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 1-8 of 8 results.

A338967 Number of achiral colorings of the 120 dodecahedral facets of the 4-D 120-cell (or 120 vertices of the 4-D 600-cell) using subsets of a set of n colors.

Original entry on oeis.org

1, 314843647550280564736, 5068890957390271123224826359979956, 11893730816857265534982913331475052373213184, 220581496716947452381892465686737251285705566406250
Offset: 1

Views

Author

Robert A. Russell, Dec 04 2020

Keywords

Comments

An achiral coloring is identical to its reflection. The Schläfli symbols of the 120-cell and 600-cell are {5,3,3} and {3,3,5} respectively. They are mutually dual. There are 7200 elements in the automorphism group of the 120-cell that are not in its rotation group. They divide into 9 conjugacy classes. The first formula is obtained by averaging the vertex (or facet) cycle indices after replacing x_i^j with n^j according to the Pólya enumeration theorem.
Count Odd Cycle Indices Count Odd Cycle Indices
60 x_1^30x_2^45 1200 x_1^2x_2^2x_6^19
60 x_1^2x_2^59 720+720 x_2^5x_5^6x_10^8
1800 x_2^2x_4^29 720+720 x_1^2x_2^4x_10^11
1200 x_2^3x_3^10x_6^14
Sequences for other elements of the 120-cell and 600-cell are not suitable for the OEIS as the first significant datum is too big. We provide formulas here.
For the 600 facets of the 600-cell (vertices of the 120-cell), the cycle indices are:
Count Odd Cycle Indices Count Odd Cycle Indices
60 x_1^60x_2^270 1200 x_2^6x_6^98
60 x_2^300 720+720 x_5^12x_10^54
1800 x_1^2x_2^1x_4^149 720+720 x_10^60
1200 x_2^6x_3^20x_6^88
The formula is (24*n^60 + 24*n^66 + 20*n^104 + 20*n^114 + 30*n^152 + n^300 + n^330) / 120.
For the 720 pentagonal faces of the 120-cell (edges of the 600-cell), the cycle indices are:
Count Odd Cycle Indices Count Odd Cycle Indices
60 x_1^72x_2^324 1200 x_6^120
60 x_2^360 720+720 x_1^2x_2^4x_5^14x_10^64
1800 x_2^4x_4^178 720+720 x_2^5x_10^71
1200 x_3^24x_6^108
The formula is (24*n^76 + 24*n^84 + 20*n^120 + 20*n^132 + 30*n^182 + n^360 + n^396) / 120.
For the 1200 edges of the 120-cell (triangular faces of the 600-cell), the cycle indices are:
Count Odd Cycle Indices Count Odd Cycle Indices
60 x_1^80x_2^560 1200 x_2^3x_6^199
60 x_2^600 720+720 x_5^16x_10^112
1800 x_2^4x_4^298 720+720 x_10^120
1200 x_1^2x_2^2x_3^26x_6^186
The formula is (24*n^120 + 24*n^128 + 20*n^202 + 20*n^216 + 30*n^302 + n^600 + n^640) / 120.

Crossrefs

Cf. A338964 (oriented), A338965 (unoriented), A338966 (chiral), A338983 (exactly n colors), A132366 (5-cell), A337955 (8-cell vertices, 16-cell facets), A337958(16-cell vertices, 8-cell facets), A338951 (24-cell).

Programs

  • Mathematica
    Table[(24n^17+24n^19+20n^23+20n^27+30n^31+n^61+n^75)/120,{n,10}]
  • PARI
    a(n)=(24*n^17+24*n^19+20*n^23+20*n^27+30*n^31+n^61+n^75)/120 \\ Charles R Greathouse IV, Jul 05 2024

Formula

a(n) = (24*n^17 + 24*n^19 + 20*n^23 + 20*n^27 + 30*n^31 + n^61 + n^75) / 120.
a(n) = Sum_{j=1..Min(n,75)} A338983(n) * binomial(n,j).
a(n) = 2*A338965(n) - A338964(n) =(A338964(n) - 2*A338966(n)) / 2 = A338965(n) - A338966(n).

