Suppose the set of primes dividing the sequence is finite instead, and number them $p_1,...,p_K.$
Associate to each term $a_n = a \cdot 2016^n + b \cdot 2017^n$ the index $i, 1 \le i \le K,$ such that biggest prime power that divides $a_n$ is $p_i^{\beta_i(n)}$.
By the pigeonhole principle, there is an index $i_0$ and $a_n,a_m,\,m>n,m-n \le K,$ such that $a_n,a_m$ are both associated to $i_0,$ for each $n.$ Therefore, letting $l = \min(\beta_{i_0}(n),\beta_{i_0}(m)),$ we get that
$$ p_{i_0}^l | a\cdot 2016^n + b\cdot 2017^n, p_{i_0}^l | a \cdot 2016^m + b\cdot 2017^m.$$
The relations above imply, on the other hand, that
$$ p_{i_0}^l | b\cdot(2016^{m-n} - 2017^{m-n}). $$
From the fact that the number of primes is bounded, we get that $p_{i_0}^{l \cdot K} \ge 2017^n.$ On the other hand, by the equation above,
$$ p_{i_0}^l \le C \cdot 2017^K.$$
As $K$ is fixed, we see that
$$ 2017^{\frac{n}{K} - K} \le C, \text{for infinitely many } n \in \mathbb{N}.$$
This leads to a contradiction by letting $n \to \infty.$
Let us prove that for arbitrary 3 points placed on a semicircle of unit radius, the sum $S$ of squares of their distances is less then or equal to 8.
Case 1: all three points on the diameter
It's easy to show that 3 arbitrary points shown on the left have smaller $S$ compared to the special case shown on the right ($AB<AB'$, $AC<AC'$, $BC<B'C'$
For the three points on the right:
$$S=x^2+(2-x)^2+2^2=x^2+4-4x+x^2+4=8-2x(2-x)$$
Obviously $x\le2$ so $S\le8$.
Case 2: Two points on the diameter, one point above on the circle.
Arbitrary case is shown on the left. For every such case it is possible to find a similar case, with one point on the diameter moved to the end of it, that has bigger $S$. For example, if ve move point $A$ to the left end of the diameter $BA'>BA$, $CA'>CA$. Now look at the picture on the right and triangles $A'BC$ and $A'BC'$. We want to prove that $S(A'BC)<S(A'BC'):$
$$S(A'BC)=c^2+a'^2+(2-x)^2=c^2+(a^2+x^2-2ax\cos\alpha)+4-4x+x^2=$$
$$S(A'BC)=c^2+a^2+4+2x^2-2ax\cos\alpha-4x=S(A'BC')-2x(2-x)-2ax\cos\alpha\le S(A'BC')$$
Note that $S(A'BC')=8$.
Case 3: Two points on circumference, one point on the diameter
For the triangle shown on the left, it is always possible to move one point to the end of the diameter and create a triangle that has a bigger $S$. For example, if you move point $A$ of triangle $ABC$ to point $A'$: $BA'>BA$, $CA'>CA$. So $S(ABC)\lt S(A'BC)$ and according to case (2), $S(A'BC)\le8$
Case 4: All three points on circumference
This case is trivial. Such triangle has smaller $S$ compared with triangle $A'BC'$ and according to case (2) $S(A'BC')=8$.
Best Answer
Since $\lim_{n \to +\infty} a_n = 0$ we have that there exist $n_0$ such that for any $n \geq n_0$ we have $a_n > ca_{n}^2$ and $1 > 2ca_n$.
Now for $n$ big enough $$a_{n+1} \geq a_n - c \cdot a_n^2 > 0$$ Consider $d>0$ such that the inequality is true until $n_0$ and $\frac{d}{n_0} \cdot 2c < 1$ and $cd < \frac{1}{3}$ (it clearly exists since there are only finitely many values of $na_n$ for $n \leq n_0$).
Suppose now $a_{n+1} < \frac{d}{n+1}$ and $a_n \geq \frac{d}{n}$ for some $n \geq n_0$.
Then, since we have $n \geq n_0$ and so the function $f(x) = x-cx^2$ is increasing in the interval we are considering, $$ \frac{d}{n+1} > \frac{d}{n} - \frac{cd^2}{n^2}$$ And so $$ dn^2 > dn(n+1) - cd^2(n+1) \iff cd(n+1) > n$$ Thus $$ 3n < n+1 \iff n < \frac{1}{2}$$ Which is absurd.
Therefore $na_n \geq d$ for any $n \in \mathbb{N}$