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.

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

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. 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.
		

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), 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).