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.

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A327084 Array read by descending antidiagonals: A(n,k) is the number of unoriented colorings of the edges of a regular n-dimensional simplex using up to k colors.

Original entry on oeis.org

1, 2, 1, 3, 4, 1, 4, 10, 11, 1, 5, 20, 66, 34, 1, 6, 35, 276, 792, 156, 1, 7, 56, 900, 10688, 25506, 1044, 1, 8, 84, 2451, 90005, 1601952, 2302938, 12346, 1, 9, 120, 5831, 533358, 43571400, 892341888, 591901884, 274668
Offset: 1

Views

Author

Robert A. Russell, Aug 19 2019

Keywords

Comments

An n-dimensional simplex has n+1 vertices and (n+1)*n/2 edges. For n=1, the figure is a line segment with one edge. For n=2, the figure is a triangle with three edges. For n=3, the figure is a tetrahedron with six edges. The Schläfli symbol, {3,...,3}, of the regular n-dimensional simplex consists of n-1 threes. Two unoriented colorings are the same if congruent; chiral pairs are counted as one.
A(n,k) is also the number of unoriented colorings of (n-2)-dimensional regular simplices in an n-dimensional simplex using up to k colors. Thus, A(2,k) is also the number of unoriented colorings of the vertices (0-dimensional simplices) of an equilateral triangle.

Examples

			Array begins with A(1,1):
  1  2   3     4     5      6       7       8        9       10 ...
  1  4  10    20    35     56      84     120      165      220 ...
  1 11  66   276   900   2451    5831   12496    24651    45475 ...
  1 34 792 10688 90005 533358 2437848 9156288 29522961 84293770 ...
  ...
For A(2,3) = 10, the nine achiral colorings are AAA, AAB, AAC, ABB, ACC, BBB, BBC, BCC, and CCC. The chiral pair is ABC-ACB.
		

Crossrefs

Cf. A327083 (oriented), A327085 (chiral), A327086 (achiral), A327088 (exactly k colors), A325000 (vertices, facets), A337884 (faces, peaks), A337408 (orthotope edges, orthoplex ridges), A337412 (orthoplex edges, orthotope ridges).
Rows 1-4 are A000027, A000292, A063842(n-1), A063843.
Cf. A063841 (k-multigraphs on n nodes).

Programs

  • Mathematica
    CycleX[{2}] = {{1,1}}; (* cycle index for permutation with given cycle structure *)
    CycleX[{n_Integer}] := CycleX[n] = If[EvenQ[n], {{n/2,1}, {n,(n-2)/2}}, {{n,(n-1)/2}}]
    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)
    CycleX[p_List] := CycleX[p] = compress[Join[CycleX[Drop[p, -1]], If[Last[p] > 1, CycleX[{Last[p]}], ## &[]], If[# == Last[p], {#, Last[p]}, {LCM[#, Last[p]], GCD[#, Last[p]]}] & /@ Drop[p, -1]]]
    pc[p_List] := 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[pc[#] j^Total[CycleX[#]][[2]] & /@ IntegerPartitions[n+1]]/(n+1)!]
    array[n_, k_] := row[n] /. j -> k
    Table[array[n,d-n+1], {d,1,10}, {n,1,d}] // Flatten
    (* Using Fripertinger's exponent per Andrew Howroyd code in A063841: *)
    pc[p_] := Module[{ci, mb}, mb = DeleteDuplicates[p]; ci = Count[p, #] &/@ mb; Total[p]!/(Times @@ (ci!) Times @@ (mb^ci))]
    ex[v_] := Sum[GCD[v[[i]], v[[j]]], {i,2,Length[v]}, {j,i-1}] + Total[Quotient[v,2]]
    array[n_,k_] := Total[pc[#]k^ex[#] &/@ IntegerPartitions[n+1]]/(n+1)!
    Table[array[n,d-n+1], {d,10}, {n,d}] // Flatten
    (* Another program (translated from Andrew Howroyd's PARI code): *)
    permcount[v_] := Module[{m=1, s=0, k=0, t}, For[i=1, i <= Length[v], i++, t = v[[i]]; k = If[i>1 && t == v[[i-1]], k+1, 1]; m *= t*k; s += t]; s!/m];
    edges[v_] := Sum[GCD[v[[i]], v[[j]]], {i, 2, Length[v]}, {j, 1, i-1}] + Total[Quotient[v, 2]];
    T[n_, k_] := Module[{s = 0}, Do[s += permcount[p]*k^edges[p], {p, IntegerPartitions[n+1]}]; s/(n+1)!];
    Table[T[n-k+1, k], {n, 1, 9}, {k, n, 1, -1}] // Flatten (* Jean-François Alcover, Jan 08 2021 *)
  • PARI
    permcount(v) = {my(m=1, s=0, k=0, t); for(i=1, #v, t=v[i]; k=if(i>1&&t==v[i-1], k+1, 1); m*=t*k; s+=t); s!/m}
    edges(v) = {sum(i=2, #v, sum(j=1, i-1, gcd(v[i], v[j]))) + sum(i=1, #v, v[i]\2)}
    T(n, k) = {my(s=0); forpart(p=n+1, s+=permcount(p)*k^edges(p)); s/(n+1)!} \\ Andrew Howroyd, Sep 06 2019
    
