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Research notes on every problem, and a library of the papers behind them. Built from the open erdos repository.

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Statement

"We conjecture that r(Kn⋅K2)=r(Kn)r(K_n\cdot K_2)=r(K_n) when n≥4n\ge4. It is not hard to see that this would follow if r(Km,Kn)≥r(Km,Kn−1)+mr(K_m,K_n)\ge r(K_m,K_{n-1})+m for all m≥n≥3m\ge n\ge3; this question in classical Ramsey theory does not seem to have been investigated. Tantalizingly, it is easy to prove that r(Km,Kn)≥r(Km,Kn−1)+m−1r(K_m,K_n)\ge r(K_m,K_{n-1})+m-1 if m≥n≥3m\ge n\ge3, but the stronger result has resisted our efforts." (printed p. 251, end of Section 3.)

Here Kn⋅K2K_n\cdot K_2 is KnK_n with one pendant edge: one new point joined by a single edge to one point of the KnK_n. It has (n2)+1\binom n2+1 edges and is the graph HH of Problem 545 for m=(n2)+1m=\binom n2+1 (t=1t=1); the conjecture asserts that this HH has the same Ramsey number as KnK_n, so that, for the problem's inequality at t=1t=1, the bound to beat is r(Kn)r(K_n) itself.

Source. S. A. Burr and P. Erdős, Extremal Ramsey theory for graphs, Utilitas Math. 9 (1976), 247--258; printed p. 251 is PDF p. 5 of the scan, read on the page image.

Read depth. Claims checked: the passage was read clause by clause on the page image. No proof is given.

Proof pointer

None; a conjecture. The paper notes only the two implications quoted above.

Dependencies

None.

Bears on

  • Problem 545: context for the case t=1t=1; a 1976 conjecture, since proved: Theorem 3 of Burr, Erdős, Faudree and Schelp (1989) (p. 117) with m=n≥4m=n\ge4 gives r(Kn,n−3∗)=r(Kn)r(K^*_{n,n-3})=r(K_n), where Kn,n−3∗K^*_{n,n-3} is KnK_n with n−3n-3 pendant edges at distinct new vertices and so contains Kn⋅K2K_n\cdot K_2; their Theorem 1 also gives r(Km,Kn)≥r(Km,Kn−1)+2m−3≥r(Km,Kn−1)+mr(K_m,K_n)\ge r(K_m,K_{n-1})+2m-3\ge r(K_m,K_{n-1})+m for m≥n≥3m\ge n\ge3, the condition from which this paper says the conjecture would follow (both specializations are made here; the 1989 paper does not mention the conjecture, and it leaves the cases m=n=4m=n=4 and {m,n}={3,5}\{m,n\}=\{3,5\} of its Theorem 3 to the reader).