Assume $(x_n)$ is such a sequence. WLOG assume $x_1 = 1$ (otherwise we divide everything by $x_1$). By induction, we prove $x_n = a^{n-1}+b\frac{a^{n-1}-1}{a-1}$. Now fix $n$ and take $m$ divisible by $n$. $$a^{n-1}+b\frac{a^{n-1}-1}{a-1} \Big| a^{m-1}+b\frac{a^{m-1}-1}{a-1} \iff a^{n-1}+b\frac{a^{n-1}-1}{a-1} \Big| b\frac{a^{m-n}-1}{a-1}$$ Choosing $m=2n$ and noticing $\frac{a^{n}-1}{a-1} \Big| \frac{a^{kn}-1}{a-1}$ for any $k$ tells us $$a^{n-1}+b\frac{a^{n-1}-1}{a-1} \Big| b\frac{a^{n}-1}{a-1}$$ is actually equivalent to the problem. But also notice that, by subtracting $a$ times the lest side, we get that having $$a^{n-1}+b\frac{a^{n-1}-1}{a-1} \Big| a^n-b$$ for all $n$ is equivalent to the problem. The right hand side is less then $a$ times the left. But $a$ is fixed, and this holds for infinitely many $n$, so the RHS is $k$ times LHS for infinitely many $n$, $k$ is fixed. By expanding and rearranging we arrive at $$a^{n-1}(a^2-(k+1)a-kb) = ba-(b+k-kb)$$ But unless $a^2-(k+1)a-kb = 0$, the left hand side grows arbitrarily large while the right hand side stays fixed. So $a^2-(k+1)a-kb = 0 \implies ba-(b+k-kb) = 0 \implies a = b = 1$ are the only positive solutions. We see that then $x_n = n$, which obviously works. Undoing the assumption $x_1=1$, we get that $(a, b, x_1) = (1, t, t)$ for a positive integer $t$ works is equivalent to having the stuff we want to hold $\blacksquare$ EDIT: to avoid technical issues, start by assuming $a > 1$ and note that $a=1$ forces $b=1$