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May 10, 2021 at 15:04 history edited Fedor Petrov CC BY-SA 4.0
deleted 81 characters in body
May 25, 2010 at 20:13 comment added Fedor Petrov Of course, but if you carry on the sharp constant in the exponent, I have to think bit more:) Now the lower estimate is $2^{n/4+o(n)}$ . But I suppose that calculating the best constant is very hard, similar to Ramsey numbers precise asymptotics.
May 25, 2010 at 17:46 comment added Joel David Hamkins Thanks very much for this excellent answer! I appreciate it very much. Could I ask kindly whether you might carry your argument through to the conclusion of an explicit lower bound?
May 25, 2010 at 16:26 history edited Benoît Kloeckner CC BY-SA 2.5
corrected a typo (m -> n)
May 25, 2010 at 15:08 history answered Fedor Petrov CC BY-SA 2.5