2 n ( n + 1 ) + a {\displaystyle 2n(n+1)+a\,}
a = { 2 n , wenn n gerade 2 ( n − 1 ) , wenn n ungerade {\displaystyle a={\begin{cases}2n,&{\text{wenn }}n{\text{ gerade}}\\2(n-1),&{\text{wenn }}n{\text{ ungerade}}\end{cases}}}
3 n ( n + 1 ) 2 + a {\displaystyle 3n(n+1)^{2}+a\,}
6 ( n − 1 ) 2 + 8 {\displaystyle 6(n-1)^{2}+8\,}