A199406 The number of inequivalent ways to color the edges of a cube using at most n colors.

1, 144, 12111, 358120, 5131650, 45528756, 288936634, 1433251296, 5887880415, 20842168600, 65402344161, 185788177224, 485443851256, 1181242399260, 2703252560100, 5864398969216, 12138503871789, 24101498435616, 46112016365155, 85335258695400, 153249227870046

Offset: 1

Author

Geoffrey Critzer, Nov 05 2011

Keywords

• nonn
• easy

Comments

Two edge colorings are equivalent if one is the mirror image of the other or the cube can be picked up and rotated in any manner to obtain the other.

The group here has order 48 (compare A060530). - N. J. A. Sloane, Aug 14 2012

Also the number of unoriented colorings of the 12 edges of a regular octahedron with n or fewer colors. The Schläfli symbols of the cube and octahedron are {4,3} and {3,4} respectively. They are mutually dual. For an unoriented coloring, chiral pairs are counted as one. - Robert A. Russell, Oct 17 2020

Links

• T. D. Noe, Table of n, a(n) for n = 1..1000
• Index entries for linear recurrences with constant coefficients, signature (13,-78,286,-715,1287,-1716,1716,-1287,715,-286,78,-13,1).

Crossrefs

Cf. A060530 (oriented), A337406 (chiral), A331351 (achiral), A128766 (cube vertices, octahedron faces), A198833 (cube faces, octahedron vertices), A063842(n-1) (tetrahedron), A337963 (dodecahedron, icosahedron).

Row 3 of A337408 (orthotope edges, orthoplex ridges) and A337412 (orthoplex edges, orthotope ridges).

Programs

Mathematica

Table[CycleIndex[KSubsetGroup[Automorphisms[CubicalGraph], Edges[CubicalGraph]],s] /. Table[s[i]->n, {i,1,6}], {n,1,15}]
Table[(8n^2+12n^3+8n^4+4n^6+12n^7+3n^8+n^12)/48, {n,20}] (* Robert A. Russell, Oct 17 2020 *)

Formula

a(n) = n^12/48 + n^8/16 + n^7/4 + n^6/12 + n^4/6 + n^3/4 + n^2/6.

Cycle index = 1/48(s_1^12+3s_1^4s_2^4+12s_1^2s_2^5+4s_2^6+8s_3^4+12s_4^3+8s_6^2).

G.f.: -x*(76*x^10 +10016*x^9 +212772*x^8 +1380453*x^7 +3384939*x^6 +3388593*x^5 +1380279*x^4 +211623*x^3 +10317*x^2 +131*x +1)/(x -1)^13. [Colin Barker, Aug 13 2012]

From Robert A. Russell, Oct 17 2020: (Start)

a(n) = A060530(n) - A337406(n) = (A060530(n) + A331351(n)) / 2 = A337406(n) + A331351(n).

a(n) = 1*C(n,1) + 142*C(n,2) + 11682*C(n,3) + 310536*C(n,4) + 3460725*C(n,5) + 19870590*C(n,6) + 65886660*C(n,7) + 133585200*C(n,8) + 168399000*C(n,9) + 128898000*C(n,10) + 54885600*C(n,11) + 9979200*C(n,12), where the coefficient of C(n,k) is the number of unoriented colorings using exactly k colors. (End)