, 2012

For $k = q$, prime, we have
$$a_{qp+1} = pa_q - 3a_p + 13 = qa_p - 3a_q + 13,$$
so $(p+3)a_q = (q+3)a_p$. It follows that for every primes $p$ and $q$,
$$\frac{a_p}{p+3} = \frac{a_q}{q+3}. \qquad (1)$$
Then, since $2011$ is a prime, we obtain
$$\frac{a_{2011}}{2014} = \frac{a_7}{10}. \qquad (2)$$
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 $a_{2011} = 2014$.