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 1-5 of 5 results.

A327087 Triangle read by rows: T(n,k) is the number of oriented colorings of the edges of a regular n-dimensional simplex using exactly k colors. Row n has (n+1)*n/2 columns.

Original entry on oeis.org

1, 1, 2, 2, 1, 10, 54, 136, 150, 60, 1, 38, 1080, 14040, 85500, 274104, 493920, 504000, 272160, 60480, 1, 182, 42111, 2848060, 70361815, 841626366, 5722670282, 24262499520, 67665563280, 127639512000, 164044520640, 141664723200, 78702624000, 25427001600, 3632428800
Offset: 1

Views

Author

Robert A. Russell, Aug 19 2019

Keywords

Comments

An n-dimensional simplex has n+1 vertices and (n+1)*n/2 edges. For n=1, the figure is a line segment with one edge. For n=2, the figure is a triangle with three edges. For n=3, the figure is a tetrahedron with six edges. The Schläfli symbol {3,...,3}, of the regular n-dimensional simplex consists of n-1 threes. Two oriented colorings are the same if one is a rotation of the other; chiral pairs are counted as two.
T(n,k) is also the number of oriented colorings of (n-2)-dimensional regular simplices in an n-dimensional simplex using exactly k colors. Thus, T(2,k) is also the number of oriented colorings of the vertices (0-dimensional simplices) of an equilateral triangle.

Examples

			Triangle begins with T(1,1):
  1
  1  2    2
  1 10   54   136   150     60
  1 38 1080 14040 85500 274104 493920 504000 272160 60480
  ...
For T(2,1)=1, all edges of the triangle are the same color. For T(2,2)=2, the edges are AAB and ABB. For T(2,3)=2, the chiral pair is ABC-ACB.
		

Crossrefs

Cf. A327088 (unoriented), A327089 (chiral), A327090 (achiral), A327083 (up to k colors), A325002 (vertices and facets), A338113 (faces and peaks), A338142 (orthotope edges, orthoplex ridges), A338146 (orthoplex edges, orthotope ridges).

Programs

  • Mathematica
    CycleX[{2}] = {{1,1}}; (* cycle index for permutation with given cycle structure *)
    CycleX[{n_Integer}] := CycleX[n] = If[EvenQ[n], {{n/2,1}, {n,(n-2)/2}}, {{n,(n-1)/2}}]
    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)
    CycleX[p_List] := CycleX[p] = compress[Join[CycleX[Drop[p,-1]], If[Last[p] > 1, CycleX[{Last[p]}], ## &[]], If[# == Last[p], {#, Last[p]}, {LCM[#, Last[p]], GCD[#, Last[p]]}] & /@ Drop[p,-1]]]
    pc[p_List] := 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[CycleX[#]][[2]], 0] & /@ IntegerPartitions[n+1]]/((n+1)!/2)]
    array[n_, k_] := row[n] /. j -> k
    Table[LinearSolve[Table[Binomial[i,j], {i,1,(n+1)n/2}, {j,1,(n+1)n/2}], Table[array[n,k], {k,1,(n+1)n/2}]], {n,1,6}] // Flatten

Formula

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.
A327083(n,k) = Sum_{j=1..(n+1)*n/2} T(n,j) * binomial(k,j).
T(n,k) = A327088(n,k) + A327089(n,k) = 2*A327088(n,k) - A327090(n,k) = 2*A327089(n,k) + A327090(n,k).

A338142 Triangle read by rows: T(n,k) is the number of oriented colorings of the edges of a regular n-D orthotope (or ridges of a regular n-D orthoplex) using exactly k colors. Row n has n*2^(n-1) columns.

Original entry on oeis.org

1, 1, 4, 9, 6, 1, 216, 22164, 613804, 6901425, 39713430, 131754420, 267165360, 336798000, 257796000, 109771200, 19958400, 1, 22409618, 9651132365418, 96038196404417832, 120785673234798359850
Offset: 1

Views

Author

Robert A. Russell, Oct 12 2020

Keywords

Comments

Each chiral pair is counted as two when enumerating oriented arrangements. A ridge is an (n-2)-face of an n-D polytope. For n=1, the figure is a line segment with one edge. For n=2, the figure is a square with 4 edges (vertices). For n=3, the figure is a cube (octahedron) with 12 edges. The number of edges (ridges) is n*2^(n-1). The Schläfli symbols for the n-D orthotope (hypercube) and the n-D orthoplex (hyperoctahedron, cross polytope) are {4,...,3,3} and {3,3,...,4} respectively, with n-2 3's in each case. The figures are mutually dual.
The algorithm used in the Mathematica program below assigns each permutation of the axes to a partition of n and then considers separate conjugacy classes for axis reversals. It uses the formulas in Balasubramanian's paper. 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(1,1):
  1
  1   4     9      6
  1 216 22164 613804 6901425 39713430 131754420 267165360 336798000
  ...
		

