+n!$, as $11|k!$ for $k\geq11$,and $11|(9!+10!)$ ($9!+10!=11\cdot9!$), you can only check them for $n=1,2,\cdots,10$, and that $a_{n}$ is divisible by 11 or not for $n\geq11$ follows from that $a_{8}$ (or $a_{10}$) is divisible by 11 or not.