  • Python
    from itertools import combinations
    from math import prod, gcd, factorial
    from fractions import Fraction
    from sympy.utilities.iterables import partitions
    def A327084_T(n,k): return int(sum(Fraction(k**(sum(p[r]*p[s]*gcd(r,s) for r,s in combinations(p.keys(),2))+sum((q>>1)*r+(q*r*(r-1)>>1) for q, r in p.items())),prod(q**r*factorial(r) for q, r in p.items())) for p in partitions(n+1))) # Chai Wah Wu, Jul 09 2024

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.
A(n,k) = Sum_{j=1..(n+1)*n/2} A327088(n,j) * binomial(k,j).
A(n,k) = A327083(n,k) - A327085(n,k) = (A327083(n,k) + A327086(n,k)) / 2 = A327085(n,k) + A327086(n,k).
A(n,k) = A063841(n+1,k-1).

A327087 Triangle read by rows: T(n,k) is the number of oriented colorings of the edges of a regular n-dimensional simplex using exactly k colors. Row n has (n+1)*n/2 columns.

Original entry on oeis.org

1, 1, 2, 2, 1, 10, 54, 136, 150, 60, 1, 38, 1080, 14040, 85500, 274104, 493920, 504000, 272160, 60480, 1, 182, 42111, 2848060, 70361815, 841626366, 5722670282, 24262499520, 67665563280, 127639512000, 164044520640, 141664723200, 78702624000, 25427001600, 3632428800
Offset: 1

Views

Author

Robert A. Russell, Aug 19 2019

Keywords

Comments

An n-dimensional simplex has n+1 vertices and (n+1)*n/2 edges. For n=1, the figure is a line segment with one edge. For n=2, the figure is a triangle with three edges. For n=3, the figure is a tetrahedron with six edges. The Schläfli symbol {3,...,3}, of the regular n-dimensional simplex consists of n-1 threes. Two oriented colorings are the same if one is a rotation of the other; chiral pairs are counted as two.
T(n,k) is also the number of oriented colorings of (n-2)-dimensional regular simplices in an n-dimensional simplex using exactly k colors. Thus, T(2,k) is also the number of oriented colorings of the vertices (0-dimensional simplices) of an equilateral triangle.

Examples

			Triangle begins with T(1,1):
  1
  1  2    2
  1 10   54   136   150     60
  1 38 1080 14040 85500 274104 493920 504000 272160 60480
  ...
For T(2,1)=1, all edges of the triangle are the same color. For T(2,2)=2, the edges are AAB and ABB. For T(2,3)=2, the chiral pair is ABC-ACB.
		