Crossrefs

Cf. A338143 (unoriented), A338144 (chiral), A338145 (achiral), A337407 (k or fewer colors), A325016 (orthotope vertices, orthoplex facets).
Cf. A327087 (simplex), A338146 (orthoplex edges, orthotope ridges).

Programs

  • Mathematica
    m=1; (* dimension of color element, here an edge *)
    Fi1[p1_] := Module[{g, h}, Coefficient[Product[g = GCD[k1, p1]; h = GCD[2 k1, p1]; (1 + 2 x^(k1/g))^(r1[[k1]] g) If[Divisible[k1, h], 1, (1+2x^(2 k1/h))^(r2[[k1]] h/2)], {k1, Flatten[Position[cs, n1_ /; n1 > 0]]}], x, n - m]];
    FiSum[] := (Do[Fi2[k2] = Fi1[k2], {k2, Divisors[per]}]; DivisorSum[per, DivisorSum[d1 = #, MoebiusMu[d1/#] Fi2[#] &]/# &]);
    CCPol[r_List] := (r1 = r; r2 = cs - r1; If[EvenQ[Sum[If[EvenQ[j3], r1[[j3]], r2[[j3]]], {j3, n}]], (per = LCM @@ Table[If[cs[[j2]] == r1[[j2]], If[0 == cs[[j2]], 1, j2], 2j2], {j2, n}]; Times @@ Binomial[cs, r1] 2^(n-Total[cs]) b^FiSum[]), 0]);
    PartPol[p_List] := (cs = Count[p, #]&/@ Range[n]; Total[CCPol[#]&/@ Tuples[Range[0, cs]]]);
    pc[p_List] := Module[{ci, mb}, mb = DeleteDuplicates[p]; ci = Count[p, #]&/@ mb; n!/(Times@@(ci!) Times@@(mb^ci))] (*partition count*)
    row[n_Integer] := row[n] = Factor[(Total[(PartPol[#] pc[#])&/@ IntegerPartitions[n]])/(n! 2^(n-1))]
    array[n_, k_] := row[n] /. b -> k
    Table[LinearSolve[Table[Binomial[i,j],{i,2^(n-m)Binomial[n,m]},{j,2^(n-m)Binomial[n,m]}], Table[array[n,k],{k,2^(n-m)Binomial[n,m]}]], {n,m,m+4}] // Flatten

Formula

A337407(n,k) = Sum_{j=1..n*2^(n-1)} T(n,j) * binomial(k,j).
T(n,k) = A338143(n,k) + A338144(n,k) = 2*A338143(n,k) - A338145(n,k) = 2*A338144(n,k) + A338145(n,k).
T(2,k) = A338146(2,k) = A325016(2,k) = A325008(2,k); T(3,k) = A338146(3,k).

A338147 Triangle read by rows: T(n,k) is the number of unoriented colorings of the edges of a regular n-D orthoplex (or ridges of a regular n-D orthotope) using exactly k colors. Row 1 has 1 column; row n>1 has 2*n*(n-1) columns.

Original entry on oeis.org

1, 1, 4, 6, 3, 1, 142, 11682, 310536, 3460725, 19870590, 65886660, 133585200, 168399000, 128898000, 54885600, 9979200, 1, 49125, 740212980, 730815102166, 151600044933990, 11420034970306170, 415777158607920585
Offset: 1

Views

Author

Robert A. Russell, Oct 12 2020

Keywords

Comments

Each chiral pair is counted as one when enumerating unoriented arrangements. A ridge is an (n-2)-face of an n-D polytope. For n=1, the figure is a line segment with one edge. For n=2, the figure is a square with 4 edges (vertices). For n=3, the figure is an octahedron (cube) with 12 edges. For n>1, the number of edges (ridges) is 2*n*(n-1). The Schläfli symbols for the n-D orthotope (hypercube) and the n-D orthoplex (hyperoctahedron, cross polytope) are {4,3,...,3,3} and {3,3,...,3,4} respectively, with n-2 3's in each case. The figures are mutually dual.
The algorithm used in the Mathematica program below assigns each permutation of the axes to a partition of n and then considers separate conjugacy classes for axis reversals. It uses the formulas in Balasubramanian's paper. 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(1,1):
  1
  1   4     6      3
  1 142 11682 310536 3460725 19870590 65886660 133585200 168399000
  ...
For T(2,3)=6, the 3 achiral colorings are ABAC, ABCB, and ACBC. The three chiral pairs are AABC-AACB, ABBC-ACBB, and ABCC-ACCB.
		

