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Weighted sum of the Simsun (Andre) permutations

Let $ c_{n,k} $ be the Simsun permutations$^1$ defined by the following relations: $\displaystyle c_{n,0} = 1, \hspace{0.1cm} (n \ge 1);$ $$ c_{n,k} = (k+1) c_{n-1,k} +(n-2k+1) c_{n-1,k-1}, \hspace{0.5cm} (1 \leq k \leq \lfloor n/2 \rfloor);$$ and $ c_{n,k} = 0, \hspace{0.1cm} ( k> \lfloor n/2 \rfloor). $

Now, let $n=2p.$ I am trying to find the value of \begin{eqnarray} A_p:=\sum_{k=0}^{p} c_{2p,k} \hspace{0.1cm} \frac{(2p-2k)!}{2^{p-k} (p-k)!}. \end{eqnarray} or at least a sharp upper bound for it. We know$^2$ that $$ \sum_{k=0}^{p} c_{2p,k} \hspace{0.1cm} 2^{2p-k} = (2p+1)!.$$

From this identity we can easily obtain the bound $$ A_p \leq 2^{(-p)} p!(2p+1)! $$ for $ A_p $ which is a big upper bound.

Additionally, we know$^2$ that $ \sum_{k=0}^{p} c_{2p,k} = T_{p+1}$, where $ T_n= \frac{2^{2n}(2^{2n}-1) |B_{2n}|}{2n} $ is the sequence of tangent numbers$^3$ (defined by the Bernoulli numbers $B_n$), appearing in the Taylor series expansion of tan($x$): $$\text{tan}(x)=\sum_{n=1}^{\infty} T_n \frac{x^{2n-1}}{(2n-1)!}.$$

Motivation: In a part of my research (in quantum statistical mechanics) I need to show convergence of a series. I have reduced the initial problem to finding the value of $ A_p $, or at least a good upper bound for $ A_p $.

Any hint or idea would be greatly appreciated! Thanks in advance!

  1. For Andre and Simsun permutations see here and here.
  2. See here for the paper "increasing trees and alternating permutations" by G. Kuztensov, I. Pak, and A.E. Postnikov. In this paper, the Andre permutations are denoted by $ d_{n,k} $.
  3. See here.
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