cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-7 of 7 results.

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

Original entry on oeis.org

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

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Author

Keywords

Comments

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)

References

  • 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).

Crossrefs

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).

Programs

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

Formula

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

Extensions

Corrected version of A006550 and A006529.
Entry revised by N. J. A. Sloane, Jan 03 2005

A000543 Number of inequivalent ways to color vertices of a cube using at most n colors.

Original entry on oeis.org

0, 1, 23, 333, 2916, 16725, 70911, 241913, 701968, 1798281, 4173775, 8942021, 17930628, 34009053, 61518471, 106823025, 179003456, 290715793, 459239463, 707740861, 1066780100, 1576090341, 2286660783, 3263156073, 4586706576
Offset: 0

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Author

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

Keywords

Comments

Here inequivalent means under the action of the rotation group of the cube, of order 24, which in its action on the vertices has cycle index (x1^8 + 9*x2^4 + 6*x4^2 + 8*x1^2*x3^2)/24.
Also the number of ways to color the faces of a regular octahedron with n colors, counting mirror images separately.
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 vertex (octahedron face) 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_1^8
Vertex rotation 8 x_1^2x_3^2
Edge rotation 6 x_2^4
Small face rotation 6 x_4^2
Large face rotation 3 x_2^4 (End)

References

  • 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).

Crossrefs

Cf. A128766 (unoriented), A337896 (chiral), A337897 (achiral).
Other elements: A060530 (edges), A047780 (cube faces, octahedron vertices).
Cf. A006008 (tetrahedron), A000545 (dodecahedron faces, icosahedron vertices), A054472 (icosahedron faces, dodecahedron vertices).
Row 3 of A325012 (orthotope vertices, orthoplex facets) and A337891 (orthoplex faces, orthotope peaks).

Programs

  • Magma
    [(1/24)*n^2*(n^6+17*n^2+6): n in [0..30]]; // Vincenzo Librandi, Apr 15 2012
  • Maple
    f:= n->(1/24)*n^2*(n^6+17*n^2+6); seq(f(n), n=0..40);
  • Mathematica
    CoefficientList[Series[x*(1+x)*(1+13*x+149*x^2+514*x^3+149*x^4+13*x^5+x^6)/(1-x)^9,{x,0,30}],x] (* Vincenzo Librandi, Apr 15 2012 *)
    Table[(n^8+17n^4+6n^2)/24,{n,0,30}] (* Robert A. Russell, Oct 08 2020 *)

Formula

a(n) = (1/24)*n^2*(n^6+17*n^2+6). (Replace all x_i's in the cycle index with n.)
G.f.: x*(1+x)*(1+13*x+149*x^2+514*x^3+149*x^4+13*x^5+x^6)/(1-x)^9. - Colin Barker, Jan 29 2012
a(n) = 1*C(n,1) + 21*C(n,2) + 267*C(n,3) + 1718*C(n,4) + 5250*C(n,5) + 7980*C(n,6) + 5880*C(n,7) + 1680*C(n,8), where the coefficient of C(n,k) is the number of oriented colorings using exactly k colors.
a(n) = A128766(n) + A337896(n) = 2*A128766(n) - A337897(n) = 2*A337896(n) + A337897(n). - Robert A. Russell, Oct 08 2020

Extensions

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

A252705 The number of ways to color the faces of a regular dodecahedron with n colors, counting mirror images as one.

Original entry on oeis.org

1, 82, 5379, 148648, 2085655, 18356514, 116081245, 574795936, 2359033605, 8345970370, 26180606287, 74354990568, 194253329803, 472634761522, 1081541381145, 2346163937920, 4856060529001, 9641643580530, 18446420258299, 34136541925480, 61303301959263
Offset: 1

Views

Author

Robert A. Russell, Dec 20 2014

Keywords

Comments

The cycle index using the full automorphism group for faces of a dodecahedron is (x1^12+15*x2^6+20*x3^4+24*x1^2*x5^2+15*x1^4*x2^4+x2^6+20*x6^2+24*x2*x10)/120.
Also the number of ways to color the vertices of a regular icosahedron with n colors, counting mirror images as one.

Examples

			For n=2, a(2)=82, the number of ways to color the faces of a regular dodecahedron with two colors, counting mirror images as the same. Of these, two use the same color for all faces, and 80 use both colors.
		

References

  • F. S. Roberts and B. Tesman, Applied Combinatorics, 2d Ed., Pearson Prentice Hall, 2005, pages 439-488.
  • J. H. van Lint and R. M. Wilson, A Course in Combinatorics, Cambridge University Press, 1992, pages 461-474.

Crossrefs

Cf. A000545 (number when mirror images are counted separately).
Cf. A000332 (tetrahedron), A198833 (cube), A128766 (octahedron), A252704 (icosahedron).

