If $p_{n}$ is the largest prime factor of $p_{n-1}+p_{n-2}+m$, then $p_{n}$ is bounded

Question

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?

Answer 1

If $m$, $p_1$ and $p_2$ are odd we will have $p_{n+1} \le (p_n + p_{n-1} + m)/3$ unless $p_n + p_{n-1} + m$ is prime. Heuristically one should expect prime values of $p_n + p_{n-1} + m$ to be very rare if the $p_n$ are large, so it should be unusual for $p_{n+1}$ to get very much larger than $\max(p_1, p_2, m)$. For example, I tried $m=1$ and odd primes $p_1, p_2 < 1000$. In each case, the sequence eventually became periodic; the largest prime that occurred was $55289$ (for $p_1 = 809$, $p_2 = 449$). However, I would find it surprising if it were possible to prove $p_n$ is always bounded.

Answer 2

Here's a thought too long for a comment, but prompted by Gerhard's comment and Robert's answer: It seems to me that when $m = 3$ (or $3k$), and $p_n, p_{n-1}$ are both not small, we cannot have the sequence "succeed"--i.e. hit a prime without having to divide--twice in a row.

That is, suppose $p_{n+1} = p_n + p_{n-1} + 3$; for all of these to be prime (and not 3) we must have $p_n \equiv p_{n-1}$ mod 3 and $p_{n+1}\equiv 2p_n$ mod 3. Then $p_{n+1} + p_n + 3\equiv 0$ mod 3, forcing $p_{n+2}\le(p_{n+1} + p_n+3)/3$.

Then it seems to me we can bound the sequence unless $p_n$ is very small very frequently. I am blanking on how to push this argument further but suggest it for anyone interested.