Decreasing pattern minimizes avoidance among alternating permutations

Let $A_n(q)$ denote the set of alternating permutations of length $n$ avoiding the pattern $q$, and let $S_k$ be the symmetric group on $k$ letters. Alternating-avoidance minimization conjecture. For all positive integers $k,n$ and all $q\in S_k$ with $q\ne k(k-1)(k-2)\cdots1$,

$$|A_n(q)|\ge |A_n(k(k-1)(k-2)\cdots1)|.$$

If $n\ge2k-2$ is even, then the inequality is strict. The claim is motivated by brute-force enumerations, is known for $k=3$, and the case $n=2k-2$ was proved in the cited corollary.