<s>[INST] Show that there exist infinitely many $i$ such that $a_i-1$ is divisible by $2^{2015}$ Let $(a_n)$ be a sequence defined by: $a_o=2, a_1=4, a_2=11$ and $\forall n \geq 3$, $$a_n = (n+6)a_{n-1}-3(2n+1)a_{n-2}+9(n-2)a_{n-3}$$
Show that there exist infinitely many $i$ such that $a_i-1$ is divisible by $2^{2015}$
 [/INST] $$a_n = (n+6)a_{n-1}-3(2n+1)a_{n-2}+9(n-2)a_{n-3}$$
can be written as
$$a_n-(n+3)a_{n-1}+3(n-1)a_{n-2}=3\left(a_{n-1}-(n+2)a_{n-2}+3(n-2)a_{n-3}\right)$$
So, we can have
$$a_n-(n+3)a_{n-1}+3(n-1)a_{n-2}=-3^{n-1}.$$
This can be written as
$$a_n-na_{n-1}+n\cdot 3^{n-1}=3\left(a_{n-1}-(n-1)a_{n-2}+(n-1)\cdot 3^{n-2}\right).$$
So, we can have
$$a_n-na_{n-1}+n\cdot 3^{n-1}=3^n.$$
This can be written as
$$a_n-3^n=n(a_{n-1}-3^{n-1}).$$
So, 
$$a_n-3^n=n!,$$
i.e.
$$a_n=3^n+n!.$$
Now $a_n-1=3^n-1+n!$. 
Note here that $3^{2^k}-1$ is divisible by $2^{k+2}$ (you can prove this by induction on $k$). Also, $n!$ is divisible by $i$ at least when $n\ge i$. 
Thus, we know that $a_i-1$ is divisible by $2^{2015}$ at least when $i=2^{2015+m}$ where $m=0,1,2,\cdots$.
Hence, it follows from this that there are infinitely many $i$ such that $a_i-1$ is divisible by $2^{2015}$.
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