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 1-6 of 6 results.

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

Views

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

A337883 Array read by descending antidiagonals: T(n,k) is the number of oriented colorings of the triangular faces of a regular n-dimensional simplex using k or fewer colors.

Original entry on oeis.org

1, 2, 1, 3, 5, 1, 4, 15, 40, 1, 5, 36, 1197, 3504, 1, 6, 75, 18592, 9753615, 13724608, 1, 7, 141, 166885, 3056311808, 19854224207910, 3574466244480, 1, 8, 245, 1019880, 264940140875, 468488921670219776, 25959704193068472575379, 106607224611810055168, 1
Offset: 2

Views

Author

Robert A. Russell, Sep 28 2020

Keywords

Comments

Each chiral pair is counted as two when enumerating oriented arrangements. An n-simplex has n+1 vertices. For n=2, the figure is a triangle with one triangular face. For n=3, the figure is a tetrahedron with 4 triangular faces. For higher n, the number of triangular faces is C(n+1,3).
Also the number of oriented colorings of the peaks of a regular n-dimensional simplex. A peak of an n-simplex is an (n-3)-dimensional simplex.

Examples

			The table begins with T(2,1):
 1    2       3          4            5              6               7 ...
 1    5      15         36           75            141             245 ...
 1   40    1197      18592       166885        1019880         4738153 ...
 1 3504 9753615 3056311808 264940140875 10156268150064 221646915632373 ...
For T(3,4)=36, the 34 achiral arrangements are AAAA, AAAB, AAAC, AAAD, AABB, AABC, AABD, AACC, AACD, AADD, ABBB, ABBC, ABBD, ABCC, ABDD, ACCC, ACCD, ACDD, ADDD, BBBB, BBBC, BBBD, BBCC, BBCD, BBDD, BCCC, BCCD, BCDD, BDDD, CCCC, CCCD, CCDD, CDDD, and DDDD. The chiral pair is ABCD-ABDC.
		

Crossrefs

Cf. A337884 (unoriented), A337885 (chiral), A337886 (achiral), A051168 (binary Lyndon words).
Other elements: A324999 (vertices), A327083 (edges).
Other polytopes: A337887 (orthotope), A337891 (orthoplex).
Rows 2-4 are A000027, A006008, A331350.

Programs

  • Mathematica
    m=2; (* dimension of color element, here a triangular face *)
    lw[n_,k_]:=lw[n, k]=DivisorSum[GCD[n,k],MoebiusMu[#]Binomial[n/#,k/#]&]/n (*A051168*)
    cxx[{a_, b_},{c_, d_}]:={LCM[a, c], GCD[a, c] b d}
    compress[x:{{, } ...}] := (s=Sort[x];For[i=Length[s],i>1,i-=1,If[s[[i,1]]==s[[i-1,1]], s[[i-1,2]]+=s[[i,2]]; s=Delete[s,i], Null]]; s)
    combine[a : {{, } ...}, b : {{, } ...}] := Outer[cxx, a, b, 1]
    CX[p_List, 0] := {{1, 1}} (* cycle index for partition p, m vertices *)
    CX[{n_Integer}, m_] := If[2m>n, CX[{n}, n-m], CX[{n},m] = Table[{n/k, lw[n/k, m/k]}, {k, Reverse[Divisors[GCD[n, m]]]}]]
    CX[p_List, m_Integer] := CX[p, m] = Module[{v = Total[p], q, r}, If[2 m > v, CX[p, v - m], q = Drop[p, -1]; r = Last[p]; compress[Flatten[Join[{{CX[q, m]}}, Table[combine[CX[q, m - j], CX[{r}, j]], {j, Min[m, r]}]], 2]]]]
    pc[p_] := Module[{ci, mb}, mb = DeleteDuplicates[p]; ci = Count[p, #] &/@ mb; Total[p]!/(Times @@ (ci!) Times @@ (mb^ci))] (* partition count *)
    row[n_Integer] := row[n] = Factor[Total[If[EvenQ[Total[1-Mod[#, 2]]], pc[#] j^Total[CX[#, m+1]][[2]], 0] & /@ IntegerPartitions[n+1]]/((n+1)!/2)]
    array[n_, k_] := row[n] /. j -> k
    Table[array[n,d+m-n], {d,8}, {n,m,d+m-1}] // Flatten

