%I
%S 1,1,12,156,888
%N Total number of coverings of a cubic board with the minimal number of knights.
%F Sequence A110214 gives minimal number of knights needed to cover an n X n X n board. This sequence gives total number of solutions using A110214(n) knights.
%e a(3) = 12, since reflections or rotations of
%e OOO KKK OOO
%e OKO OKO OKO
%e OOO OOO OOO generate twelve different coverings.
%Y This sequence is a 3dimensional analog of A103315. a(n) = A110219(n, n, n). A102215 gives the number of inequivalent solutions.
%K hard,nonn
%O 1,3
%A Nikolaus Meyberg (Nikolaus.Meyberg(AT)tonline.de), Jul 17 2005
