Conjecture

Let $m$ be give postive integer numbers,a sequence $p_{1},p_{2},\cdots $of primes satisfies the following condition:for $n\ge 3$,$p_{n}$ is the greatest prime divisor $p_{n-1}+p_{n-2}+m$, Prove that the sequence is bounded

I have solve when $m$ is even number,But for $m$ is odd I can't sove it

when $m$ is even,Let $b_{n}=\max\{p_{n},p_{n+1}\}$ for $n\ge 1$

Lemma: for all $n$ have $$b_{n+2}\le b_{n}+m+2$$  Proof:Certainly $p_{n+1}\le b_{n}$, so it suffices to show that $p_{n+2}\le b_{n}+m+2$.if either $p_{n}$ or $p_{n+1}$ equals $2$,then we have $$p_{n+2}\le p_{n}+p_{n+1}+m=b_{n}+m+2$$ otherwis，$p_{n}$ and $p_{n+1}$ are both odd,so $p_{n}+p_{n+1}+m$ is even,Because $p_{n+2}\neq 2$ divides this number,we have $$p_{n+2}\le\dfrac{1}{2}(p_{n}+p_{n+1}+m)\le b_{n}+\frac{m}{2}$$ This prove the claim.

choose $k$ large enough so that $b_{1}\le k\cdot (m+3)!+1$.we prove by induction that $b_{n}\le k\cdot (m+3)!+1$ for all $n$, if this statement holds for some $n$, then $$b_{n+1}\le b_{n}+m+2\le k\cdot (m+3)!+m+3$$ if $b_{n+1}>k\cdot (m+3)!+1$ ,then let $q=b_{n+1}-k\cdot (m+3)!$,we have $1<q\le m+3$,so we have $q|(m+3)!$,hence,$q$ is proper divisor of $k\cdot (m+3)!+q=b_{n+1}$,which is impossible,because $b_{n+1}$ is prime,

Thus $p_{n}\le b_{n}\le k\cdot(m+3)!+1$ for all $n$.

But for $m$ be odd number,then $p_{n}+p_{n+1}+m$ be odd number,so we can't have following inequality $$p_{n+2}\le\dfrac{1}{2}(p_{n}+p_{n+1}+m)!$$,so this case How to prove，Thanks ?

Conjecture

Let $m$ be give positive integer numbers,a sequence $p_{1},p_{2},\cdots $of primes satisfies the following condition:for $n\ge 3$,$p_{n}$ is the greatest prime divisor $p_{n-1}+p_{n-2}+m$, Prove that the sequence is bounded.

I have solved it when $m$ is even number, but for $m$ is odd, I can't solve it.

When $m$ is even, let $b_{n}=\max\{p_{n},p_{n+1}\}$ for $n\ge 1$

Lemma: for all $n$ we have $$b_{n+2}\le b_{n}+m+2$$ 

Proof: certainly $p_{n+1}\le b_{n}$, so it suffices to show that $p_{n+2}\le b_{n}+m+2$. If either $p_{n}$ or $p_{n+1}$ equals $2$, then we have $$p_{n+2}\le p_{n}+p_{n+1}+m=b_{n}+m+2$$ otherwise, $p_{n}$ and $p_{n+1}$ are both odd, so $p_{n}+p_{n+1}+m$ is even. Because $p_{n+2}\neq 2$ divides this number, we have $$p_{n+2}\le\dfrac{1}{2}(p_{n}+p_{n+1}+m)\le b_{n}+\frac{m}{2}$$ This proves the claim.

Choose $k$ large enough so that $b_{1}\le k\cdot (m+3)!+1$. We prove by induction that $b_{n}\le k\cdot (m+3)!+1$ for all $n$. If this statement holds for some $n$, then $$b_{n+1}\le b_{n}+m+2\le k\cdot (m+3)!+m+3$$ if $b_{n+1}>k\cdot (m+3)!+1$ ,then let $q=b_{n+1}-k\cdot (m+3)!$, we have $1<q\le m+3$, so we have $q|(m+3)!$, hence,  $q$ is proper divisor of $k\cdot (m+3)!+q=b_{n+1}$, which is impossible, because $b_{n+1}$ is prime.

Thus $p_{n}\le b_{n}\le k\cdot(m+3)!+1$ for all $n$.

But if $m$ is an odd number, then $p_{n}+p_{n+1}+m$ is an odd number, so we can't have the following inequality: $$p_{n+2}\le\dfrac{1}{2}(p_{n}+p_{n+1}+m)!$$.

So how to prove this case?