Formula

The algorithm used in the Mathematica program below assigns each permutation of the vertices to a partition of n+1. It then determines the number of permutations for each partition and the cycle index for each partition using a formula for binary Lyndon words. If the value of m is increased, one can enumerate colorings of higher-dimensional elements beginning with T(m,1).
T(n,k) = A337884(n,k) + A337885(n,k) = 2*A337884(n,k) - A337886(n,k) = 2*A337885(n,k) + A337886(n,k).

A337887 Array read by descending antidiagonals: T(n,k) is the number of oriented colorings of the square faces of a regular n-dimensional orthotope (hypercube) using k or fewer colors.

Original entry on oeis.org

1, 2, 1, 3, 10, 1, 4, 57, 90054, 1, 5, 240, 1471640157, 629648865588086369152, 1, 6, 800, 1466049174160, 76983765319971901895960429658208179, 76686070519895153193719509580895099970955878067526648007224125292544, 1
Offset: 2

Views

Author

Robert A. Russell, Sep 28 2020

Keywords

Comments

Each chiral pair is counted as two when enumerating oriented arrangements. Each face is a square bounded by four edges. For n=2, the figure is a square with one face. For n=3, the figure is a cube with 6 faces. For n=4, the figure is a tesseract with 24 faces. The number of faces is 2^(n-2)*C(n,2).
Also the number of oriented colorings of peaks of an n-dimensional orthoplex. A peak is an (n-3)-dimensional simplex.

Examples

			Array begins with T(2,1):
 1     2          3             4               5                 6 ...
 1    10         57           240             800              2226 ...
 1 90054 1471640157 1466049174160 310441584462375 24679078461920106 ...
		

Crossrefs

Cf. A337888 (unoriented), A337889 (chiral), A337890 (achiral).
Other elements: A325012 (vertices), A337407 (edges).
Other polytopes: A337883 (simplex), A337891 (orthoplex).
Rows 2-4 are A000027, A047780, A331354.

Programs

  • Mathematica
    m = 2;(* dimension of color element, here a square face *)
    Fi1[p1_] := Module[{g, h}, Coefficient[Product[g = GCD[k1, p1]; h = GCD[2 k1, p1]; (1 + 2 x^(k1/g))^(r1[[k1]] g) If[Divisible[k1, h], 1, (1+2x^(2 k1/h))^(r2[[k1]] h/2)], {k1, Flatten[Position[cs, n1_ /; n1 > 0]]}], x, n - m]];
    FiSum[] := (Do[Fi2[k2] = Fi1[k2], {k2, Divisors[per]}];DivisorSum[per, DivisorSum[d1 = #, MoebiusMu[d1/#] Fi2[#] &]/# &]);
    CCPol[r_List] := (r1 = r; r2 = cs - r1; If[EvenQ[Sum[If[EvenQ[j3], r1[[j3]], r2[[j3]]], {j3,n}]], (per = LCM @@ Table[If[cs[[j2]] == r1[[j2]], If[0 == cs[[j2]],1,j2], 2j2], {j2,n}]; Times @@ Binomial[cs, r1] 2^(n-Total[cs]) b^FiSum[]),0]);
    PartPol[p_List] := (cs = Count[p, #]&/@ Range[n]; Total[CCPol[#]&/@ Tuples[Range[0,cs]]]);
    pc[p_List] := Module[{ci, mb}, mb = DeleteDuplicates[p]; ci = Count[p, #]&/@ mb; n!/(Times@@(ci!) Times@@(mb^ci))] (*partition count*)
    row[n_Integer] := row[n] = Factor[(Total[(PartPol[#] pc[#])&/@ IntegerPartitions[n]])/(n! 2^(n-1))]
    array[n_, k_] := row[n] /. b -> k
    Table[array[n,d+m-n], {d,6}, {n,m,d+m-1}] // Flatten

