Prove that the sequence $p_i$ is bounded

Let $p_1,p_2,...$ be a sequence of natural numbers. $p_1$ and $p_2$ are prime and $p_n$ for $n\ge 3$ is the largest prime divisor of $p_{n-1}+p_{n-2}+2014$. Prove that $(p_n)$ is bounded.

• Wow! Starting from $p_1=2=p_2-1$ it becomes eventually periodic. In general ???? I wan't to see how to approach such a proof. – Pp.. Jan 23 '15 at 14:58
• @Pp..: Of course "eventually periodic" and "bounded" are identical for this sequence. – Charles Jan 23 '15 at 15:02
• @Charles What is the "Of course"? Just need to say something, no matter what? – Pp.. Jan 23 '15 at 15:50

Let $P_n = \max(p_1,\ldots,p_n)$. We show that $p_{n+1} \leq P_n + 2016$.

If $p_n=2$ or $p_{n-1}=2$ this follows from $p_{n+1} \leq p_n + p_{n-1} + 2014$. If $p_n$ and $p_{n-1}$ are both odd, then $p_n + p_{n-1} + 2014$ is even so we have $p_{n+1} \leq \frac12(p_n + p_{n-1} + 2014) < P_n + 2016$ (note that this is also true when $p_n + p_{n-1} + 2014$ happens to be a power of $2$).

So to reach a large prime $Q$, we first have to reach a prime in the interval $[Q, Q-2016]$. However, there are arbitrarily long stretches of arbitrarily large composite numbers (for instance, $N!+2$, $N!+3$, ..., $N!+N$ gives $N-1$ consecutive composite numbers) so our sequence of prime numbers will be bounded.

This is a partial solution, for sequences that don't have $2$ in them.
If $2$ does not appear, then $p_n$ are all odd. If $p_n$ is also unbounded, then to get beyond 100000, there must be several terms in a row with $p_{n+2}=(p_n+p_{n+1}+2014)/2$, $p_{n+3}=(p_{n+1}+p_{n+2}+2014)/2$. The remainders when you divide by $5$ must follow one of the following sequences $$\{1,1,3,4,3,3,0\}, \{1,2,1,1\}, \{1,4,2,0\}, \{2,3,2,2,4,0\}, \{3,1,4\}, \{3,3,0\}, \{4,4,1,2\}$$
so one of the $p_n$ is a multiple of $5$. This contradicts $p_n$ being prime and greater than $90000$. So we can't have ten odd numbers in a row without some $p_{n+2}\leq(p_n+p_{n+1}+2014)/4$.

If $2$ does appear, then it must reappear infinitely often, otherwise the first half of this proof applies. Let $q_m=p_{i_m}$ be the sequence of $p_n$ immediately before each $2$. Suppose $q_{k+1}>q_k$ and $q_k$ large enough. We can't afford to divide even numbers by more than 2, or odd numbers by more than 1. This forces the following sequence: $$q_m,2,q_m+2016,q_m+4032,q_m+4031$$
But $q_m+4031$ is even, and we have a contradiction, so $2$ must reappear instead of $q_m+4031$. So the only way that $q_{m+1}>q_m$, if $q_m$ is large enough, is if $q_{m+1}=q_m+4032$.
If $q_{m+1}<q_m$, then there must have been a reduction by a factor $2$ at some point, with at most two increases by $2016$ and ten increases by $2014/2$. To ensure that $q_m$ increase unboundedly, we need several increases of $q_m$ with no reduction in between. But then we will have a sequence $q_m+4032k, (k=1,2,3,4,5)$, one of which is a multiple of $5$, and we have a contradiction.