A000545 Number of ways of n-coloring a dodecahedron.

1, 96, 9099, 280832, 4073375, 36292320, 230719293, 1145393152, 4707296613, 16666924000, 52307593239, 148602435840, 388302646355, 944900450144, 2162441849625, 4691253854208, 9710376716137, 19280531603808, 36888593841475, 68266682784000, 122597146773927

Offset: 1

Author

Clint. C. Williams (Clintwill(AT)aol.com)

Keywords

• nonn
• easy

Comments

More explicitly, a(n) is the number of colorings with at most n colors of the faces of a regular dodecahedron, inequivalent under the action of the rotation group of the dodecahedron. It is also the number of inequivalent colorings of the vertices of a regular icosahedron using at most n colors. - José H. Nieto S., Jan 19 2012

From Robert A. Russell, Oct 03 2020: (Start)

Each chiral pair is counted as two when enumerating oriented arrangements. The Schläfli symbols for the regular icosahedron and regular dodecahedron are {3,5} and {5,3} respectively. They are mutually dual. There are 60 elements in the rotation group of the regular dodecahedron/icosahedron. They divide into five conjugacy classes. The first formula is obtained by averaging the dodecahedron face (icosahedron vertex) cycle indices after replacing x_i^j with n^j according to the Pólya enumeration theorem.

Conjugacy Class Count Even Cycle Indices

Identity 1 x_1^12

Edge rotation 15 x_2^6

Vertex rotation 20 x_3^4

Small face rotation 12 x_1^2x_5^2

Large face rotation 12 x_1^2x_5^2 (End)

Links

• T. D. Noe, Table of n, a(n) for n = 1..1000
• Eric Weisstein's World of Mathematics, Polyhedron Coloring
• Index entries for linear recurrences with constant coefficients, signature (13, -78, 286, -715, 1287, -1716, 1716, -1287, 715, -286, 78, -13, 1).

Crossrefs

Cf. A252705 (unoriented), A337961 (chiral), A337962 (achiral).

Other elements: A054472 (dodecahedron vertices, icosahedron faces), A282670 (edges).

Other polyhedra: A006008 (tetrahedron), A047780 (cube faces, octahedron vertices), A000543 (octahedron faces, cube vertices).

Programs

Maple

(1/60)*n^12+(1/4)*n^6+(11/15)*n^4;

Mathematica

Table[n^12/60+n^6/4+11 n^4/15,{n,20}] (* or *) CoefficientList[Series[ -(((1+x) (1+x (82+x (7847+x (161900+x (943640+x (1764740+x (943640+x (161900+x (7847+x (82+x)))))))))))/(x-1)^13),{x,0,20}],x] (* Harvey P. Dale, Apr 25 2011 *)

Formula

G.f.: x*((1+x)*(1+x*(82+x*(7847+x*(161900+x*(943640+x*(1764740+x*(943640+x*(161900+x*(7847+x*(82+x)))))))))))/(1-x)^13. - Harvey P. Dale, Apr 25 2011

From Robert A. Russell, Oct 03 2020: (Start)

a(n) = (n^12 + 15*n^6 + 44*n^4) / 60.

a(n) = 1*C(n,1) + 94*C(n,2) + 8814*C(n,3) + 245008*C(n,4) + 2759250*C(n,5) + 15884004*C(n,6) + 52701264*C(n,7) + 106866144*C(n,8) + 134719200*C(n,9) + 103118400*C(n,10) + 43908480*C(n,11) + 7983360*C(n,12), where the coefficient of C(n,k) is the number of oriented colorings using exactly k colors.

a(n) = A252705(n) + A337961(n) = 2*A252705(n) - A337962(n) = 2*A337961(n) + A337962(n). (End)