Formula

The algorithm used in the Mathematica program below assigns each permutation of the axes to a partition of n and then considers separate conjugacy classes for axis reversals. It uses the formulas in Balasubramanian's paper. If the value of m is increased, one can enumerate colorings of higher-dimensional elements beginning with T(m,1).
T(n,k) = A337888(n,k) + A337889(n,k) = 2*A337888(n,k) - A337890(n,k) = 2*A337889(n,k) + A337890(n,k).

A337893 Array read by descending antidiagonals: T(n,k) is the number of chiral pairs of colorings of the faces of a regular n-dimensional orthoplex (cross polytope) using k or fewer colors.

Original entry on oeis.org

0, 0, 0, 0, 1, 0, 0, 66, 11158298, 0, 0, 920, 4825452718593, 314824333015938998688, 0, 0, 6350, 48038354542204960, 38491882659300767730994725249684096, 31716615393292685397985382790580028572676096, 0
Offset: 2

Views

Author

Robert A. Russell, Sep 28 2020

Keywords

Comments

Each member of a chiral pair is a reflection, but not a rotation, of the other. For n=2, the figure is a square with one square face. For n=3, the figure is an octahedron with 8 triangular faces. For higher n, the number of triangular faces is 8*C(n,3).
Also the number of chiral pairs of colorings of the peaks of an n-dimensional orthotope (hypercube). A peak is an (n-3)-dimensional orthotope.

Examples

			Table begins with T(2,1):
 0        0             0                 0                    0 ...
 0        1            66               920                 6350 ...
 0 11158298 4825452718593 48038354542204960 60632976384183154375 ...
		

Crossrefs

Cf. A337891 (oriented), A337892 (unoriented), A337894 (achiral).
Other elements: A325006 (vertices), A337413 (edges).
Other polytopes: A337885 (simplex), A337889 (orthotope).
Rows 2-4 are A000004, A337896, A331360.

Programs

  • Mathematica
    m=2; (* dimension of color element, here a face *)
    Fi1[p1_] := Module[{g, h}, Coefficient[Product[g = GCD[k1, p1]; h = GCD[2 k1, p1]; (1 + 2 x^(k1/g))^(r1[[k1]] g) If[Divisible[k1, h], 1, (1+2x^(2 k1/h))^(r2[[k1]] h/2)], {k1, Flatten[Position[cs, n1_ /; n1 > 0]]}], x, m+1]];
    FiSum[] := (Do[Fi2[k2] = Fi1[k2], {k2, Divisors[per]}];DivisorSum[per, DivisorSum[d1 = #, MoebiusMu[d1/#] Fi2[#] &]/# &]);
    CCPol[r_List] := (r1 = r; r2 = cs - r1; per = LCM @@ Table[If[cs[[j2]] == r1[[j2]], If[0 == cs[[j2]],1,j2], 2j2], {j2,n}]; If[EvenQ[Sum[If[EvenQ[j3], r1[[j3]], r2[[j3]]], {j3,n}]],1,-1]Times @@ Binomial[cs, r1] 2^(n-Total[cs]) b^FiSum[]);
    PartPol[p_List] := (cs = Count[p, #]&/@ Range[n]; Total[CCPol[#]&/@ Tuples[Range[0,cs]]]);
    pc[p_List] := Module[{ci, mb}, mb = DeleteDuplicates[p]; ci = Count[p, #]&/@ mb; n!/(Times@@(ci!) Times@@(mb^ci))] (*partition count*)
    row[n_Integer] := row[n] = Factor[(Total[(PartPol[#] pc[#])&/@ IntegerPartitions[n]])/(n! 2^n)]
    array[n_, k_] := row[n] /. b -> k
    Table[array[n,d+m-n], {d,6}, {n,m,d+m-1}] // Flatten

Formula

The algorithm used in the Mathematica program below assigns each permutation of the axes to a partition of n and then considers separate conjugacy classes for axis reversals. It uses the formulas in Balasubramanian's paper. If the value of m is increased, one can enumerate colorings of higher-dimensional elements beginning with T(m,1).
T(n,k) = A337891(n,k) - A337892(n,k) = (A337891(n,k) - A337894(n,k)) / 2 = A337892(n,k) - A337894(n,k).