A132366 Partial sum of centered tetrahedral numbers A005894.

Original entry on oeis.org

1, 6, 21, 56, 125, 246, 441, 736, 1161, 1750, 2541, 3576, 4901, 6566, 8625, 11136, 14161, 17766, 22021, 27000, 32781, 39446, 47081, 55776, 65625, 76726, 89181, 103096, 118581, 135750, 154721, 175616, 198561, 223686, 251125, 281016, 313501, 348726, 386841
Offset: 0

Views

Author

Jonathan Vos Post, Nov 09 2007

Keywords

Comments

From Robert A. Russell, Oct 09 2020: (Start)
a(n-1) is the number of achiral colorings of the 5 tetrahedral facets (or vertices) of a regular 4-dimensional simplex using n or fewer colors. An achiral arrangement is identical to its reflection. The 4-dimensional simplex is also called a 5-cell or pentachoron. Its Schläfli symbol is {3,3,3}.
There are 60 elements in the automorphism group of the 4-dimensional simplex that are not in its rotation group. Each is an odd permutation of the vertices and can be associated with a partition of 5 based on the conjugacy class of the permutation. The first formula for a(n-1) is obtained by averaging their cycle indices after replacing x_i^j with n^j according to the Pólya enumeration theorem.
Partition Count Odd Cycle Indices
41 30 x_1x_4^1
32 20 x_2^1x_3^1
2111 10 x_1^3x_2^1 (End)

Crossrefs

Cf. A337895 (oriented), A000389(n+4) (unoriented), A000389 (chiral), A331353 (5-cell edges, faces), A337955 (8-cell vertices, 16-cell facets), A337958 (16-cell vertices, 8-cell facets), A338951 (24-cell), A338967 (120-cell, 600-cell).
a(n-1) = A325001(4,n).

Programs

  • Mathematica
    Do[Print[n, " ", (n^4 + 4 n^3 + 11 n^2 + 14 n + 6)/6 ], {n, 0, 10000}]
    Accumulate[Table[(2n+1)(n^2+n+3)/3,{n,0,40}]] (* or *) LinearRecurrence[ {5,-10,10,-5,1},{1,6,21,56,125},40] (* Harvey P. Dale, Feb 26 2020 *)

Formula

a(n) = (n^4 + 4*n^3 + 11*n^2 + 14*n + 6)/6 = (n^2+2*n+6)*(n+1)^2/6.
G.f.: -(x+1)*(x^2+1) / (x-1)^5. - Colin Barker, May 04 2013
From Robert A. Russell, Oct 09 2020: (Start)
a(n-1) = n^2 * (5 + n^2) / 6.
a(n-1) = binomial(n+4,5) - binomial(n,5) = A000389(n+4) - A000389(n).
a(n-1) = 1*C(n,1) + 4*C(n,2) + 6*C(n,3) + 4*C(n,4), where the coefficient of C(n,k) is the number of achiral colorings using exactly k colors.
a(n-1) = 2*A000389(n+4) - A337895(n) = A337895(n) - 2*A000389(n) .
G.f. for a(n-1): x * (x+1) * (x^2+1) / (1-x)^5. (End)
From Amiram Eldar, Feb 14 2023: (Start)
Sum_{n>=0} 1/a(n) = Pi^2/5 + 3/25 - 3*Pi*coth(sqrt(5)*Pi)/(5*sqrt(5)).
Sum_{n>=0} (-1)^n/a(n) = Pi^2/10 - 3/25 + 3*Pi*cosech(sqrt(5)*Pi)/(5*sqrt(5)). (End)
a(n) = A006007(n) + A006007(n+1) = A002415(n) + A002415(n+2). - R. J. Mathar, Jun 05 2025

Extensions

Corrected offset, Mathematica program by Tomas J. Bulka (tbulka(AT)rodincoil.com), Sep 02 2009

A234249 Number of ways to choose 4 points in an n X n X n triangular grid.

