Saudi Arabia Mathematical Competitions 2012

For $k=q$, prime, we have $$a_{qp+1}=p{a}_q-3a_p+13=q{a}_p-3a_q+13,$$ so $(p+3)a_q=(q+3)a_p$. It follows that for every primes $p$ and $q$, $$\frac{a_p}{p+3}=\frac{a_q}{q+3}.(1)$$ Then, since $2011$ is a prime, we obtain $$\frac{a_{2011}}{2014}=\frac{a_7}{10}.(2)$$ On the other hand, $a_7=a_{2\cdot 3+1}=2a_3-3a_2+13$. Hence, using (1), $$\frac{a_7}{10}=\frac{a_3}{6}=\frac{a_2}{5}=\frac{a_7-2a_3+3a_2}{10-12+15}=\frac{13}{13}=1.$$ From (2) we get $a_{2011}=2014$.