{\displaystyle S_{n}^{(2m)}=\left(S_{n}^{(2)}\right)^{m}+\sum _{k=1}^{\left\lfloor m/2\right\rfloor }{\frac {1}{2^{k}}}{\binom {m}{2k}}{\binom {2k}{k}}\left(S_{n}^{(4)}-\left(S_{n}^{(2)}\right)^{2}\right)^{k}\left(S_{n}^{(2)}\right)^{m-2k}}