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.

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

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. 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.
		

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), 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).