\[E_{2n}=\sum_{k=1}^{2n}\frac{(-1)^{k}}{2^{k-1}}{2n+1\choose k+1}\*\sum_{j=0}^{%
\left\lfloor\tfrac{1}{2}k-\tfrac{1}{2}\right\rfloor}{k\choose j}(k-2j)^{2n}.\]