Programs

  • Mathematica
    Table[n^2(n^2+1)(n^8-n^6+16n^4+44)/120,{n,1,30}]
  • PARI
    vector(60, n, n^2*(n^2+1)*(n^8-n^6+16*n^4+44)/120) \\ Michel Marcus, Dec 21 2014

Formula

a(n) = n^2*(n^2+1)*(n^8-n^6+16*n^4+44)/120.
G.f.: x*(x+1)*(x^10+68*x^9+4323*x^8+80508*x^7+469548*x^6+886944*x^5+469548*x^4 +80508*x^3+4323*x^2+68*x+1)/(1-x)^13.
a(n) = C(n,1)+80*C(n,2)+5136*C(n,3)+127620*C(n,4)+1395390*C(n,5)+7965948*C(n,6) +26368272*C(n,7)+53438112*C(n,8)+67359600*C(n,9)+51559200*C(n,10)+21954240*C(n,11)+3991680*C(n,12). Each term indicates the number of ways to use n colors to color the dodecahedron with exactly 1, 2, 3, ..., 10, 11, or 12 colors.

A054472 Number of ways to color faces of an icosahedron using at most n colors.

Original entry on oeis.org

0, 1, 17824, 58130055, 18325477888, 1589459765875, 60935989677984, 1329871177501573, 19215358684143616, 202627758536996445, 1666666669200004000, 11212499922098481787, 63895999889747261952, 316749396282749868607, 1394470923827552301472, 5542094550277768379625
Offset: 0

Views

Author

Vladeta Jovovic, May 20 2000

Keywords

Comments

More explicitly, a(n) is the number of colorings with at most n colors of the faces of a regular icosahedron, inequivalent under the action of the rotation group of the icosahedron. It is also the number of inequivalent colorings of the vertices of a regular dodecahedron using at most n colors. - José H. Nieto S., Jan 19 2012
From Robert A. Russell, Oct 19 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 icosahedron face (dodecahedron 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_1^20
Vertex rotation 20 x_1^2x_3^6
Edge rotation 15 x_2^10
Small face rotation 12 x_5^4
Large face rotation 12 x_5^4 (End)

Crossrefs

Cf. A252704 (unoriented), A337959 (chiral), A337960 (achiral), A282670 (edges), A000545 (dodecahedron faces, icosahedron vertices), A006008 (tetrahedron), A047780 (cube faces, octahedron vertices), A000543 (octahedron faces, cube vertices).

Programs

  • Maple
    A054472:=n->(n^20 + 15*n^10 + 20*n^8 + 24*n^4)/60; seq(A054472(n), n=0..15); # Wesley Ivan Hurt, Jan 28 2014
  • Mathematica
    Table[(n^20+15n^10+20n^8+24n^4)/60,{n,0,15}] (* Harvey P. Dale, Nov 04 2011 *)
    LinearRecurrence[{21,-210,1330,-5985,20349,-54264,116280,-203490,293930,-352716,352716,-293930,203490,-116280,54264,-20349,5985,-1330,210,-21,1},{0,1,17824,58130055,18325477888,1589459765875,60935989677984,1329871177501573,19215358684143616,202627758536996445,1666666669200004000,11212499922098481787,63895999889747261952,316749396282749868607,1394470923827552301472,5542094550277768379625,20148763660520129167360,67737190111299199134361,212470603607497593076128,626499557627304397693519,1747626666669235200064000},20] (* Harvey P. Dale, Aug 11 2021 *)

Formula

a(n) = (1/60)*(n^20+15*n^10+20*n^8+24*n^4).
G.f.: -x*(x +1)*(x^18 +17802*x^17 +57738159*x^16 +17050750284*x^15 +1199757591558*x^14 +30128721042672*x^13 +329847884196810*x^12 +1749288479932404*x^11 +4727182539811968*x^10 +6598854419308684*x^9 +4727182539811968*x^8 +1749288479932404*x^7 +329847884196810*x^6 +30128721042672*x^5 +1199757591558*x^4 +17050750284*x^3 +57738159*x^2 +17802*x +1) / (x -1)^21. - Colin Barker, Jul 13 2013
a(n) = 1*C(n,1) + 17822*C(n,2) + 58076586*C(n,3) + 18093064608*C(n,4) + 1498413498750*C(n,5) + 51672950917308*C(n,6) + 936058547290608*C(n,7) + 10194866756893728*C(n,8) + 72644237439379200*C(n,9) + 357895538663241600*C(n,10) + 1264592451488446080*C(n,11) + 3281293750348373760*C(n,12) + 6337930306906598400*C(n,13) + 9157388718839961600*C(n,14) + 9858321678965760000*C(n,15) + 7794071905639219200*C(n,16) + 4394429252269056000*C(n,17) + 1672620130621440000*C(n,18) + 385209484627968000*C(n,19) + 40548366802944000*C(n,20), where the coefficient of C(n,k) is the number of oriented colorings using exactly k colors. - Robert A. Russell, Dec 03 2014
a(n) = A252704(n) + A337959(n) = 2*A252704(n) - A337960(n) = 2*A337959(n) + A337960(n). - Robert A. Russell, Oct 19 2020

Extensions

More terms from James Sellers, May 23 2000
More terms from Colin Barker, Jul 12 2013

A337962 Number of achiral colorings of the 12 pentagonal faces of a regular dodecahedron or the 12 vertices of a regular icosahedron using n or fewer colors.