Crossrefs

Cf. A327088 (unoriented), A327089 (chiral), A327090 (achiral), A327083 (up to k colors), A325002 (vertices and facets), A338113 (faces and peaks), A338142 (orthotope edges, orthoplex ridges), A338146 (orthoplex edges, orthotope ridges).

Programs

  • Mathematica
    CycleX[{2}] = {{1,1}}; (* cycle index for permutation with given cycle structure *)
    CycleX[{n_Integer}] := CycleX[n] = If[EvenQ[n], {{n/2,1}, {n,(n-2)/2}}, {{n,(n-1)/2}}]
    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)
    CycleX[p_List] := CycleX[p] = compress[Join[CycleX[Drop[p,-1]], If[Last[p] > 1, CycleX[{Last[p]}], ## &[]], If[# == Last[p], {#, Last[p]}, {LCM[#, Last[p]], GCD[#, Last[p]]}] & /@ Drop[p,-1]]]
    pc[p_List] := 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[CycleX[#]][[2]], 0] & /@ IntegerPartitions[n+1]]/((n+1)!/2)]
    array[n_, k_] := row[n] /. j -> k
    Table[LinearSolve[Table[Binomial[i,j], {i,1,(n+1)n/2}, {j,1,(n+1)n/2}], Table[array[n,k], {k,1,(n+1)n/2}]], {n,1,6}] // 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.
A327083(n,k) = Sum_{j=1..(n+1)*n/2} T(n,j) * binomial(k,j).
T(n,k) = A327088(n,k) + A327089(n,k) = 2*A327088(n,k) - A327090(n,k) = 2*A327089(n,k) + A327090(n,k).

A327089 Triangle read by rows: T(n,k) is the number of chiral pairs of colorings of the edges of a regular n-dimensional simplex using exactly k colors. Row n has (n+1)*n/2 columns.

Original entry on oeis.org

0, 0, 0, 1, 0, 1, 18, 62, 75, 30, 0, 6, 387, 6320, 41350, 135792, 246540, 252000, 136080, 30240, 0, 28, 17070, 1347200, 34546670, 418081188, 2854567996, 12121240320, 33824042280, 63815598000, 82021428720, 70832361600, 39351312000, 12713500800, 1816214400
Offset: 1

Views

Author

Robert A. Russell, Aug 19 2019

Keywords

Comments

An n-dimensional simplex has n+1 vertices and (n+1)*n/2 edges. For n=1, the figure is a line segment with one edge. For n-2, the figure is a triangle with three edges. For n=3, the figure is a tetrahedron with six edges. The Schläfli symbol, {3,...,3}, of the regular n-dimensional simplex consists of n-1 threes. The chiral colorings of its edges come in pairs, each the reflection of the other.
T(n,k) is also the number of chiral pairs of colorings of (n-2)-dimensional regular simplices in an n-dimensional simplex using exactly k colors. Thus, T(2,k) is also the number of chiral pairs of colorings of the vertices (0-dimensional simplices) of an equilateral triangle.

Examples

			Triangle begins with T(1,1):
0
0 0   1
0 1  18   62    75     30
0 6 387 6320 41350 135792 246540 252000 136080 30240
For T(2,3)=2, the chiral pair is ABC-ACB.
		

Crossrefs

Cf. A327087 (oriented), A327088 (unoriented), A327090 (achiral), A327085 (exactly k colors).

