|
| |
|
|
A338114
|
|
Triangle read by rows: T(n,k) is the number of unoriented 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, 1, 1, 32, 693, 7720, 44150, 138312, 247380, 252000, 136080, 30240, 1, 2134, 4971504, 1513872568, 124978335900, 4307090369304, 78010256156784, 849590196841344, 6053725780061400, 29824685516682000, 105382759395846240, 273441179492268480
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
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. Two unoriented colorings are the same if they are congruent; chiral pairs are counted as one.
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).
|
|
|
LINKS
|
|
|
|
FORMULA
|
A337884(n,k) = Sum_{j=1..C(n+1,3)} T(n,j) * binomial(k,j).
|
|
|
EXAMPLE
|
Triangle begins with T(2,1):
1
1 3 3 1
1 32 693 7720 44150 138312 247380 252000 136080 30240
...
For T(3,2)=3, the tetrahedron has one, two, or three faces (vertices) of one color. For T(3,4)=1, each of the four tetrahedron faces (vertices) is a different color.
|
|
|
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[pc[#] j^Total[CX[#, m+1]][[2]] & /@ IntegerPartitions[n+1]]/(n+1)!]
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
|
|
|
|
KEYWORD
|
nonn,tabf
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
