cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

A338113 Triangle read by rows: T(n,k) is the number of oriented colorings of the faces (and peaks) of a regular n-dimensional simplex using exactly k colors. Row n has C(n+1,3) columns.

Original entry on oeis.org

1, 1, 3, 3, 2, 1, 38, 1080, 14040, 85500, 274104, 493920, 504000, 272160, 60480, 1, 3502, 9743106, 3017318368, 249756082950, 8612276962188, 156010151929968, 1699145259725088, 12107373916276800, 59649257217110400
Offset: 2

Views

Author

Robert A. Russell, Oct 10 2020

Keywords

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 oriented colorings are the same if one is a rotation of the other; chiral pairs are counted as two.
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).

Examples

			Triangle begins with T(2,1):
  1
  1  3    3     2
  1 38 1080 14040 85500 274104 493920 504000 272160 60480
  ...
For T(3,2)=3, the tetrahedron has one, two, or three faces (vertices) of one color.
		

Crossrefs

Cf. A338114 (unoriented), A338115 (chiral), A338116 (achiral), A337883 (k or fewer colors), A325002 (vertices and facets), A327087 (edges and ridges).

Programs

  • 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[EvenQ[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

Formula

A337883(n,k) = Sum_{j=1..C(n+1,3)} T(n,j) * binomial(k,j).
T(n,k) = A338114(n,k) + A338115(n,k) = 2*A338114(n,k) - A338116(n,k) = 2*A338115(n,k) + A338116(n,k).
T(3,k) = A325002(3,k); T(4,k) = A327087(4,k).

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.

Original entry on oeis.org

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
Offset: 2

Views

Author

Robert A. Russell, Oct 10 2020

Keywords

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).

Examples

			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.
		

Crossrefs

Cf. A338113 (oriented), A338115 (chiral), A338116 (achiral), A337884 (k or fewer colors), A007318(n,k-1) (vertices and facets), A327088 (edges and ridges).

Programs

  • 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

Formula

A337884(n,k) = Sum_{j=1..C(n+1,3)} T(n,j) * binomial(k,j).
T(n,k) = A338113(n,k) - A338115(n,k) = (A338113(n,k) + A338116(n,k)) / 2 = A338115(n,k) + A338116(n,k).
T(3,k) = A007318(3,k-1); T(4,k) = A327088(4,k).

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.

Original entry on oeis.org

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

Views

Author

Robert A. Russell, Oct 10 2020

Keywords

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).

Examples

			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.
		

Crossrefs

Cf. A338113 (oriented), A338114 (unoriented), A338115 (chiral), A337886 (k or fewer colors), A325003 (vertices and facets), A327090 (edges and ridges).

Programs

  • 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

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).
Showing 1-3 of 3 results.