A006550 -id:A006550 - cp's OEIS frontend

A047780 Number of inequivalent ways to color faces of a cube using at most n colors.

0, 1, 10, 57, 240, 800, 2226, 5390, 11712, 23355, 43450, 76351, 127920, 205842, 319970, 482700, 709376, 1018725, 1433322, 1980085, 2690800, 3602676, 4758930, 6209402, 8011200, 10229375, 12937626, 16219035, 20166832, 24885190, 30490050

Offset: 0

Jud McCranie

Here inequivalent means under the action of the rotation group of the cube, of order 24, which in its action on the faces has cycle index (x1^6 + 3*x1^2*x2^2 + 6*x1^2*x4 + 6*x2^3 + 8*x3^2)/24.
a(n) is also the number of inequivalent colorings of the vertices of a regular octahedron using at most n colors. - José H. Nieto S., Jan 19 2012
From Robert A. Russell, Oct 08 2020: (Start)
Each chiral pair is counted as two when enumerating oriented arrangements. The Schläfli symbols for the regular octahedron and cube are {3,4} and {4,3} respectively. They are mutually dual.
There are 24 elements in the rotation group of the regular octahedron/cube. They divide into five conjugacy classes. The first formula is obtained by averaging the cube face (octahedron 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_1^6
  Vertex rotation       8     x_3^2
  Edge rotation         6     x_2^3
  Small face rotation   6     x_1^2x_4^1
  Large face rotation   3     x_1^2x_2^2 (End)

L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 254 (corrected).
N. G. De Bruijn, Polya's theory of counting, in E. F. Beckenbach, ed., Applied Combinatorial Mathematics, Wiley, 1964, pp. 144-184 (see p. 147).
M. Gardner, New Mathematical Diversions from Scientific American. Simon and Schuster, NY, 1966, p. 246 (the formula given is incorrect but was corrected in the second printing).
J.-P. Delahaye, 'Le miraculeux "lemme de Burnside"','Le coloriage du cube' p. 147 in 'Pour la Science' (French edition of 'Scientific American') No.350 December 2006 Paris.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Eric Weisstein's World of Mathematics, Polyhedron Coloring
Index entries for linear recurrences with constant coefficients, signature (7,-21,35,-35,21,-7,1)

Cf. A198833 (unoriented), A093566(n+1) (chiral), A337898 (achiral).
Other elements: A060530 (edges), A000543 (cube vertices, octahedron faces).
Cf. A006008 (tetrahedron), A000545 (dodecahedron faces, icosahedron vertices), A054472 (icosahedron faces, dodecahedron vertices).
Row 3 of A325004 (orthoplex vertices, orthotope facets) and A337887 (orthotope faces, orthoplex peaks).

[(n^6 + 3*n^4 + 12*n^3 + 8*n^2)/24: n in [1..30]]; // Vincenzo Librandi, Apr 27 2012

CoefficientList[Series[x*(1+3*x+8*x^2+16*x^3+2*x^4)/(1-x)^7,{x,0,33}],x] (* Vincenzo Librandi, Apr 27 2012 *)

a(n) = (n^6 + 3*n^4 + 12*n^3 + 8*n^2)/24 = n+8*C(n, 2)+30*C(n, 3)+68*C(n, 4)+75*C(n, 5)+30*C(n, 6). Each term of the RHS indicates the number of ways to use n colors to color the cube faces (octahedron vertices) with exactly 1, 2, 3, 4, 5, or 6 colors.
G.f.: x*(1+3*x+8*x^2+16*x^3+2*x^4)/(1-x)^7. - Colin Barker, Jan 29 2012
a(n) = A198833(n) + A093566(n+1) = 2*A198833(n) - A337898(n) = 2*A093566(n+1) + A337898(n). - Robert A. Russell, Oct 08 2020

Corrected version of A006550 and A006529.

Entry revised by N. J. A. Sloane, Jan 03 2005

A006008 Number of inequivalent ways to color vertices of a regular tetrahedron using <= n colors.

