, 2011

Consider the sequence of integers $(b_n)$ determined by
$$b_0 = 1,\ b_1 = -1 \text{ and } b_n = 6b_{n-1} + 2016b_{n-2} \text{ for all } n \ge 2.$$
Clearly, for all $n \ge 0$, we have $a_n \equiv b_n \pmod{2011}$. (*)
The characteristic equation of sequence $(b_n)$: $x^2 - 6x - 2016 = 0$, or $(x - 48)(x + 42) = 0$.
Consequently, the general term of $(b_n)$ has the form: $b_n = C_1(-42)^n + C_2 48^n$.
By the initial conditions for sequence $(b_n)$, we obtain
$$\begin{cases} C_1 + C_2 = 1 \\ 42C_1 - 48C_2 = 1. \end{cases}$$
Hence $C_1 = \frac{49}{90}$ and $C_2 = \frac{41}{90}$. Thus $b_n = \frac{49(-42)^n + 41 \cdot 48^n}{90} \quad \forall n \ge 0$.
Since $2011$ is a prime, according to the little Fermat theorem we have:
$$(-42)^{2010} \equiv 48^{2010} \equiv 1 \pmod{2011}.$$
Hence $90b_{2012} = 49 \cdot (-42)^{2012} + 41 \cdot 48^{2012} = 49 \cdot (-42)^2 + 41 \cdot 48^2 = 90b_2 \pmod{2011}$.
Consequently $b_{2012} \equiv b_2 \pmod{2011}$ (since $(90, 2011) = 1$).
But $b_2 = 6b_1 + 2016b_0 = 2010$, hence $b_{2012} = 2010 \pmod{2011}$.
Thus $a_{2012} = 2010 \pmod{2011}$ (by means of (*)).