3 deleted 21 characters in body

The

Originally I thought that the answer to your question is yes. This can could be easily deduced from results in a recent paper written by Sara Billey, Chris Burdzy, and myself, arXiv:1209.0693. Clearly Theorem 1.1 in that paper gives an enumeration result for the number of permutations you are considering is less than the number, a_n, in S_n which have interior a given set of peaks (what you are calling interior local maxima) at only odd indices and valleys (what you are calling interior local minima) anywheremaxima). A simple computation using Unfortunately, the Theorem 1.1 theorem only applies if the number of elements in the cited paper shows that a_n peak set is at most p(n) 2^{2n} for some polynomial p(n)constant with respect to n, which is not true in this case. But since (2n+1)! grows faster than [(2n+1)/e]^{2n+1}, the ratio clearly goes It would be very interesting to zerofind an analogue of this theorem where the size of the peak set varies with n.

Cheers, Bruce Sagan

PS Thanks to Richard Stanley for pointing out this post to us.

2 added 67 characters in body

The answer to your question is yes. This can be easily deduced from results in a recent paper written by Sara Billey, Chris Burdzy, and myself, arXiv:1209.0693. Clearly the number of permutations you are considering is less than the number, a_n, which have interior peaks (what you are calling interior local maxima) at only odd indices and valleys (what you are calling interior local minima) anywhere. A simple computation using the Theorem 1.1 in the cited paper shows that a_n is at most p(n) 2^{2n} for some polynomial p(n). But since (2n+1)! grows faster than [(2n+1)/e]^{2n+1}, the ratio clearly goes to zero.

Cheers, Bruce Sagan

PS Thanks to Richard Stanley for pointing out this post to us.

1

The answer to your question is yes. This can be easily deduced from results in a recent paper written by Sara Billey, Chris Burdzy, and myself, arXiv:1209.0693. Clearly the number of permutations you are considering is less than the number, a_n, which have interior peaks (what you are calling interior local maxima) at only odd indices and valleys (what you are calling interior local minima) anywhere. A simple computation using the Theorem 1.1 in the cited paper shows that a_n is at most p(n) 2^{2n} for some polynomial p(n). But since (2n+1)! grows faster than [(2n+1)/e]^{2n+1}, the ratio clearly goes to zero.

Cheers, Bruce Sagan