Original entry on oeis.org

15, 210, 1365, 5985, 20475, 58905, 148995, 341055, 720720, 1426425, 2672670, 4780230, 8214570, 13633830, 21947850, 34389810, 52602165, 78738660, 115584315, 166695375, 236561325, 330791175, 456326325, 621682425, 837222750, 1115465715, 1471429260, 1923014940
Offset: 3

Views

Author

Heinrich Ludwig, Feb 02 2014

Keywords

Comments

Sequence is column #5 of A084546: a(n) = A084546(n+1, 4).
All elements of the sequence are multiples of 15.
a(n-1) is the number of chiral pairs of colorings of the 8 cubic facets of a tesseract (hypercube) with Schläfli symbol {4,3,3} or of the 8 vertices of a hyperoctahedron with Schläfli symbol {3,3,4}. Both figures are regular 4-D polyhedra and they are mutually dual. Each member of a chiral pair is a reflection, but not a rotation, of the other. - Robert A. Russell, Oct 20 2020

Crossrefs

Cf. A084546, A050534 (number of ways to choose 2 points), A093566 (3 points), A231653.
Cf. A337956 (oriented), A337956 (unoriented), A337956 (achiral) colorings, A331356 (hyperoctahedron edges, tesseract faces), A331360 (hyperoctahedron faces, tesseract edges), A337954 (hyperoctahedron facets, tesseract vertices).
Other polychora: A000389 (5-cell), A338950 (24-cell), A338966 (120-cell, 600-cell).
Row 4 of A325006 (orthotope facets, orthoplex vertices).

Programs

  • Maple
    A234249:=n->n*(n + 1)*(n - 1)*(n + 2)*(n - 2)*(n + 3)*(n^2 + n - 4)/384: seq(A234249(n), n=3..40); # Wesley Ivan Hurt, Jan 10 2017
  • Mathematica
    Table[Binomial[Binomial[n,2],4], {n,4,30}] (* Robert A. Russell, Oct 20 2020 *)
  • PARI
    Vec(-15*x^3*(x^2+5*x+1)/(x-1)^9 + O(x^100)) \\ Colin Barker, Feb 02 2014

Formula

a(n) = n*(n + 1)*(n - 1)*(n + 2)*(n - 2)*(n + 3)*(n^2 + n - 4)/384.
a(n) = C(C(n + 1, 2), 4).
G.f.: -15*x^3*(x^2+5*x+1) / (x-1)^9. - Colin Barker, Feb 02 2014
From Robert A. Russell, Oct 20 2020: (Start)
a(n-1) = 15*C(n,4) + 135*C(n,5) + 330*C(n,6) + 315*C(n,7) + 105*C(n,8), where the coefficient of C(n,k) is the number of chiral pairs of colorings using exactly k colors.
a(n-1) = A337956(n) - A337957(n) = (A337956(n) - A337958(n)) / 2 = A337957(n) - A337958(n).
a(n-1) = A325006(4,n). (End)

A338951 Number of achiral colorings of the 24 octahedral facets (or 24 vertices) of the 4-D 24-cell using subsets of a set of n colors.

Original entry on oeis.org

1, 6504, 8416440, 1455789440, 80139247500, 2125945744776, 34026498820524, 376045864704000, 3131319814422255, 20854395850585000, 115919421344402676, 554976171149122944, 2343894146343268610, 8896568181794053320
Offset: 1

