SAUDI ARABIAN MATHEMATICAL COMPETITIONS

Put $k=2016$ and $a_n=\max \{p_n,p_{n+1}\}$ for each $n⩾1$. We shall show that $a_{n+1}⩽a_n+k+2$. In fact, since $p_{n+1}⩽a_n$, it remains to show that $p_{n+2}⩽a_n+k+2$. If $p_{n+2}=2$ then the inequality is trivially true. It suffices to consider the case $p_{n+2}>2$. If either $p_n=2$ or $p_{n+1}=2$ then $p_{n+2}⩽p_n+p_{n+1}+k=a_n+k+2$. Otherwise, both primes $p_n$ and $p_{n+1}$ are odd, in particular $p_n+p_{n+1}+k$ is even. But $p_{n+2}>2$, we get $$p_{n+2}⩽\frac{p_n+p_{n+1}+k}{2}<a_n+k+2.$$ It is clear that there is a positive integer $s$ such that $a_1⩽s(k+3)!+1$. We show, by induction that $a_n⩽s(k+3)!+1$. Indeed, suppose this is true for $n$, then $$a_{n+1}⩽a_n+k+2⩽s(k+3)!+k+3.$$ If $a_{n+1}>s(k+3)!+1$, then $1<m⩽k+3$ with $m=a_{n+1}-q(k+3)!$. Hence, $m$ must be a divisor of $(k+3)!$, that is $m$ is a proper divisor of $s(k+3)!+m=a_{n+1}$. However, this cannot happen since $a_{n+1}$ is a prime. Thus, $a_n⩽s(k+3)!+1$, and this completes the solution. $\square$