Stirling Binomial Polynomial Let $\{\cdot\}$ denote Stirling Numbers of the second kind. Let $(\cdot)$ denote the usual binomial coefficients. It is known that $$\sum_{j=k}^n {n\choose j} \left\{\begin{matrix} j \\ k \end{matrix}\right\} = \left\{\begin{matrix} n+1 \\ k+1 \end{matrix}\right\}.$$ Note: The indexes for $j$ aren't really needed since the terms are zero when $j>n$ or $j<k$. 
How do I calculate $$\sum_{j=k}^n 4^j{n\choose j} \left\{\begin{matrix} j \\ k \end{matrix}\right\} = ?$$ I have been trying to think of this sum as some special polynomial (maybe a Bell polynomial of some kind) that has been evaluated at 4. 
I have little knowledge of Stirling Numbers in the context of polynomials. Any help would be appreciated; even a reference to a comprehensive book on Stirling Numbers and polynomials. 

One can continue vadim123's approach with the ideas presented here (A sum with binomial coefficients).