A337894 Array read by descending antidiagonals: T(n,k) is the number of achiral colorings of the faces of a regular n-dimensional orthoplex (cross polytope) using k or fewer colors.

Original entry on oeis.org

1, 2, 1, 3, 21, 1, 4, 201, 93024, 1, 5, 1076, 294157089, 199556208371776, 1, 6, 4025, 91983927296, 1370366433970979158839987, 346179533768149850758531729588224, 1
Offset: 2

Views

Author

Robert A. Russell, Sep 28 2020

Keywords

Comments

An achiral arrangement is identical to its reflection. For n=2, the figure is a square with one square face. For n=3, the figure is an octahedron with 8 triangular faces. For higher n, the number of triangular faces is 8*C(n,3).
Also the number of achiral colorings of the peaks of an n-dimensional orthotope (hypercube). A peak is an (n-3)-dimensional orthotope.

Examples

			Table begins with T(2,1):
1     2         3           4             5               6 ...
1    21       201        1076          4025           11901 ...
1 93024 294157089 91983927296 7960001890625 304914963625056 ...
		

Crossrefs

Cf. A337891 (oriented), A337892 (unoriented), A337893 (chiral).
Other elements: A325007 (vertices), A337414 (edges).
Other polytopes: A337886 (simplex), A337890 (orthotope).
Rows 2-4 are A000027, A337897, A331361.

Programs

  • Mathematica
    m=2; (* dimension of color element, here a face *)
    Fi1[p1_] := Module[{g, h}, Coefficient[Product[g = GCD[k1, p1]; h = GCD[2 k1, p1]; (1 + 2 x^(k1/g))^(r1[[k1]] g) If[Divisible[k1, h], 1, (1+2x^(2 k1/h))^(r2[[k1]] h/2)], {k1, Flatten[Position[cs, n1_ /; n1 > 0]]}], x, m+1]];
    FiSum[] := (Do[Fi2[k2] = Fi1[k2], {k2, Divisors[per]}];DivisorSum[per, DivisorSum[d1 = #, MoebiusMu[d1/#] Fi2[#] &]/# &]);
    CCPol[r_List] := (r1 = r; r2 = cs - r1; If[EvenQ[Sum[If[EvenQ[j3], r1[[j3]], r2[[j3]]], {j3,n}]],0,(per = LCM @@ Table[If[cs[[j2]] == r1[[j2]], If[0 == cs[[j2]],1,j2], 2j2], {j2,n}]; Times @@ Binomial[cs, r1] 2^(n-Total[cs]) b^FiSum[])]);
    PartPol[p_List] := (cs = Count[p, #]&/@ Range[n]; Total[CCPol[#]&/@ Tuples[Range[0,cs]]]);
    pc[p_List] := Module[{ci, mb}, mb = DeleteDuplicates[p]; ci = Count[p, #]&/@ mb; n!/(Times@@(ci!) Times@@(mb^ci))] (*partition count*)
    row[m]=b;
    row[n_Integer] := row[n] = Factor[(Total[(PartPol[#] pc[#])&/@ IntegerPartitions[n]])/(n! 2^(n-1))]
    array[n_, k_] := row[n] /. b -> k
    Table[array[n,d+m-n], {d,7}, {n,m,d+m-1}] // Flatten

Formula

The algorithm used in the Mathematica program below assigns each permutation of the axes to a partition of n and then considers separate conjugacy classes for axis reversals. It uses the formulas in Balasubramanian's paper. If the value of m is increased, one can enumerate colorings of higher-dimensional elements beginning with T(m,1).
T(n,k) = 2*A337892(n,k) - A337891(n,k) = A337891(n,k) - 2*A337893(n,k) = A337892(n,k) - A337893(n,k).

