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.

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

Author

Robert A. Russell, Aug 19 2019

Keywords

• nonn
• tabf

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.
		

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.

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).