$$a_1 = 1, \quad a_n = 1 + k + k^2 + \dots + k^{n-1}, \quad n = 2, \ldots, 2016.$$
We have to find such $k$, for which $a_n$ is not divisible by $2016$.

Let us suggest that $(k, 2016) = d > 1$. As we see, every coordinate after the jump has residue $1$ modulo $d$ thus cannot be divisible by $2016$. So all such $k$ are the answers.

If $(k, 2016) = 1$, let us look at all the residues of $a_n$ modulo $2016$. If at least one value has residue $0$ then $k$ does not satisfy the condition. Let us suggest that there are no zeroes, then at least two residues are the same. Let them be $a_m$ and $a_l$, $m > l$. Their difference $a_m - a_l$ is divisible by $2016$, but then
$$a_m - a_l = (1 + k + k^2 + \dots + k^{m-1}) - (1 + k + k^2 + \dots + k^{l-1}) =$$
$$k^l (1 + k + k^2 + \dots + k^{m-l-1}) = k^l a_{m-l}.$$
Thus $a_{m-l}$ is divisible by $2016$, and we have a contradiction.