easily we get $a_i$ is prime for $i \ge 3$
if $a_2=a_3=2$ we get $a_i=2$ for all $i$ so we are done
now if we have $a_3>2$ we get
$a_3= odd$ , so $a_4=odd$ aswell but $a_5=2$ so $a_{3k+2}=2$ we prove the sequance is bounded , if not , lets get the biggest prime $q$ in the first $10^{10}$ numbers , so $q>2$ if
$q=a_n$
then
$a_{n-1}+a_{n-2}=a_n$
now take a look at this structure
$2,p_1,q_1,2,p_2,q_2,2$ now if $q_1=p_1+2$ and $q_1>7$ we have $p_2 \le \frac{2+q_1}{3}$ and
$q_2 \le \frac{q_1+8}{3}$ so any time we have $q_1=p_1+2$ we go low fast , so we must have the set of distinct $q_i$ such that
$q_i=p_i+2$ is finite , now for $p_i$
if $p_2=q_1+2$ , if $p_2>7$ we have
$q_2 \le \frac{p_2 +2}{3}$ so $p_3 \le \frac{p_2+8}{3}$ so we are done here as well , so $q_i,p_i$ is bounded , so its perodic from somewhere forward and done !