Let {$x_n$} be such a recurrence relations that obeys the following:
For fixed naturals $a,b$, $x_ {n+1}$ is the least prime divisor of $ax_n+b$.
Calculations showed that{$x_n$} appears to be eventually periodic. For example, if $a=6,b=7$ and $x_1=2$, {$x_n$} follows thus: $2,19,11,73,5,37,229,1381,8293,5,37,229,1381,8293, \dots$
How does one prove $x_{n}$ is eventually periodic?
Attempt
Assume that $x_n$ is not eventually periodic. This implies that no $x_i$ is the same, and can grow infinitely large.
Let $x_i$ be the smallest value in $x_i$ which does not divide $ab$ and $i \neq 1$.
Then there must not exist for all $j>i$ $x_j \equiv x_ {i-1}$ or $0 \pmod {x_i}$ because ${x_i}$ is the least possible prime.
By $PHP$, there must be some $x_j \equiv x_k \pmod {x_i}$ if $i-1 \le j,k \le i+x_i$. However, from our assumption $i<j,k\le i+x_i$.
I have no idea how to proceed from here. Any help would be appreciated.
There are some observations, not an answer. Everything that follows is very loose and nothing even near a proof.
If $ax_n+b$ is not a prime, then its least prime factor is at most $\sqrt{ax_n+b}$. This means that it will take a large number of consecutive primes before $x_n$ will be reached again.
More concretely, if there are $m$ consecutive primes, the $x_k$ will be about $a^m$ times as large. So, if $v = ax_n+b$, we want $a^m\sqrt{v} \ge v$ or $a^m \ge v^{1/2}$ or $m \ge \frac{\ln v}{2\ln a}$.
However, the probability of any particular $ax_n+b$ being prime is about $\frac1{\ln(ax_n+b)} \approx \frac1{\ln(ax_n)} $. The probability of $m$ consecutive iterations being prime is thus
$\begin{array}\\ P_m &\approx \prod_{k=1}^m \frac1{\ln(a^kx_n)}\\ &\approx \prod_{k=1}^m \frac1{k \ln a+\ln(x_n)}\\ &\approx \frac1{m!}\prod_{k=1}^m \frac1{\ln a+\ln(x_n)/k}\\ &< \frac1{m!(\ln a)^m}\\ \end{array} $
Since $m! > (m/e)^m $, for $c \approx \ln a$,
$\begin{array}\\ P_m &< \frac1{(m/e)^m c^m}\\ &= \left(\frac{e}{cm}\right)^m\\ &< \left(\frac{e}{c\frac{\ln v}{2\ln a}}\right)^{\frac{\ln v}{2\ln a}}\\ &\approx \left(\frac{2e}{\ln v}\right)^{\frac{\ln v}{2\ln a}}\\ \end{array} $
If $v$ is quite large, then this probability is extremely small.
I would be interested in any actual computations to see how large $x_n$ gets. According to this, large values would be quite rare.