Let $ c_{n,k} $ be a sequence 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). $

[We can derive some precise formulas from here. Example: $ c_{n,1}=2^n - n-1 .$]

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 that $$ \sum_{k=0}^{p} c_{2p,k} \hspace{0.1cm} 2^{2p-k} = (2p+1)!.$$

Note that $\frac{(2n)!}{2^n n!} $is the product of odd numbers $ 1,3,...,(2n-1). $

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

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

PS. There is a table of values for $ c_{n,k} $ at http://oeis.org/A094503.