Programs

  • Mathematica
    CycleX[{2}] = {{1,1}}; (* cycle index for permutation with given cycle structure *)
    CycleX[{n_Integer}] := CycleX[n] = If[EvenQ[n], {{n/2,1}, {n,(n-2)/2}}, {{n,(n-1)/2}}]
    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)
    CycleX[p_List] := CycleX[p] = compress[Join[CycleX[Drop[p, -1]], If[Last[p] > 1, CycleX[{Last[p]}], ## &[]], If[# == Last[p], {#, Last[p]}, {LCM[#, Last[p]], GCD[#, Last[p]]}] & /@ Drop[p, -1]]]
    pc[p_List] := 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]]], 1, -1] pc[#] j^Total[CycleX[#]][[2]] & /@ IntegerPartitions[n+1]]/(n+1)!]
    array[n_, k_] := row[n] /. j -> k
    Table[LinearSolve[Table[Binomial[i,j], {i,1,(n+1)n/2}, {j,1,(n+1)n/2}], Table[array[n,k], {k,1,(n+1)n/2}]], {n,1,6}] // 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.
A327085(n,k) = Sum_{j=1..(n+1)*n/2} T(n,j) * binomial(k,j).
A(n,k) = A327087(n,k) - A327088(n,k) = (A327087(n,k) - A327090(n,k)) / 2 = A327088(n,k) - A327090(n,k).

A327090 Triangle read by rows: T(n,k) is the number of achiral colorings of the edges of a regular n-dimensional simplex using exactly k colors. Row n has (n+1)*n/2 columns.

Original entry on oeis.org

1, 1, 2, 0, 1, 8, 18, 12, 0, 0, 1, 26, 306, 1400, 2800, 2520, 840, 0, 0, 0, 1, 126, 7971, 153660, 1268475, 5463990, 13534290, 20018880, 17478720, 8316000, 1663200, 0, 0, 0, 0
Offset: 1

Views

Author

Robert A. Russell, Aug 19 2019

Keywords

Comments

An n-dimensional simplex has n+1 vertices and (n+1)*n/2 edges. For n=1, the figure is a line segment with one edge. For n-2, the figure is a triangle with three edges. For n=3, the figure is a tetrahedron with six edges. The Schläfli symbol, {3,...,3}, of the regular n-dimensional simplex consists of n-1 threes. An achiral coloring is identical to its reflection.
T(n,k) is also the number of achiral colorings of (n-2)-dimensional regular simplices in an n-dimensional simplex using exactly k colors. Thus, T(2,k) is also the number of achiral colorings of the vertices (0-dimensional simplices) of an equilateral triangle.

Examples

			Triangle begins with T(1,1):
1
1  2   0
1  8  18   12    0    0
1 26 306 1400 2800 2520 840 0 0 0
For T(2,2) = 2, the two colorings of the triangle edges are AAB and ABB.
		

Crossrefs

Cf. A327087 (oriented), A327088 (unoriented), A327089 (chiral), A327086 (up to k colors), A325003 (vertices).

Programs

  • Mathematica
    CycleX[{2}] = {{1,1}}; (* cycle index for permutation with given cycle structure *)
    CycleX[{n_Integer}] := CycleX[n] = If[EvenQ[n], {{n/2,1}, {n,(n-2)/2}}, {{n,(n-1)/2}}]
    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)
    CycleX[p_List] := CycleX[p] = compress[Join[CycleX[Drop[p, -1]], If[Last[p] > 1, CycleX[{Last[p]}], ## &[]], If[# == Last[p], {#, Last[p]}, {LCM[#, Last[p]], GCD[#, Last[p]]}] & /@ Drop[p, -1]]]
    pc[p_List] := 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]]], 0, pc[#] j^Total[CycleX[#]][[2]]] & /@ IntegerPartitions[n+1]]/((n+1)!/2)]
    array[n_, k_] := row[n] /. j -> k
    Table[LinearSolve[Table[Binomial[i,j], {i,1,(n+1)n/2}, {j,1,(n+1)n/2}], Table[array[n,k], {k,1,(n+1)n/2}]], {n,1,6}] // 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. The last n-1 entries in row n are zero.
A327086(n,k) = Sum_{j=1..(n+1)*n/2} T(n,j) * binomial(k,j).
A(n,k) = 2*A327084(n,k) - A327083(n,k) = A327083(n,k) - 2*A327085(n,k) = A327084(n,k) - A327085(n,k).

A338143 Triangle read by rows: T(n,k) is the number of unoriented colorings of the edges of a regular n-D orthotope (or ridges of a regular n-D orthoplex) using exactly k colors. Row n has n*2^(n-1) columns.