Original entry on oeis.org

1, 68, 1659, 16464, 97935, 420708, 1443197, 4198720, 10770597, 25016740, 53619335, 107545296, 204013251, 369072900, 640912665, 1074021632, 1744341865, 2755557252, 4246675123, 6401066960, 9457144599, 13720858404
Offset: 1

Views

Author

Robert A. Russell, Oct 03 2020

Keywords

Comments

An achiral coloring is identical to its reflection. 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 automorphism group of the regular dodecahedron/icosahedron that are not in the rotation group. 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 Odd Cycle Indices
Inversion 1 x_2^6
Edge rotation* 15 x_1^4x_2^4 Asterisk indicates that the
Vertex rotation* 20 x_6^2 operation is followed by an
Small face rotation* 12 x_2^1x_10^1 inversion.
Large face rotation* 12 x_2^1x_10^1

Crossrefs

Cf. A000545 (oriented), A252705 (unoriented), A337961 (chiral).
Other elements: A337960 (dodecahedron vertices, icosahedron faces), A337953 (edges).
Other polyhedra: A006003 (tetrahedron), A337898 (cube faces, octahedron vertices), A337897 (octahedron faces, cube vertices).

Programs

  • Mathematica
    Table[(15n^8+n^6+44n^2)/60,{n,30}]

Formula

a(n) = n^2 * (15*n^6 + n^4 + 44)/60.
a(n) = 1*C(n,1) + 66*C(n,2) + 1458*C(n,3) + 10232*C(n,4) + 31530*C(n,5) + 47892*C(n,6) + 35280*C(n,7) + 10080*C(n,8), where the coefficient of C(n,k) is the number of achiral colorings using exactly k colors.
a(n) = 2*A252705(n) - A000545(n) = A000545(n) - 2*A337961(n) = A252705(n) - A337961(n).
From Stefano Spezia, Oct 04 2020: (Start)
G.f.: x*(1+59*x+1083*x^2+3897*x^3+3087*x^4+1083*x^5+59*x^6+x^7)/(1-x)^9.
a(n) = 9*a(n-1)-36*a(n-2)+84*a(n-3)-126*a(n-4)+126*a(n-5)-84*a(n-6)+36*a(n-7)-9*a(n-8)+a(n-8) for n > 8.
(End)

A282670 Number of inequivalent ways to color the edges of a dodecahedron using at most n colors.

Original entry on oeis.org

0, 1, 17912448, 3431529649899, 19215359484207104, 15522042948408209375, 3684565329384186949248, 375655671519845961645597, 20632333988160040350515200, 706519304587399981447927557, 16666666666669166670000400000, 290823371148118276083759139095
Offset: 0

Views

Author

David Nacin, Feb 20 2017

Keywords

Comments

Cycle index of symmetry group A5 acting on the 30 edges of the dodecahedron is (24s(5)^6 + 20s(3)^10 + 15s(2)^14*s(1)^2 + s(1)^30)/60.
Also the number of inequivalent ways to color the edges of the icosahedron using at most n colors.
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 edge 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_1^30
Edge rotation 15 x_1^2x_2^14
Vertex rotation 20 x_3^10
Small face rotation 12 x_5^6
Large face rotation 12 x_5^6 (End)

Examples

			There are a(2) = 17912448 inequivalent ways to color the edges of the dodecahedron using at most two colors.
		

Crossrefs

Other elements: A054472 (dodecahedron vertices, icosahedron faces), A000545 (dodecahedron faces, icosahedron vertices).
Other polyhedra: A046023 (tetrahedron), A060530 (cube/octahedron).
Cf. A337963 (unoriented), A337964 (chiral), A337953 (achiral).

