Consider the double sequence $A(n,k)$ which is recursively defined by $$A(n,n)=1 \text{ for } n=0,1,2,\dots \text{ and }$$ $$A(n,k)=2\sum_{l=1}^{k+1} \binom{2n+1}{2l} A(n-l,k+1-l) \text{ for }0\leq k < n.$$ Furthermore let $B(n,k)$ be such that $$\sum_{k=0}^m A(n,k)B(n,m-k)=\begin{cases}1 & \text{ if }m=0 \newline 0 &\text{ otherwise}.\end{cases}.$$ Prove that $B(n,k)$ is alternating for fixed $n$, i.e. $B(n,k) \cdot (-1)^k>0$ for $0 \leq k \leq n$.

Remark: The sequence $A(n,k)$ occurs when we expand the Eulerian Polynomial $A_{2n+1}$ as follows $$A_{2n+1}(x)=\sum_{k=0}^n A(n,k) x^{n-k}(x-1)^{2k}.$$