Original entry on oeis.org

1, 1, 4, 6, 3, 1, 142, 11682, 310536, 3460725, 19870590, 65886660, 133585200, 168399000, 128898000, 54885600, 9979200, 1, 11251320, 4825713121719, 48019143606137456, 60392840368910627325
Offset: 1

Views

Author

Robert A. Russell, Oct 12 2020

Keywords

Comments

Each chiral pair is counted as one when enumerating unoriented arrangements. A ridge is an (n-2)-face of an n-D polytope. For n=1, the figure is a line segment with one edge. For n=2, the figure is a square with 4 edges (vertices). For n=3, the figure is a cube (octahedron) with 12 edges. The number of edges (ridges) is n*2^(n-1). The Schläfli symbols for the n-D orthotope (hypercube) and the n-D orthoplex (hyperoctahedron, cross polytope) are {4,...,3,3} and {3,3,...,4} respectively, with n-2 3's in each case. The figures are mutually dual.
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).

Examples

			Triangle begins with T(1,1):
  1
  1   4     6      3
  1 142 11682 310536 3460725 19870590 65886660 133585200 168399000
  ...
		

Crossrefs

Cf. A338142 (oriented), A338144 (chiral), A338145 (achiral), A337408 (k or fewer colors), A325017 (orthotope vertices, orthoplex facets).
Cf. A327088 (simplex), A338147 (orthoplex edges, orthotope ridges).

Programs

  • Mathematica
    m=1; (* dimension of color element, here an edge *)
    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; 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[n_Integer] := row[n] = Factor[(Total[(PartPol[#] pc[#])&/@ IntegerPartitions[n]])/(n! 2^n)]
    array[n_, k_] := row[n] /. b -> k
    Table[LinearSolve[Table[Binomial[i,j],{i,2^(n-m)Binomial[n,m]},{j,2^(n-m)Binomial[n,m]}], Table[array[n,k],{k,2^(n-m)Binomial[n,m]}]], {n,m,m+4}] // Flatten

Formula

A337408(n,k) = Sum_{j=1..n*2^(n-1)} T(n,j) * binomial(k,j).
T(n,k) = A338142(n,k) - A338144(n,k) = (A338142(n,k) + A338145(n,k)) / 2 = A338144(n,k) + A338145(n,k).
T(2,k) = A338147(2,k) = A325017(2,k) = A325009(2,k); T(3,k) = A338147(3,k).

A338147 Triangle read by rows: T(n,k) is the number of unoriented colorings of the edges of a regular n-D orthoplex (or ridges of a regular n-D orthotope) using exactly k colors. Row 1 has 1 column; row n>1 has 2*n*(n-1) columns.

Original entry on oeis.org

1, 1, 4, 6, 3, 1, 142, 11682, 310536, 3460725, 19870590, 65886660, 133585200, 168399000, 128898000, 54885600, 9979200, 1, 49125, 740212980, 730815102166, 151600044933990, 11420034970306170, 415777158607920585
Offset: 1

Views

Author

Robert A. Russell, Oct 12 2020

Keywords

Comments

Each chiral pair is counted as one when enumerating unoriented arrangements. A ridge is an (n-2)-face of an n-D polytope. For n=1, the figure is a line segment with one edge. For n=2, the figure is a square with 4 edges (vertices). For n=3, the figure is an octahedron (cube) with 12 edges. For n>1, the number of edges (ridges) is 2*n*(n-1). The Schläfli symbols for the n-D orthotope (hypercube) and the n-D orthoplex (hyperoctahedron, cross polytope) are {4,3,...,3,3} and {3,3,...,3,4} respectively, with n-2 3's in each case. The figures are mutually dual.
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).

Examples

			Triangle begins with T(1,1):
  1
  1   4     6      3
  1 142 11682 310536 3460725 19870590 65886660 133585200 168399000
  ...
For T(2,3)=6, the 3 achiral colorings are ABAC, ABCB, and ACBC. The three chiral pairs are AABC-AACB, ABBC-ACBB, and ABCC-ACCB.
		

