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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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Claim. Theorem 4 (p. 656) of Xuemei Zhang, Yaojun Chen and T.C. Edwin Cheng, Polarity graphs and Ramsey numbers for C4C_4 versus stars, Discrete Math. 340 (2017), no. 4, 655--660: "Let qq be an odd prime power. Then R(C4,K1,q2−t)=q2+q−(t−1)R(C_4,K_{1,q^2-t})=q^2+q-(t-1) if 1≤t≤2⌈q4⌉1\le t\le2\lceil\frac q4\rceil and t≠2⌈q4⌉−1t\ne2\lceil\frac q4\rceil-1." These are values of the function f(n)=R(C4,Sn)f(n)=R(C_4,S_n) of Problem 552, each equal to n+⌈n⌉+1n+\lceil\sqrt n\rceil+1. The paper recalls Parsons's 1976 family, the same formula for even tt when qq is odd, so the new values are those with odd tt; its summary records f(24)=30f(24)=30 and f(48)=56f(48)=56 as new. Its Ramsey graphs for odd tt add one edge to a subgraph of the polarity graph. The statement is recorded on the result page Theorem 4 of the library home zhang_2017_polarity_graphs_ramsey_numbers_c_4_versus_stars.

Covers. The value of f(n)f(n) at n=q2−tn=q^2-t for every odd prime power qq and every tt with 1≤t≤2⌈q/4⌉1\le t\le2\lceil q/4\rceil, t≠2⌈q/4⌉−1t\ne2\lceil q/4\rceil-1. The value at every other nn, and the second question, whether f(n)≤n+n−cf(n)\le n+\sqrt n-c for infinitely many nn, are not settled by it; the paper's Question 1 (p. 656) asks whether every value is n+⌈n⌉n+\lceil\sqrt n\rceil or one more.

Depends on. Nothing in this wiki; the theorem rests on the paper's own constructions and lemmas.

Acceptance. Refereed: the paper is a journal publication in Discrete Mathematics, volume 340, number 4 (April 2017). This is the refereed evidence. The paper was available online on 3 January 2017, its earliest public date, which dates this page. The erdosproblems.com curator refers to this paper in the commentary on exact values. But the erdosproblems.com label OPEN settles neither the problem nor a declared part of it, so reviewed is not listed. The statement is checked against the publisher's text, and the proof is read for structure only.