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 Prove: \begin{eqnarray} A_p:=\sum_{k=0}^{p} c_{2p,k} \hspace{0.1cm} \frac{(2p-2k)!}{2^{p-k} (p-k)!} \leq \frac{p+7}{6} (2p)!. \end{eqnarray} There is evidence that for $ p < 10 $ the upper bound can be taken $ (2p)! $. However, for $ p \ge 10 $ there is a steady increase in $ A_p $ beyond $ (2p)! $. [With trial and error and using a table of values for $ c_{n,k} $ and evaluating $ A_p $, we can see that any upper bound for $ A_p $ seems to include $ (2p)! $.]\

If we apply induction on $p$ and use recursive definition of $ c_{2p+2,k} $, we obtain the following. \begin{eqnarray} A_{p+1}=\sum_{k=0}^{p} f_p(k) \hspace{0.1cm} c_{2p,k} \hspace{0.1cm} \frac{(2p-2k)!}{2^{p-k} (p-k)!}, \end{eqnarray} where $ f_p(k)= (k^2 +4k+5)(2p-2k)+(k+1)(k+2) $. Or, in the expanded form, $ f_p(k)= -2k^3 +(2p-7) k^2 +(8p-7)k+10p+2 $. To use induction assumption we need to maximize $f_p (k)$ and maximizing turns out to cause new problem. The maximum of $f_p(k) $ occurs at $ \lfloor 2p/3 \rfloor $. I have shown that $$ f_p (k) \leq \frac{p+8}{6} (2p+1)(2p+2). $$

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 for proving this inequality or even correcting the upper bound for a better bound would be appreciated! Thanks in advance!\

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