more detail is in a 70kb Excel spreadsheet : http://users.pandora.be/Wouter.Meeussen/DodecahedralVectorSum.xls for 2 vertices, there are 5 (not six) different sets: {10 pairs with norm^2 of sum = 0.000} {30 pairs with 1.000} {60, 2.618} {60, 5.236} {30, 6.854} the norm^2 is taken with the side of the pentagons =1. And of course 10+30+60+60+30 = 190 = 20 choose 2 For the icosahedron, I get for k=1..6 1, 3, 5, 8, 8, 12 remark that the two 'penultimate ones' (resp k=4 & 5 for icosahedral and k=8 & 9 for dodecahedral) are equal in count. (??) W. ----- Original Message ----- From: "N. J. A. Sloane" <njas@research.att.com> To: <seqfan@ext.jussieu.fr>; <math-fun@mailman.xmission.com> Cc: <wouter.meeussen@vandemoortele.com>; <wouter.meeussen@pandora.be> Sent: Tuesday, December 28, 2004 5:26 PM Subject: Possibly new sequence related to dodecahedron The following is an interesting question raised by Wouter Meeussen <wouter.meeussen@pandora.be> on Dec 27 2004: He said:
Based on the symmetry point-group, how many 'different' sets of 1, 2, .., 10 vertices exist on a dodecahedron?
If I count different norms of sums of 1, 2, .., 10 unit vectors, I find: 1,5,12,22,34,50,65,78,78,86 but there might be 'different' sets that accidentaly sum to the same resultant's norm. So the given integers are a lower bound.
We can rephrase the question as follows. Let G be the full icosahedral group, of order 120. Let v_1, ..., v_20 be the vertices of the dodecahedron. Let S(n) be the set of vectors v_{i_1} + v_{i_2} + ... + v_{i_n} where 1 <= i_1 <= i_2 <= ... <= i_n <= 20. Then what is s(n), the number of orbits of G on S(n)? (so s(1) = 1, s(2) = 6, ...) Presumably this is different from A039742: ... %S A039742 1,6,50,475,4881,52835,593382,6849415,80757819,968400940,11773656517, %T A039742 144791296055,1797935761182 %N A039742 Lattice animals in the fcc lattice (12 nearest neighbors), connected rhombic dodecahedra, edge-connected cubes. ... Repeat for icosahedron, etc.! NJAS