Rudolph conjectures that for permutations $p$ and $q$ of the same length, $A_n(p) \le A_n(q)$ for all $n$ if and only if the spine structure of $T(p)$ is less than or equal to the spine structure of $T(q)$ in refinement order. We prove one direction of this conjecture, by showing that if the spine structure of $T(p)$ is less than or equal to the spine structure of $T(q)$, then $A_n(p) \le A_n(q)$ for all $n$. We disprove the opposite direction by giving a counterexample, and hence disprove the conjecture.