There must be something missing, since this is trivial. We don't even need to know $a_2=2$ since it follows from the definition. Assume $a_k=k$ for all $1\leq k \leq n$. Then since $\gcd(a_n, n+1) = \gcd(n,n+1)=1$ it means we must take $a_{n+1} = n+1$. Maybe, since the first two terms are given, the condition is that $a_{n+1}$ is the least $m$ not yet in the sequence, and such that $\gcd(a_{n-1},m)=1$. Indeed, all primes will clearly belong to the sequence, and so any skipped number will become coprime with some second-to-the-left, and thus become eligible for the sequence.

Actually, it's supposed to say $\gcd(m,a_n) \neq 1$. Edit: Just for the record, all Tournament of Town problems in English from 2001 onward can be found on this site: http://www.math.toronto.edu/oz/turgor/archives.php

Sorry for that dreadful typo@mavropnevma. And thank you for the link but I am posting the problems for the resource page @sabbasi.

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=57&t=575422