Crossrefs

Cf. A338146 (oriented), A338148 (chiral), A338149 (achiral), A337412 (k or fewer colors), A325009 (orthoplex vertices, orthotope facets).
Cf. A327088 (simplex), A338143 (orthotope edges, orthoplex ridges).

Programs

  • Mathematica
    m=1; (* dimension of color element, here an edge *)
    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
    Join[{{1}},Table[LinearSolve[Table[Binomial[i,j],{i,2^(m+1)Binomial[n,m+1]},{j,2^(m+1)Binomial[n,m+1]}], Table[array[n,k],{k,2^(m+1)Binomial[n,m+1]}]], {n,m+1,m+4}]] // Flatten

Formula

For n>1, A337412(n,k) = Sum_{j=1..2*n*(n-1)} T(n,j) * binomial(k,j).
T(n,k) = A338146(n,k) - A338148(n,k) = (A338146(n,k) + A338149(n,k)) / 2 = A338148(n,k) + A338149(n,k).
T(2,k) = A338143(2,k) = A325017(2,k) = A325009(2,k); T(3,k) = A338143(3,k).

A338114 Triangle read by rows: T(n,k) is the number of unoriented colorings of the faces (and peaks) of a regular n-dimensional simplex using exactly k colors. Row n has C(n+1,3) columns.

Original entry on oeis.org

1, 1, 3, 3, 1, 1, 32, 693, 7720, 44150, 138312, 247380, 252000, 136080, 30240, 1, 2134, 4971504, 1513872568, 124978335900, 4307090369304, 78010256156784, 849590196841344, 6053725780061400, 29824685516682000, 105382759395846240, 273441179492268480
Offset: 2

Views

Author

Robert A. Russell, Oct 10 2020

Keywords

Comments

An n-dimensional simplex has n+1 vertices, C(n+1,3) faces, and C(n+1,3) peaks, which are (n-3)-dimensional simplexes. For n=2, the figure is a triangle with one face. For n=3, the figure is a tetrahedron with four triangular faces and four peaks (vertices). For n=4, the figure is a 4-simplex with ten triangular faces and ten peaks (edges). The Schläfli symbol {3,...,3}, of the regular n-dimensional simplex consists of n-1 3's. Two unoriented colorings are the same if they are congruent; chiral pairs are counted as one.
The algorithm used in the Mathematica program below assigns each permutation of the vertices to a cycle-structure partition of n+1. It then determines the number of permutations for each partition and the cycle index for each partition. If the value of m is increased, one can enumerate colorings of higher-dimensional elements beginning with T(m,1).

Examples

			Triangle begins with T(2,1):
  1
  1  3   3    1
  1 32 693 7720 44150 138312 247380 252000 136080 30240
  ...
For T(3,2)=3, the tetrahedron has one, two, or three faces (vertices) of one color. For T(3,4)=1, each of the four tetrahedron faces (vertices) is a different color.
		

Crossrefs

Cf. A338113 (oriented), A338115 (chiral), A338116 (achiral), A337884 (k or fewer colors), A007318(n,k-1) (vertices and facets), A327088 (edges and ridges).

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[pc[#] j^Total[CX[#, m+1]][[2]] & /@ IntegerPartitions[n+1]]/(n+1)!]
    array[n_, k_] := row[n] /. j -> k
    Table[LinearSolve[Table[Binomial[i,j],{i,Binomial[n+1,m+1]},{j,Binomial[n+1,m+1]}], Table[array[n,k],{k,Binomial[n+1,m+1]}]], {n,m,m+4}] // Flatten

Formula

A337884(n,k) = Sum_{j=1..C(n+1,3)} T(n,j) * binomial(k,j).
T(n,k) = A338113(n,k) - A338115(n,k) = (A338113(n,k) + A338116(n,k)) / 2 = A338115(n,k) + A338116(n,k).
T(3,k) = A007318(3,k-1); T(4,k) = A327088(4,k).
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