\[\sum_{k=0}^{\left\lfloor\ifrac{n}{2}\right\rfloor}{n\choose 2k}\left(\frac{5}{%
9}\right)^{k}B_{2k}u_{n-2k}=\frac{n}{6}v_{n-1}+\frac{n}{3^{n}}v_{2n-2},\]