Stirling numbers satisfy the following orthogonality condition. For $k<n$: $$ \sum_{i=k}^n S(n,i) s(i,k)=0,$$ where $s(i,k)$ and $S(n,i)$ are Stirling numbers of the first and second kind, respectively.
My question: Is there a way to simplify the following $$ \sum_{i=k}^n \frac{S(n,i) s(i,k)}{2^i}.$$