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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A327086 Array read by descending antidiagonals: A(n,k) is the number of achiral 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, 9, 10, 1, 5, 16, 45, 28, 1, 6, 25, 136, 387, 128, 1, 7, 36, 325, 2784, 8352, 792, 1, 8, 49, 666, 13125, 186304, 382563, 7620, 1, 9, 64, 1225, 46836, 2117750, 36507008, 44526672, 124344
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.
A(n,k) is also the number of achiral 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 achiral 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      11      12 ...
  1  4   9   16    25    36     49     64     81     100     121     144 ...
  1 10  45  136   325   666   1225   2080   3321    5050    7381   10440 ...
  1 28 387 2784 13125 46836 137543 349952 797769 1667500 3248971 5973408 ...
  ...
For A(2,3) = 9, the colorings are AAA, AAB, AAC, ABB, ACC, BBB, BBC, BCC, and CCC.
		

Crossrefs

Cf. A327083 (oriented), A327084 (unoriented), A327085 (chiral), A327090 (exactly k colors), A325001 (vertices, facets), A337886 (faces, peaks), A337410 (orthotope edges, orthoplex ridges), A337414 (orthoplex edges, orthotope ridges).
Rows 1-4 are A000027, A000290, A037270, A331353.

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[array[n,d-n+1], {d,1,10}, {n,1,d}] // Flatten
    (* Using Fripertinger's exponent per Andrew Howroyd's 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[If[OddQ[Total[1-Mod[#,2]]], pc[#]k^ex[#], 0] &/@ IntegerPartitions[n+1]]/((n+1)!/2)
    Table[array[n,d-n+1], {d,10}, {n,d}] // 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.
A(n,k) = Sum_{j=1..(n+1)*n/2} A327090(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).

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).

A327088 Triangle read by rows: T(n,k) is the number of unoriented 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, 1, 1, 9, 36, 74, 75, 30, 1, 32, 693, 7720, 44150, 138312, 247380, 252000, 136080, 30240, 1, 154, 25041, 1500860, 35815145, 423545178, 2868102286, 12141259200, 33841521000, 63823914000, 82023091920, 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. Two unoriented colorings are the same if congruent; chiral pairs are counted as one.
T(n,k) is also the number of unoriented 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 unoriented colorings of the vertices (0-dimensional simplices) of an equilateral triangle.

Examples

			Triangle begins with T(1,1):
1
1  2   1
1  9  36   74    75     30
1 32 693 7720 44150 138312 247380 252000 136080 30240
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. A327087 (oriented), A327089 (chiral), A327090 (achiral), A327084 (exactly k colors), A007318(n,k-1) (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[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.
A327084(n,k) = Sum_{j=1..(n+1)*n/2} T(n,j) * binomial(k,j).
A(n,k) = A327087(n,k) - A327089(n,k) = (A327087(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).

A338145 Triangle read by rows: T(n,k) is the number of achiral 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, 3, 0, 1, 68, 1200, 7268, 20025, 27750, 18900, 5040, 0, 0, 0, 0, 1, 93022, 293878020, 90807857080, 7503022894800, 258528829444320, 4681671089961600, 50981530073846400, 363246007692204000
Offset: 1

Views

Author

Robert A. Russell, Oct 12 2020

Keywords

Comments

An achiral coloring is identical to its reflection. 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    3    0
  1 68 1200 7268 20025 27750 18900 5040 0 0 0 0
  ...
For T(2,2)=4, the achiral colorings are AAAB, AABB, ABAB, and ABBB. For T(2,3)=3, the achiral colorings are ABAC, ABCB, and ACBC.
		

Crossrefs

Cf. A338142 (oriented), A338143 (unoriented), A338144 (chiral), A337410 (k or fewer colors), A325019 (orthotope vertices, orthoplex facets).
Cf. A327090 (simplex), A338149 (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; 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[n_Integer] := row[n] = Factor[(Total[(PartPol[#] pc[#])&/@ IntegerPartitions[n]])/(n! 2^(n-1))]
    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

A337410(n,k) = Sum_{j=1..n*2^(n-1)} T(n,j) * binomial(k,j).
T(n,k) = 2*A338143(n,k) - A338142(n,k) = A338142(n,k) - 2*A338144(n,k) = A338143(n,k) - A338144(n,k).
T(2,k) = A338149(2,k) = A325019(2,k) = A325011(2,k); T(3,k) = A338149(3,k).

A338149 Triangle read by rows: T(n,k) is the number of achiral 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, 3, 0, 1, 68, 1200, 7268, 20025, 27750, 18900, 5040, 0, 0, 0, 0, 1, 8198, 9055962, 1467050480, 74035775370, 1679679306420, 20864180531565, 159341117375160, 804216787965360, 2808560520334800, 6981656802951600
Offset: 1

Views

Author

Robert A. Russell, Oct 12 2020

Keywords

Comments

An achiral coloring is identical to its reflection. 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       3          0
  1   68    1200       7268       20025         27750 18900 5040 0 0 0 0
  1 8198 9055962 1467050480 74035775370 1679679306420 ...
  ...
For T(2,2)=4, the achiral colorings are AAAB, AABB, ABAB, and ABBB. For T(2,3)=3, the 3 achiral colorings are ABAC, ABCB, and ACBC.
		

Crossrefs

Cf. A338146 (oriented), A338147 (unoriented), A338148 (chiral), A337414 (k or fewer colors), A325011 (orthoplex vertices, orthotope facets).
Cf. A327090 (simplex), A338145 (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; 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
    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, A337414(n,k) = Sum_{j=1..2*n*(n-1)} T(n,j) * binomial(k,j).
T(n,k) = 2*A338147(n,k) - A338146(n,k) = A338146(n,k) - 2*A338148(n,k) = A338147(n,k) - A338148(n,k).
T(2,k) = A338145(2,k) = A325019(2,k) = A325011(2,k); T(3,k) = A338145(3,k).

A338116 Triangle read by rows: T(n,k) is the number of achiral 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, 0, 1, 26, 306, 1400, 2800, 2520, 840, 0, 0, 0, 1, 766, 199902, 10426768, 200588850, 1903776420, 10360383600, 35133957600, 77643846000, 113816253600, 109880971200, 67199932800, 23610787200, 3632428800, 0, 0, 0, 0, 0, 0
Offset: 2

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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. An achiral coloring is identical to its reflection.
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        0
  1  26    306     1400      2800       2520         840           0   0   0
  1 766 199902 10426768 200588850 1903776420 10360383600 35133957600 ...
  ...
For T(3,3)=3, one of the three colors appears on two faces (vertices) of the tetrahedron.
		

Crossrefs

Cf. A338113 (oriented), A338114 (unoriented), A338115 (chiral), A337886 (k or fewer colors), A325003 (vertices and facets), A327090 (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[If[OddQ[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[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

A337886(n,k) = Sum_{j=1..C(n+1,3)} T(n,j) * binomial(k,j).
T(n,k) = 2*A338114(n,k) - A338113(n,k) = A338113(n,k) - 2*A338115(n,k) = A338114(n,k) - A338115(n,k).
T(3,k) = A325003(3,k); T(4,k) = A327090(4,k).
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