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.

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

1,3

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.

Links

Robert A. Russell, Table of n, a(n) for n = 1..220 First 10 rows.

E. M. Palmer and R. W. Robinson, Enumeration under two representations of the wreath product, Acta Math., 131 (1973), 123-143.

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).

Example

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.

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

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).

Sequence in context: A184251 A262348 A307093 * A088876 A346078 A014846

Adjacent sequences: A327084 A327085 A327086 * A327088 A327089 A327090

Keyword

nonn,tabf

Author

Robert A. Russell, Aug 19 2019

Status

approved