Recursion equation and finding odd prime factor I was randomly browsing on a site which contains questions from various topics of science and mathematics and I found this problem:-

Then I tried to solve it but could only proceed till the general solution of the sequence which is,
$a_{n}=\frac{1}{2}\{(2+\sqrt{3})^n+(2-\sqrt{3})^n\}$,Is there any way to lead from here.
Thanks.
 A: The sequence $a_n =\{1,2,7,26,97\cdots\}$ is $A001075$ in the OEIS.   

For each $n$, $a_0$ divides $a_n$, $a_1$ divides $a_{2n+1}$, $a_2$ divides $a_{4n+2}$, $a_3$ divides $a_{6n+3}$ and so on. In general, $a_{k}$ divides $a_{(2n+1)k}$.You can find a proof of this statement here in problem $A2$.  But, I will give a small outline below.   
Proof: You have proceeded correctly that  $a_n =\frac{(2+\sqrt{3})^n +(2-\sqrt{3})^n}{2} = \frac{a^n + b^n}{2} \forall n$ where $a = 2+\sqrt{3}$ and $b =2-\sqrt{3}$. If $k$ is odd then $$\frac{a_{kn}}{a_n} = \frac{a^{kn} + b^{kn}}{a^n + b^n} = a^{(k-1)n}-a^{(k-2)n}b^n \cdots + b^{(k-1)n}$$ This expression is both rational (because $a_n$ and $a_{kn}$ are integers) and of the form $\alpha +\beta\sqrt{3}$ for some integers $\alpha, \beta$ by the expressions for $a,b$; it follows that it must be an integer, and so $a_{kn}$ is divisible by $a_n$.
In our problem, we can write $a_{2015} = a_{(2\times 201+1)5}$. Thus, $a_5$ divides $a_{2015}$. As we know, $a_{5} = 362$, the largest prime factor that divides $a_{2015}$ is thus $181$. Hope it helps.
