(Question was suggested and studied by a friend) Consider the following sequence: $$ a_n = \text{least prime divisor of } (2a_{n - 1} + 1) $$ For example, if $a_1 = 5$, we have $a_2 = 11, a_3 = 23, a_4 = 47, a_5 = 5$ and on. The question is, for every initial value of $a_1$, is the sequence $a_n$ bounded?
Here are some observations made during the attempts at a solution:
If the sequence decreases at a point, then it decreases at least at the order of $x^{1/2}$; this is attained at the case $x = p^2$ for some prime $p$.
Considering $a, 2a + 1, 4a + 3, \dots $ modulo $p$ for prime $p$ (this is periodic with period $ p - 1$), for some initial values of $p$, the sequence attains $0$ modulo $p$. For the class of primes $p$ where $2$ has multiplicative order $p - 1$ (modulo $p$), it is possible to prove that $x$ must be $p - 1$ modulo $p$, however as it is not known even whether this class is infinite or not, this is impossible to use.
If the length of the increasing part (a Cunningham chain) is asympotically $O(\log p)$ with initial value $p$, then the number increases polynomially, and with some refining of bounds, it is possible that the descending part of the sequence may dominate the ascending part, and force the sequence into a cycle. However, the only result known is that the length is $O(p)$. A preprint paper has a conditional result: https://arxiv.org/pdf/2205.07650.pdf.
Intuition suggests, and numerical experiments further shows that the least prime divisor operation is much more rapidly decreasing than $x^{1/2}$; however because of the number-theoretical nature of the function, it is very difficult to gain any good estimates. A study on factors of numbers of the form $2p + 1$ with $p$ prime may be fruitful.