Views

Author

Robert A. Russell, Nov 17 2020

Keywords

Comments

An achiral coloring is identical to its reflection. The Schläfli symbol of the 24-cell is {3,4,3}. It is self-dual. There are 576 elements in the automorphism group of the 24-cell that are not in its rotation group. They divide into 10 conjugacy classes. The first formula is obtained by averaging the vertex (or facet) cycle indices after replacing x_i^j with n^j according to the Pólya enumeration theorem.
Count Odd Cycle Indices Count Odd Cycle Indices
12 x_1^12x_2^6 72 x_2^2x_4^5
12 x_1^6x_2^9 96 x_1^2x_2^2x_6^3
12 x_1^2x_2^11 96 x_2^3x_3^2x_6^2
12 x_2^12 96 x_3^4x_6^2
72 x_1^2x_2^1x_4^5 96 x_6^4

Crossrefs

Cf. A338948 (oriented), A338949 (unoriented), A338950 (chiral), A338955 (edges, faces), A132366 (5-cell), A337955 (8-cell vertices, 16-cell facets), A337958 (16-cell vertices, 8-cell facets), A338967 (120-cell, 600-cell).

Programs

  • Mathematica
    Table[(8n^4+8n^6+22n^7+6n^8+n^12+n^13+n^15+n^18)/48,{n,15}]

Formula

a(n) = (8*n^4 + 8*n^6 + 22*n^7 + 6*n^8 + n^12 + n^13 + n^15 + n^18) / 48.
a(n) = 1*C(n,1) + 6502*C(n,2) + 8396931*C(n,3) + 1422162700*C(n,4) + 72944399665*C(n,5) + 1666778870130*C(n,6) + 20777144613015*C(n,7) + 158973991255800*C(n,8) + 803196369526320*C(n,9) + 2806639981714800*C(n,10) + 6979192091902800*C(n,11) + 12538220293368000*C(n,12) + 16327662245294400*C(n,13) + 15272334392515200*C(n,14) + 10003736158416000*C(n,15) + 4357170994176000*C(n,16) + 1133753677056000*C(n,17) + 133382785536000*C(n,18), where the coefficient of C(n,k) is the number of achiral colorings using exactly k colors.
a(n) = 2*A338949(n) - A338948(n) = A338948(n) - 2*A338950(n) = A338949(n) - A338950(n).

A337955 Number of achiral colorings of the 16 tetrahedral facets of a hyperoctahedron or of the 16 vertices of a tesseract.

Original entry on oeis.org

1, 308, 34128, 1056576, 15303750, 136236276, 865711763, 4296782848, 17656466751, 62510672500, 196174554026, 557301826368, 1456216515468, 3543525156276, 8109415963125, 17592637669376, 36414622551373
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 tesseract and the hyperoctahedron are {4,3,3} and {3,3,4} respectively. Both figures are regular 4-D polyhedra and they are mutually dual. There are 192 elements in the automorphism group of the tesseract that are not in its rotation group. Each involves a permutation of the axes that can be associated with a partition of 4 based on the conjugacy class of the permutation. This table shows the hyperoctahedron facet (tesseract vertex) cycle indices for each member of such a class. The first formula is obtained by averaging these cycle indices after replacing x_i^j with n^j according to the Pólya enumeration theorem.
Partition Count Odd Cycle Indices
4 6 8x_1^2x_2^1x_4^3
31 8 8x_2^2x_6^2
22 3 8x_4^4
211 6 2x_1^8x_2^4 + 2x_2^8 + 4x_4^4
1111 1 8x_2^8

Crossrefs

Cf. A337952 (oriented), A128767 (unoriented), A337954 (chiral).
Other elements: A331361 (tesseract edges, hyperoctahedron faces), A331357 (tesseract faces, hyperoctahedron edges), A337958 (tesseract facets, hyperoctahedron vertices).
Other polychora: A132366(n-1) (4-simplex facets/vertices), A338951 (24-cell), A338967 (120-cell, 600-cell).
Row 4 of A325015 (orthoplex facets, orthotope vertices).

