Counting Permutations by Alternating Descents

Abstract

We find the exponential generating function for permutations with all valleys even and all peaks odd, and use it to determine the asymptotics for its coefficients, answering a question posed by Liviu Nicolaescu. The generating function can be expressed as the reciprocal of a sum involving Euler numbers: \[\left(1-E_1x+E_{3}\frac{x^{3}}{3!}-E_{4}\frac{x^{4}}{4!}+E_{6}\frac{x^{6}}{6!}-E_{7}\frac{x^{7}}{7!}+\cdots\right)^{-1},\tag{$*$}\]where $\sum_{n=0}^\infty E_n x^n\!/n! = \sec x + \tan x$. We give two proofs of this formula. The first uses a system of differential equations whose solution gives the generating function\begin{equation*}\frac{3\sin\left(\frac{1}{2}x\right)+3\cosh\left(\frac{1}{2}\sqrt{3}x\right)}{3\cos\left(\frac{1}{2}x\right)-\sqrt{3}\sinh\left(\frac{1}{2}\sqrt{3}x\right)},\end{equation*} which we then show is equal to $(*)$. The second proof derives $(*)$ directly from general permutation enumeration techniques, using noncommutative symmetric functions. The generating function $(*)$ is an "alternating" analogue of David and Barton's generating function \[\left(1-x+\frac{x^{3}}{3!}-\frac{x^{4}}{4!}+\frac{x^{6}}{6!}-\frac{x^{7}}{7!}+\cdots\right)^{-1},\]for permutations with no increasing runs of length 3 or more. Our general results give further alternating analogues of permutation enumeration formulas, including results of Chebikin and Remmel.