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Triangle read by rows: T(n,k) is the number of chiral pairs of 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

%I #10 Oct 14 2020 10:38:13

%S 0,0,0,0,1,0,6,387,6320,41350,135792,246540,252000,136080,30240,0,

%T 1368,4771602,1503445800,124777747050,4305186592884,77999895773184,

%U 849555062883744,6053648136215400,29824571700428400

%N Triangle read by rows: T(n,k) is the number of chiral pairs of colorings of the faces (and peaks) of a regular n-dimensional simplex using exactly k colors. Row n has C(n+1,3) columns.

%C 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. Each member of a chiral pair is a reflection, but not a rotation, of the other.

%C 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).

%H G. Royle, <a href="http://teaching.csse.uwa.edu.au/units/CITS7209/partition.pdf">Partitions and Permutations</a>

%F A337885(n,k) = Sum_{j=1..C(n+1,3)} T(n,j) * binomial(k,j).

%F T(n,k) = A338113(n,k) - A338114(n,k) = (A338113(n,k) - A338116(n,k)) / 2 = A338114(n,k) - A338116(n,k).

%F T(3,k) = [k==4]; T(4,k) = A327089(4,k).

%e Triangle begins with T(2,1):

%e 0

%e 0 0 0 1

%e 0 6 387 6320 41350 135792 246540 252000 136080 30240

%e ...

%e For T(4,4)=1, each of the four tetrahedron faces (vertices) is a different color.

%t m=2; (* dimension of color element, here a triangular face *)

%t lw[n_, k_]:=lw[n, k]=DivisorSum[GCD[n, k], MoebiusMu[#]Binomial[n/#, k/#]&]/n (*A051168*)

%t cxx[{a_, b_}, {c_, d_}]:={LCM[a, c], GCD[a, c] b d}

%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 combine[a : {{_, _} ...}, b : {{_, _} ...}] := Outer[cxx, a, b, 1]

%t CX[p_List, 0] := {{1, 1}} (* cycle index for partition p, m vertices *)

%t 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]]]}]]

%t 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]]]]

%t pc[p_] := 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]]], 1, -1] pc[#] j^Total[CX[#, m+1]][[2]] & /@ IntegerPartitions[n+1]]/(n+1)!]

%t array[n_, k_] := row[n] /. j -> k

%t 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

%Y Cf. A338113 (oriented), A338114 (unoriented), A338116 (achiral), A337885 (k or fewer colors), [k==n+1] (vertices and facets), A327089 (edges and ridges).

%K nonn,tabf

%O 2,7

%A _Robert A. Russell_, Oct 10 2020