Programs

  • Mathematica
    Table[(3n^12+5n^8+12n^6+28n^4)/48,{n,30}]

Formula

a(n) = n^4 * (3*n^8 + 5*n^4 + 12*n^2 + 28) / 48.
a(n) = 1*C(n,1) + 306*C(n,2) + 33207*C(n,3) + 921908*C(n,4) + 10359075*C(n,5) + 59584470*C(n,6) + 197644440*C(n,7) + 400752240*C(n,8) + 505197000*C(n,9) + 386694000*C(n,10) + 164656800*C(n,11) + 29937600*C(n,12), where the coefficient of C(n,k) is the number of achiral colorings using exactly k colors.
a(n) = 2*A128767(n) - A337952(n) = A337952(n) - 2*A337954(n) = A128767(n) - A337954(n).

A337956 Number of oriented colorings of the 8 cubic facets of a tesseract or of the 8 vertices of a hyperoctahedron.

Original entry on oeis.org

1, 15, 126, 730, 3270, 11991, 37450, 102726, 253485, 573265, 1205556, 2384460, 4475926, 8031765, 13858860, 23106196, 37372545, 58837851, 90421570, 135971430, 200486286, 290376955, 413769126, 580852650, 804281725
Offset: 1

Views

Author

Robert A. Russell, Oct 03 2020

Keywords

Comments

Each chiral pair is counted as two when enumerating oriented arrangements. The Schläfli symbols for the tesseract and the hyperoctahedron are {4,3,3} and {3,3,4} respectively. Both figures are regular 4-D polyhedra and they are mutually dual.

Crossrefs

Cf. A337957 (unoriented), A234249(n+1) (chiral), A337958 (achiral).
Other elements: A331354 (hyperoctahedron edges, tesseract faces), A331358 (hyperoctahedron faces, tesseract edges), A337952 (hyperoctahedron facets, tesseract vertices).
Other polychora: A337895 (5-cell), A338948 (24-cell), A338964 (120-cell, 600-cell).
Row 4 of A325004 (orthotope facets, orthoplex vertices).

Programs

  • Mathematica
    Table[Binomial[Binomial[n+1,2]+3,4] + Binomial[Binomial[n,2],4],{n,30}]

Formula

a(n) = binomial(binomial(n+1,2)+3,4) + binomial(binomial(n,2),4).
a(n) = n * (n+1) * (n^6 - n^5 + 7*n^4 + 29*n^3 + 16*n^2 - 4*n + 48) / 192.
a(n) = 1*C(n,1) + 13*C(n,2) + 84*C(n,3) + 312*C(n,4) + 735*C(n,5) + 1020*C(n,6) + 735*C(n,7) + 210*C(n,8), where the coefficient of C(n,k) is the number of oriented colorings using exactly k colors.
a(n) = A337957(n) + A234249(n+1) = 2*A337957(n) - A337958(n) = 2*A234249(n+1) + A337958(n).
From Stefano Spezia, Oct 04 2020: (Start)
G.f.: x*(1 + 6*x + 27*x^2 + 52*x^3 + 102*x^4 + 21*x^5 + x^6)/(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)

A337957 Number of unoriented colorings of the 8 cubic facets of a tesseract or of the 8 vertices of a hyperoctahedron.

Original entry on oeis.org

1, 15, 126, 715, 3060, 10626, 31465, 82251, 194580, 424270, 864501, 1663740, 3049501, 5359095, 9078630, 14891626, 23738715, 36890001, 56031760, 83369265, 121747626, 174792640, 247073751, 344291325, 473490550, 643304376
Offset: 1

Views

Author

Robert A. Russell, Oct 03 2020

Keywords

Comments

Each chiral pair is counted as one when enumerating unoriented arrangements. The Schläfli symbols for the tesseract and the hyperoctahedron are {4,3,3} and {3,3,4} respectively Both figures are regular 4-D polyhedra and they are mutually dual.

Crossrefs

Cf. A337956 (oriented), A234249(n+1) (chiral), A337958 (achiral).
Other elements: A331355 (hyperoctahedron edges, tesseract faces), A331359 (hyperoctahedron faces, tesseract edges), A128767 (hyperoctahedron facets, tesseract vertices).
Other polychora: A000389(n+4) (5-cell), A338949 (24-cell), A338965 (120-cell, 600-cell).
Row 4 of A325005 (orthotope facets, orthoplex vertices).

