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 $n \in \mathbb{N}$
We will prove
$$
(1)\qquad\qquad\qquad\qquad\sigma(A)\le \min_{i\le d}\{\operatorname{dist}\left(r(i),\operatorname{span}\{r(j):\ j\neq i\}\right)\}
\le \sqrt d\ \sigma(A). \qquad\qquad\qquad\qquad\qquad\qquad\qquad
$$
We only treat the case $\ $ rank$(A)=d$. In that case we set $W_0=\langle r(i)\rangle_{i=1,\dots ,d}$. Moreover $K=$ Ker $(A)$ satisfies $K\bot W_0$ and $K\oplus W_0=\Bbb R^n$. So $\inf _{v\in W_0,\|v\|=1}\|Av\|>0$ and we first prove that
$$
(2)\qquad\qquad\qquad\qquad\qquad\qquad\qquad
\sigma(A)= \inf _{v\in W_0,\|v\|=1}\|Av\|.\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad
$$
For this let $W$ be a vector space of dimension $d$. If $K\cap W\ne \{0\}$, then
$\inf _{v\in W,\|v\|=1}\|Av\|=0$. Else $K\oplus W=\Bbb R^n$ and we find an isomorphism of vector spaces $\varphi:W\to W_0$ given by $\varphi(v)= v_0$, where $v=v_k+v_0$ with $v_k\in K$, $v_0\in W_0$. This induces a bijection $\widetilde \varphi:\partial B(0,1)\cap W\to \partial B(0,1)\cap W_0$ given by $\widetilde\varphi (v)=\frac{\varphi(v)}{\|\varphi(v)\|}$.
Moreover $\|\varphi(v)\|\le \|v\|$, and so for any vector $v$ in $\partial B(0,1)\cap W$ we have
$$
\|Av\|=\|Av_0\|=\|A(\varphi(v))\|\le\frac{\|A(\varphi(v))\|}{\|\varphi(v)\|}=\|A(\widetilde\varphi(v))\|
$$
It follows that
$$
\inf _{v\in W_0,\|v\|=1}\|Av\|\ge \inf _{v\in W,\|v\|=1}\|Av\|
$$
for any vector space $W\subset \Bbb{R}^n$ of dimension $d$, and so
$$
\sigma(A)= \inf _{v\in W_0,\|v\|=1}\|Av\|.
$$
For each $j=1,\dots, d$ there exists a unique vector $s_j\in \Bbb R^n$ such that
$s_j\bot r(i)$ for $i\ne j$,
$s_j\in W_0$, or, equivalently, $s_j\bot K$,
$\langle s_j,r(j)\rangle=\|s_j\|^2$.
In fact, take any basis $k_1,\dots,k_{n-d}$ of $K$, and take the generalized cross product $t_j=k_1\times k_2\times \dots \times k_{n-d}\times r(1)\times\dots \times \widehat{r(j)} \times \dots \times r(d)$,
(where $\widehat{r(j)}$ means as usual that we delete $r(j)$), and then take
$$
s_j=\text{proj}_{t_j}r(j)=\frac{\langle r(j),t_j\rangle}{\langle t_j,t_j\rangle} t_j.
$$
Since $A(v)=\sum_{i=1}^d \langle r(j),v\rangle e_j$, where $\{e_j\}$ is the canonical basis of $\Bbb R^d$, we have $A(s_j)=\|s_j\|^2 e_j$. But then we have a basis $\{u_j\}_{j=1,\dots,d}$ of $W_0$ with $u_j=\frac{s_j}{\|s_j\|}$ and $A(u_j)=\|s_j\| e_j$.
On one hand we have $\operatorname{dist}\left(r(i),\operatorname{span}\{r(j):\ j\neq i\}\right)=\|s_i\|$, and so
$$
(3)\qquad\qquad\qquad\qquad\qquad\qquad\qquad\min_{i\le d}\{\operatorname{dist}\left(r(i),\operatorname{span}\{r(j):\ j\neq i\}\right)\}= \min_{i\le d}\{\|s_i\|\}\qquad\qquad\qquad\qquad\qquad\qquad\qquad
$$
On the other hand we can define $A^{-1}:\Bbb R^d \to W_0$ given by
$A^{-1}(e_j)=\frac{1}{\|s_j\|}u_j$. Then we have
$$(4)\qquad\qquad\qquad\qquad
\sigma(A)= \inf _{v\in W_0,\|v\|=1}\|Av\|=\min _{v\in W_0,\|v\|=1}\|Av\|=\frac{1}{\displaystyle\max_{v\in \Bbb R^d,\|v\|=1}\|A^{-1}(v)\|}. \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad
$$
Clearly
$$
\max_{v\in \Bbb R^d,\|v\|=1}\|A^{-1}(v)\|\ge \max_{j\le d}\|A^{-1}(e_j)\|=
\max_{j\le d}\frac{1}{\|s_j\|}
$$
and so
$$(5)\qquad\qquad\qquad\qquad\qquad\qquad\qquad
\sigma(A)\le \min_{j\le d}\{\|s_j\|\}.\qquad\qquad\qquad\qquad\qquad\qquad\qquad
$$
Finally consider the matrix $B$ which implements $A^{-1}$, that is, the $j$th row is the vector $\frac{1}{\|s_j\|}u_j\in \Bbb{R}^n$. Then
$\max_{v\in \Bbb R^d,\|v\|=1}\|A^{-1}(v)\|=\|B\|_2$ is just the 2-operator norm of $B$,
given by $\sqrt{\lambda_1}$, where $\lambda_1\ge \lambda_2\ge \dots \ge \lambda_d$ are the singular (positive) values of $B^*B$. In particular
$$
\lambda_1+\dots+\lambda_d=Tr(B^*B)=\sum_{j=1}^d (B^*B)_{jj}.
$$
But
$$
(B^*B)_{jj}=\sum_{k=1}^n(B^*)_{jk}B_{kj}=\sum_{k=1}^n \overline{B}_{kj}B_{kj}=\sum_{k=1}^n|B_{kj}|^2=\|B(e_j)\|^2=\frac{1}{\|s_j\|^2},
$$
and so
$$
\lambda_1\le Tr(B^*B) \le d\max_{j\le d}\{(B^*B)_{jj}\}=d\max_{j\le d}\frac{1}{\|s_j\|^2},
$$
hence
$$
\max_{v\in \Bbb R^d,\|v\|=1}\|A^{-1}(v)\|=\sqrt{\lambda_1}\le \sqrt{d}\max_{j\le d}\frac{1}{\|s_j\|}.
$$
It follows that
$$
\min_{j\le d}\{\|s_j\|\}=\frac{1}{\displaystyle\max_{j\le d}\frac{1}{\|s_j\|}}\le
\frac{\sqrt{d}}{\displaystyle\max_{v\in \Bbb R^d,\|v\|=1}\|A^{-1}(v)\|}=\sqrt{d}\ \sigma(A)
$$
which together with $(3)$ and $(5)$ proves $(1)$.
Best Answer
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 $n \to \infty.$