A Combinatorial Proof of a Cubic Congruence for the \(q\)-Secant Inversion Enumerator

Abstract

Let \(\mathcal A(2n)\) be the set of up--down alternating permutations
\[\sigma_1<\sigma_2>\sigma_3<\sigma_4>\cdots<\sigma_{2n},\]
and let
\[E_{2n}(q)=\sum_{\sigma\in \mathcal A(2n)}q^{\operatorname{inv}(\sigma)}.\]
We give a self-contained combinatorial proof of the congruence
\[E_{2n}(q)\equiv q^{2n(n-1)}-\binom n2(1+q)^2\pmod{(1+q)^3}.\]
This refines the Andrews-Foata congruence modulo \((1+q)^2\) for the \(q\)-secant numbers. The proof decomposes alternating permutations into orbits generated by switches of natural pairs \(\{2i-1,2i\}\). Three free natural pairs produce an orbit of type \((\mathbb Z/2\mathbb Z)^3\), and hence a factor \((1+q)^3\). The remaining contribution comes from permutations with exactly two free natural pairs; after forced outside blocks are removed, a sign-reversing sliding involution leaves one unpaired core, giving the correction term \(-\binom n2(1+q)^2\).