Programs

  • Mathematica
    Table[(24n^6+20n^10+15n^16+n^30)/60, {n, 0, 16}]

Formula

a(n) = n^6 (n^24 + 15 n^10 + 20 n^4 + 24)/60.
G.f.: x*(1 + x)*(1 + 17912416*x + 3430956452060*x^2 + 19105559437892000*x^3 + 14908856825730677891*x^4 + 3197392859155796794496*x^5 + 265368238349945588707496*x^6 + 10365795256050146806088576*x^7 + 215154060506484358838662001*x^8 + 2568188846096433625477331936*x^9 + 18582986600475456162494990756*x^10 + 84400699070086923625163495456*x^11 + 245956255494355672481225103371*x^12 + 465612713610802763378946154496*x^13 + 575747234318647571242943474096*x^14 + 465612713610802763378946154496*x^15 + 245956255494355672481225103371*x^16 + 84400699070086923625163495456*x^17 + 18582986600475456162494990756*x^18 + 2568188846096433625477331936*x^19 + 215154060506484358838662001*x^20 + 10365795256050146806088576*x^21 + 265368238349945588707496*x^22 + 3197392859155796794496*x^23 + 14908856825730677891*x^24 + 19105559437892000*x^25 + 3430956452060*x^26 + 17912416*x^27 + x^28) / (1 - x)^31. - Colin Barker, Mar 30 2019
From Robert A. Russell, Oct 03 2020: (Start)
a(n) = 1*C(n,1) + 17912446*C(n,2) + 3431475912558*C(n,3) + 19201633473082192*C(n,4) + 15426000466104548370*C(n,5) + 3591721233455676488292*C(n,6) + 350189004698594439734160*C(n,7) + 17729388555701917767855840*C(n,8) + 534044352737570253478824960*C(n,9) + 10485619820879148545218980480*C(n,10) + 143066535726280748444739676800*C(n,11) + 1420876074163106703694904352000*C(n,12) + 10631861498419617103267350931200*C(n,13) + 61515486939441778743810979468800*C(n,14) + 280711222366395106969585943040000*C(n,15) + 1025499893865270227589218761728000*C(n,16) + 3032858772294885663526454593536000*C(n,17) + 7319173455487770465200322686976000*C(n,18) + 14487618384525410959295952691200000*C(n,19) + 23580333216029318427870396825600000*C(n,20) + 31555723729541430372276884520960000*C(n,21) + 34619561317726617824610327429120000*C(n,22) + 30946535969611314628728933580800000*C(n,23) + 22311118596400512968549479219200000*C(n,24) + 12771433990957347267674112000000000*C(n,25) + 5668281691036644651075462758400000*C(n,26) + 1879979643918904128084836352000000*C(n,27) + 438404032189593555246120960000000*C(n,28) + 64102774454612839170441216000000*C(n,29) + 4420880996869850977271808000000*C(n,30), where the coefficient of C(n,k) is the number of oriented colorings using exactly k colors.
a(n) = A337963(n) + A337964(n) = 2*A337963(n) - A337953(n) = 2*A337964(n) + A337953(n). (End)

A337961 Number of chiral pairs of colorings of the 12 pentagonal faces of a regular dodecahedron or the 12 vertices of a regular icosahedron using n or fewer colors.

Original entry on oeis.org

0, 14, 3720, 132184, 1987720, 17935806, 114638048, 570597216, 2348263008, 8320953630, 26126986952, 74247445272, 194049316552, 472265688622, 1080900468480, 2345089916288, 4854316187136, 9638888023278, 18442173583176
Offset: 1

Views

Author

Robert A. Russell, Oct 03 2020

Keywords

Comments

Each member of a chiral pair is a reflection, but not a rotation, of the other. The Schläfli symbols for the regular icosahedron and regular dodecahedron are {3,5} and {5,3} respectively. They are mutually dual.

Crossrefs

Cf. A000545 (oriented), A252705 (unoriented), A337962 (achiral).
Other elements: A337959 (dodecahedron vertices, icosahedron faces), A337964 (edges).
Other polyhedra: A000332 (tetrahedron), A093566(n+1) (cube faces, octahedron vertices), A337896 (octahedron faces, cube vertices).

Programs

  • Mathematica
    Table[(n^12-15n^8+14n^6+44n^4-44n^2)/120,{n,30}]
    LinearRecurrence[{13,-78,286,-715,1287,-1716,1716,-1287,715,-286,78,-13,1},{0,14,3720,132184,1987720,17935806,114638048,570597216,2348263008,8320953630,26126986952,74247445272,194049316552},20] (* Harvey P. Dale, Nov 17 2024 *)

Formula

a(n) = (n-1) * n^2 * (n+1) * (n^8 + n^6 - 14*n^4 + 44) / 120.
a(n) = 14*C(n,2) + 3678*C(n,3) + 117388*C(n,4) + 1363860*C(n,5) + 7918056*C(n,6) + 26332992*C(n,7) + 53428032*C(n,8) + 67359600*C(n,9) + 51559200*C(n,10) + 21954240*C(n,11) + 3991680*C(n,12), where the coefficient of C(n,k) is the number of chiral pairs of colorings using exactly k colors.
a(n) = A000545(n) - A252705(n) = (A000545(n) - A337962(n)) / 2 = A252705(n) - A337962(n).
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