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Array read by descending antidiagonals: T(n,k) is the number of unoriented colorings of the triangular faces of a regular n-dimensional simplex using k or fewer colors.
1, 2, 1, 3, 5, 1, 4, 15, 34, 1, 5, 35, 792, 2136, 1, 6, 70, 10688, 4977909, 7013320, 1, 7, 126, 90005, 1533771392, 9930666709494, 1788782616656, 1, 8, 210, 533358, 132597435125, 234249157811872000, 12979877431438089379035, 53304527811667897248, 1
2,2
Each chiral pair is counted as one when enumerating unoriented arrangements. An n-simplex has n+1 vertices. For n=2, the figure is a triangle with one triangular face. For n=3, the figure is a tetrahedron with 4 triangular faces. For higher n, the number of triangular faces is C(n+1,3).
Also the number of unoriented colorings of the peaks of a regular n-dimensional simplex. A peak of an n-simplex is an (n-3)-dimensional simplex.
E. M. Palmer and R. W. Robinson, <a href="https://doi.org/10.1007/BF02392038">Enumeration under two representations of the wreath product</a>, Acta Math., 131 (1973), 123-143.
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 using a formula for binary Lyndon words. If the value of m is increased, one can enumerate colorings of higher-dimensional elements beginning with T(m,1).
T(n,k) = A337883(n,k) - A337885(n,k) = (A337883(n,k) + A337886(n,k)) / 2 = A337885(n,k) + A337886(n,k).
Table begins with T(2,1):
1 2 3 4 5 6 7 ...
1 5 15 35 70 126 210 ...
1 34 792 10688 90005 533358 2437848 ...
1 2136 4977909 1533771392 132597435125 5079767935320 110837593383153 ...
For T(3,4)=35, the 34 achiral arrangements are AAAA, AAAB, AAAC, AAAD, AABB, AABC, AABD, AACC, AACD, AADD, ABBB, ABBC, ABBD, ABCC, ABDD, ACCC, ACCD, ACDD, ADDD, BBBB, BBBC, BBBD, BBCC, BBCD, BBDD, BCCC, BCCD, BCDD, BDDD, CCCC, CCCD, CCDD, CDDD, and DDDD. The chiral pair is ABCD-ABDC.
m=2; (* dimension of color element, here a triangular face *)
lw[n_, k_]:=lw[n, k]=DivisorSum[GCD[n, k], MoebiusMu[#]Binomial[n/#, k/#]&]/n (*A051168*)
cxx[{a_, b_}, {c_, d_}]:={LCM[a, c], GCD[a, c] b d}
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)
combine[a : {{_, _} ...}, b : {{_, _} ...}] := Outer[cxx, a, b, 1]
CX[p_List, 0] := {{1, 1}} (* cycle index for partition p, m vertices *)
CX[{n_Integer}, m_] := If[2m>n, CX[{n}, n-m], CX[{n}, m] = Table[{n/k, lw[n/k, m/k]}, {k, Reverse[Divisors[GCD[n, m]]]}]]
CX[p_List, m_Integer] := CX[p, m] = Module[{v = Total[p], q, r}, If[2 m > v, CX[p, v - m], q = Drop[p, -1]; r = Last[p]; compress[Flatten[Join[{{CX[q, m]}}, Table[combine[CX[q, m - j], CX[{r}, j]], {j, Min[m, r]}]], 2]]]]
pc[p_] := 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[CX[#, m+1]][[2]] & /@ IntegerPartitions[n+1]]/(n+1)!]
array[n_, k_] := row[n] /. j -> k
Table[array[n, d+m-n], {d, 8}, {n, m, d+m-1}] // Flatten
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Robert A. Russell, Sep 28 2020
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