Period of a Recurrence Relation 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. 
 A: 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.