Original entry on oeis.org

0, 1, 5, 15, 36, 75, 141, 245, 400, 621, 925, 1331, 1860, 2535, 3381, 4425, 5696, 7225, 9045, 11191, 13700, 16611, 19965, 23805, 28176, 33125, 38701, 44955, 51940, 59711, 68325, 77841, 88320, 99825, 112421, 126175, 141156, 157435, 175085, 194181, 214800, 237021, 260925, 286595, 314116, 343575

Offset: 0

N. J. A. Sloane, Clint. C. Williams (Clintwill(AT)aol.com)

Here "inequivalent" refers to the rotation group of the tetrahedron, of order 12, with cycle index (x1^4 + 8*x1*x3 + 3*x2^2)/12, which is also the alternating group A_4.
Equivalently, number of distinct tetrahedra that can be obtained by painting its faces using at most n colors. - Lekraj Beedassy, Dec 29 2007
Equals row sums of triangle A144680. - Gary W. Adamson, Sep 19 2008

J.-P. Delahaye, 'Le miraculeux "lemme de Burnside"', 'Le coloriage du tetraedre' pp 147 in 'Pour la Science' (French edition of 'Scientific American') No.350 December 2006 Paris.
Martin Gardner, New Mathematical Diversions from Scientific American. Simon and Schuster, NY, 1966, p. 246.
S. M. Losanitsch, Die Isomerie-Arten bei den Homologen der Paraffin-Reihe, Chem. Ber. 30 (1897), 1917-1926.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
R. Gugisch, A. Kerber, R. Laue, M. Meringer and C. Ruecker, Kombinatorische Chemie, eine Herausforderung für Mathematik und Infomatik, Spektrum 1/02, 64-67, 2002.
S. M. Losanitsch, Die Isomerie-Arten bei den Homologen der Paraffin-Reihe, Chem. Ber. 30 (1897), 1917-1926. (Annotated scanned copy)
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992.
Eric Weisstein's World of Mathematics, Polyhedron Coloring.
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A006550, A060529.
Cf. A144680. - Gary W. Adamson, Sep 19 2008
Cf. A000332(n+3) (unoriented), A000332 (chiral), A006003 (achiral).
Row 3 of A324999.

[(n^4 + 11*n^2 )/12: n in [0..40]]; // Vincenzo Librandi, Aug 12 2011

A006008 := n->1/12*n^2*(n^2+11);
A006008:=-z*(z+1)*(z**2-z+1)/(z-1)**5; # conjectured by Simon Plouffe in his 1992 dissertation

Table[(n^4+11n^2)/12,{n,0,40}] (* or *) LinearRecurrence[{5,-10,10,-5,1},{0,1,5,15,36},40] (* Harvey P. Dale, Aug 11 2011 *)

apply( {A006008(n)=(n^4+11*n^2)/12}, [0..50]) \\ M. F. Hasler, Jan 26 2020

a(n) = (n^4 + 11*n^2)/12. (Replace all x_i's in the cycle index with n.)
Binomial transform of [1, 4, 6, 5, 2, 0, 0, 0, ...]. - Gary W. Adamson, Apr 23 2008
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5), with a(0)=0, a(1)=1, a(2)=5, a(3)=15, a(4)=36. - Harvey P. Dale, Aug 11 2011
a(n) = C(n,1) + 3C(n,2) + 3C(n,3) + 2C(n,4). Each term indicates the number of tetrahedra with exactly 1, 2, 3, or 4 colors. - Robert A. Russell, Dec 03 2014
a(n) = binomial(n+3,4) + binomial(n,4). - Collin Berman, Jan 26 2016
a(n) = A000332(n+3) + A000332(n) = 2*A000332(n+3) - A006003(n) = 2*A000332(n) + A006003(n).
a(n) = A324999(3,n).
E.g.f.: (1/12)*exp(x)*x*(12 + 18*x + 6*x^2 + x^3). - Stefano Spezia, Jan 26 2020
Sum_{n>=1} 1/a(n) = (6 + 22*Pi^2 - 6*sqrt(11)*Pi*coth(sqrt(11)*Pi))/121. - Amiram Eldar, Aug 23 2022

cp's OEIS Frontend

A047780 Number of inequivalent ways to color faces of a cube using at most n colors.

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Magma

Mathematica

Formula

Extensions