Programs

  • Mathematica
    Table[Binomial[Binomial[n+1,2]+3,4],{n,30}]

Formula

a(n) = binomial(binomial(n+1,2)+3,4).
a(n) = n * (n+1) * (n^2 + n + 2) * (n^2 + n + 4) * (n^2 + n + 6) / 384.
a(n) = 1*C(n,1) + 13*C(n,2) + 84*C(n,3) + 297*C(n,4) + 600*C(n,5) + 690*C(n,6) + 420*C(n,7) + 105*C(n,8), where the coefficient of C(n,k) is the number of unoriented colorings using exactly k colors.
a(n) = A337956(n) - A234249(n+1) = (A337956(n) + A337958(n)) / 2 = A234249(n+1) + A337958(n).
From Stefano Spezia, Oct 04 2020: (Start)
G.f.: x*(1 + 6*x + 27*x^2 + 37*x^3 + 27*x^4 + 6*x^5 + x^6)/(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)

A338983 Number of chiral pairs of colorings of the 120 dodecahedral facets of the 4-D 120-cell (or 120 vertices of the 4-D 600-cell) using exactly n colors.

Original entry on oeis.org

0, 1, 314843647550280564734, 5068890957389326592282175518285751, 11893730796581701705423717900461048616681772, 220581437248293418784474364671733389683204494492535
Offset: 0

Views

Author

Robert A. Russell, Dec 13 2020

Keywords

Comments

An achiral coloring is identical to its reflection. The Schläfli symbols of the 120-cell and 600-cell are {5,3,3} and {3,3,5} respectively. They are mutually dual. For n>75, a(n) = 0.
Sequences for other elements of the 120-cell and 600-cell are not suitable for the OEIS as the first significant datum is too big. We provide generating functions here using bp(j) = Sum_{k=1..j} k! * S2(j,k) * x^k.
For the 600 facets of the 600-cell (vertices of the 120-cell), the generating function is bp(60)/5 + bp(66)/5 + bp(104)/6 + bp(114)/6 + bp(152)/4 + bp(300)/120 + bp(330)/120.
For the 720 pentagonal faces of the 120-cell (edges of the 600-cell), the generating function is bp(76)/5 + bp(84)/5 + bp(120)/6 + bp(132)/6 + bp(182)/4 + bp(360)/120 + bp(396)/120.
For the 1200 edges of the 120-cell (triangular faces of the 600-cell), the generating function is bp(120)/5 + bp(128)/5 + bp(202)/6 + bp(216)/6 + bp(302)/4 + bp(600)/120 + bp(640)/120.

Crossrefs

Cf. A338980 (oriented), A338981 (unoriented), A338982 (chiral), A338967 (up to n colors), A132366 (5-cell), A337955 (8-cell vertices, 16-cell facets), A337958 (16-cell vertices, 8-cell facets), A338951 (24-cell).

Programs

  • Mathematica
    bp[j_] := Sum[k! StirlingS2[j, k] x^k, {k, j}] (*binomial series*)
    CoefficientList[bp[17]/5+bp[19]/5+bp[23]/6+bp[27]/6+bp[31]/4+bp[61]/120+bp[75]/120,x]

Formula

A338967(n) = Sum_{j=1..Min(n,75)} a(n) * binomial(n,j).
a(n) = 2*A338981(n) - A338980(n) = A338980(n) - 2*A338982(n) = A338981(n) - A338982(n).
G.f.: bp(17)/5 + bp(19)/5 + bp(23)/6 + bp(27)/6 + bp(31)/4 + bp(61)/120 + bp(75)/120, where bp(j) = Sum_{k=1..j} k! * S2(j,k) * x^k and S2(j,k) is the Stirling subset number, A008277.
Showing 1-8 of 8 results.