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A338116
Triangle read by rows: T(n,k) is the number of achiral colorings of the faces (and peaks) of a regular n-dimensional simplex using exactly k colors. Row n has C(n+1,3) columns.
3
1, 1, 3, 3, 0, 1, 26, 306, 1400, 2800, 2520, 840, 0, 0, 0, 1, 766, 199902, 10426768, 200588850, 1903776420, 10360383600, 35133957600, 77643846000, 113816253600, 109880971200, 67199932800, 23610787200, 3632428800, 0, 0, 0, 0, 0, 0
OFFSET
2,3
COMMENTS
An n-dimensional simplex has n+1 vertices, C(n+1,3) faces, and C(n+1,3) peaks, which are (n-3)-dimensional simplexes. For n=2, the figure is a triangle with one face. For n=3, the figure is a tetrahedron with four triangular faces and four peaks (vertices). For n=4, the figure is a 4-simplex with ten triangular faces and ten peaks (edges). The Schläfli symbol {3,...,3}, of the regular n-dimensional simplex consists of n-1 3's. An achiral coloring is identical to its reflection.
The algorithm used in the Mathematica program below assigns each permutation of the vertices to a cycle-structure partition of n+1. It then determines the number of permutations for each partition and the cycle index for each partition. If the value of m is increased, one can enumerate colorings of higher-dimensional elements beginning with T(m,1).
FORMULA
A337886(n,k) = Sum_{j=1..C(n+1,3)} T(n,j) * binomial(k,j).
T(n,k) = 2*A338114(n,k) - A338113(n,k) = A338113(n,k) - 2*A338115(n,k) = A338114(n,k) - A338115(n,k).
T(3,k) = A325003(3,k); T(4,k) = A327090(4,k).
EXAMPLE
Triangle begins with T(2,1):
1
1 3 3 0
1 26 306 1400 2800 2520 840 0 0 0
1 766 199902 10426768 200588850 1903776420 10360383600 35133957600 ...
...
For T(3,3)=3, one of the three colors appears on two faces (vertices) of the tetrahedron.
MATHEMATICA
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[If[OddQ[Total[1-Mod[#, 2]]], pc[#] j^Total[CX[#, m+1]][[2]], 0] & /@ IntegerPartitions[n+1]]/((n+1)!/2)]
array[n_, k_] := row[n] /. j -> k
Table[LinearSolve[Table[Binomial[i, j], {i, Binomial[n+1, m+1]}, {j, Binomial[n+1, m+1]}], Table[array[n, k], {k, Binomial[n+1, m+1]}]], {n, m, m+4}] // Flatten
CROSSREFS
Cf. A338113 (oriented), A338114 (unoriented), A338115 (chiral), A337886 (k or fewer colors), A325003 (vertices and facets), A327090 (edges and ridges).
Sequence in context: A141947 A216804 A010607 * A325018 A118522 A179119
KEYWORD
nonn,tabf
AUTHOR
Robert A. Russell, Oct 10 2020
STATUS
approved