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%I #8 Jun 09 2021 02:31:55
%S 1,1,2,0,1,8,18,12,0,0,1,26,306,1400,2800,2520,840,0,0,0,1,126,7971,
%T 153660,1268475,5463990,13534290,20018880,17478720,8316000,1663200,0,
%U 0,0,0
%N 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.
%C 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.
%C 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.
%H Robert A. Russell, <a href="/A327090/b327090.txt">Table of n, a(n) for n = 1..220</a> First 10 rows.
%H E. M. Palmer and R. W. Robinson, <a href="https://projecteuclid.org/euclid.acta/1485889789">Enumeration under two representations of the wreath product</a>, Acta Math., 131 (1973), 123-143.
%F 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.
%F A327086(n,k) = Sum_{j=1..(n+1)*n/2} T(n,j) * binomial(k,j).
%F A(n,k) = 2*A327084(n,k) - A327083(n,k) = A327083(n,k) - 2*A327085(n,k) = A327084(n,k) - A327085(n,k).
%e Triangle begins with T(1,1):
%e 1
%e 1 2 0
%e 1 8 18 12 0 0
%e 1 26 306 1400 2800 2520 840 0 0 0
%e For T(2,2) = 2, the two colorings of the triangle edges are AAB and ABB.
%t CycleX[{2}] = {{1,1}}; (* cycle index for permutation with given cycle structure *)
%t CycleX[{n_Integer}] := CycleX[n] = If[EvenQ[n], {{n/2,1}, {n,(n-2)/2}}, {{n,(n-1)/2}}]
%t 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)
%t 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]]]
%t pc[p_List] := Module[{ci, mb}, mb = DeleteDuplicates[p]; ci = Count[p, #] & /@ mb; Total[p]!/(Times @@ (ci!) Times @@ (mb^ci))] (* partition count *)
%t 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)]
%t array[n_, k_] := row[n] /. j -> k
%t 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
%Y Cf. A327087 (oriented), A327088 (unoriented), A327089 (chiral), A327086 (up to k colors), A325003 (vertices).
%K nonn,tabf
%O 1,3
%A _Robert A. Russell_, Aug 19 2019
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