cp's OEIS Frontend

If more than one configuration with maximal volume exists for a given n, the one with the largest symmetry group is chosen. Berman and Hanes give optimality proofs for n<=8. Higher terms are only conjectures. An independent verification of the results by Hardin, Sloane and Smith has been performed by Pfoertner in 1992 for n<28. An archive of the results with improvements for n=23,24 added in 2003 is available at link. A conjectured continuation of the sequence starting with n=28 is: 12,6,2,6,120,2,4,4,2,20,4,12,24,12,20,4,8,2,2,2,4,1,24

Berman and Hanes (see link, page 81) proved in 1970 that an arrangement of 8 points on the surface of a sphere with 4 points with node degree 4 and 4 points with node degree 5 is the one with a maximum volume of their convex hull.

The polyhedron (see linked illustration) has vertices at the poles and two square rings of vertices rotated by Pi/4 against each other, with a polar angle of approx. +-62.89908285 degrees against the poles. The polyhedron is completely described by this angle and its order 16 symmetry. It would be desirable to know a closed formula representation of this angle and the volume.

The polyhedron (see linked illustration) with a symmetry group of order 4 has a vertex in the north pole on its axis of symmetry. The remaining 10 vertices are diametrically opposite in pairs relative to this axis of symmetry. The polar vertex has vertex degree 6. 8 vertices have vertex degree 5. 2 vertices have vertex degree 4.

This allocation seems to be the best possible approximation of a medial distribution of the vertex degrees, which is a known necessary condition for maximum volume. Of the 25 possible triangulations with vertex degree >= 4, all the others have more than 2 vertices with vertex degree 4, which leads to more pointed corners and therefore smaller volumes.