A337892 Array read by descending antidiagonals: T(n,k) is the number of unoriented colorings of the faces of a regular n-dimensional orthoplex (cross polytope) using k or fewer colors.

Original entry on oeis.org

1, 2, 1, 3, 22, 1, 4, 267, 11251322, 1, 5, 1996, 4825746875682, 314824532572147370464, 1, 6, 10375, 48038446526132256, 38491882660671134164965704408524083, 31716615393638864931753532641338560302264320, 1
Offset: 2

Views

Author

Robert A. Russell, Sep 28 2020

Keywords

Comments

Each chiral pair is counted as one when enumerating unoriented arrangements. For n=2, the figure is a square with one square face. For n=3, the figure is an octahedron with 8 triangular faces. For higher n, the number of triangular faces is 8*C(n,3).
Also the number of unoriented colorings of the peaks of an n-dimensional orthotope (hypercube). A peak is an (n-3)-dimensional orthotope.

Examples

			Array begins with T(2,1):
 1        2             3                 4                    5 ...
 1       22           267              1996                10375 ...
 1 11251322 4825746875682 48038446526132256 60632984344185045000 ...
		

Crossrefs

Cf. A337891 (oriented), A337893 (chiral), A337894 (achiral).
Other elements: A325005 (vertices), A337412 (edges).
Other polytopes: A337884 (simplex), A337888 (orthotope).
Rows 2-4 are A000027, A128766, A331359

Programs

  • Mathematica
    m=2; (* dimension of color element, here a face *)
    Fi1[p1_] := Module[{g, h}, Coefficient[Product[g = GCD[k1, p1]; h = GCD[2 k1, p1]; (1 + 2 x^(k1/g))^(r1[[k1]] g) If[Divisible[k1, h], 1, (1+2x^(2 k1/h))^(r2[[k1]] h/2)], {k1, Flatten[Position[cs, n1_ /; n1 > 0]]}], x, m+1]];
    FiSum[] := (Do[Fi2[k2] = Fi1[k2], {k2, Divisors[per]}];DivisorSum[per, DivisorSum[d1 = #, MoebiusMu[d1/#] Fi2[#] &]/# &]);
    CCPol[r_List] := (r1 = r; r2 = cs - r1; per = LCM @@ Table[If[cs[[j2]] == r1[[j2]], If[0 == cs[[j2]],1,j2], 2j2], {j2,n}]; Times @@ Binomial[cs, r1] 2^(n-Total[cs]) b^FiSum[]);
    PartPol[p_List] := (cs = Count[p, #]&/@ Range[n]; Total[CCPol[#]&/@ Tuples[Range[0,cs]]]);
    pc[p_List] := Module[{ci, mb}, mb = DeleteDuplicates[p]; ci = Count[p, #]&/@ mb; n!/(Times@@(ci!) Times@@(mb^ci))] (*partition count*)
    row[m]=b;
    row[n_Integer] := row[n] = Factor[(Total[(PartPol[#] pc[#])&/@ IntegerPartitions[n]])/(n! 2^n)]
    array[n_, k_] := row[n] /. b -> k
    Table[array[n,d+m-n], {d,6}, {n,m,d+m-1}] // Flatten

Formula

The algorithm used in the Mathematica program below assigns each permutation of the axes to a partition of n and then considers separate conjugacy classes for axis reversals. It uses the formulas in Balasubramanian's paper. If the value of m is increased, one can enumerate colorings of higher-dimensional elements beginning with T(m,1).
T(n,k) = A337891(n,k) - A337893(n,k) = (A337891(n,k) + A337894(n,k)) / 2 = A337893(n,k) + A337894(n,k).
Showing 1-6 of 6 results.