Crossrefs

Cf. A338146 (oriented), A338148 (chiral), A338149 (achiral), A337412 (k or fewer colors), A325009 (orthoplex vertices, orthotope facets).
Cf. A327088 (simplex), A338143 (orthotope edges, orthoplex ridges).

Programs

  • Mathematica
    m=1; (* dimension of color element, here an edge *)
    Fi1[p1_] := Module[{g, h}, Coefficient[Product[g = GCD[k1, p1]; h = GCD[2 k1, p1]; (1 + 2 x^(k1/g))^(r1[[k1]] g) If[Divisible[k1, h], 1, (1+2x^(2 k1/h))^(r2[[k1]] h/2)], {k1, Flatten[Position[cs, n1_ /; n1 > 0]]}], x, m+1]];
    FiSum[] := (Do[Fi2[k2] = Fi1[k2], {k2, Divisors[per]}]; DivisorSum[per, DivisorSum[d1 = #, MoebiusMu[d1/#] Fi2[#] &]/# &]);
    CCPol[r_List] := (r1 = r; r2 = cs - r1; per = LCM @@ Table[If[cs[[j2]] == r1[[j2]], If[0 == cs[[j2]], 1, j2], 2j2], {j2, n}]; Times @@ Binomial[cs, r1] 2^(n-Total[cs]) b^FiSum[]);
    PartPol[p_List] := (cs = Count[p, #]&/@ Range[n]; Total[CCPol[#]&/@ Tuples[Range[0, cs]]]);
    pc[p_List] := Module[{ci, mb}, mb = DeleteDuplicates[p]; ci = Count[p, #]&/@ mb; n!/(Times@@(ci!) Times@@(mb^ci))] (*partition count*)
    row[m]=b; row[n_Integer] := row[n] = Factor[(Total[(PartPol[#] pc[#])&/@ IntegerPartitions[n]])/(n! 2^n)]
    array[n_, k_] := row[n] /. b -> k
    Join[{{1}},Table[LinearSolve[Table[Binomial[i,j],{i,2^(m+1)Binomial[n,m+1]},{j,2^(m+1)Binomial[n,m+1]}], Table[array[n,k],{k,2^(m+1)Binomial[n,m+1]}]], {n,m+1,m+4}]] // Flatten

Formula

For n>1, A337412(n,k) = Sum_{j=1..2*n*(n-1)} T(n,j) * binomial(k,j).
T(n,k) = A338146(n,k) - A338148(n,k) = (A338146(n,k) + A338149(n,k)) / 2 = A338148(n,k) + A338149(n,k).
T(2,k) = A338143(2,k) = A325017(2,k) = A325009(2,k); T(3,k) = A338143(3,k).

A338148 Triangle read by rows: T(n,k) is the number of chiral pairs of colorings of the edges of a regular n-D orthoplex (or ridges of a regular n-D orthotope) using exactly k colors. Row 1 has 1 column; row n>1 has 2*n*(n-1) columns.

Original entry on oeis.org

0, 0, 0, 3, 3, 0, 74, 10482, 303268, 3440700, 19842840, 65867760, 133580160, 168399000, 128898000, 54885600, 9979200, 0, 40927, 731157018, 729348051686, 151526009158620, 11418355290999750, 415756294427389020, 8643340000393019040
Offset: 1

Views

Author

Robert A. Russell, Oct 12 2020

Keywords

Comments

Chiral colorings come in pairs, each the reflection of the other. A ridge is an (n-2)-face of an n-D polytope. For n=1, the figure is a line segment with one edge. For n=2, the figure is a square with 4 edges (vertices). For n=3, the figure is an octahedron (cube) with 12 edges. For n>1, the number of edges (ridges) is 2*n*(n-1). The Schläfli symbols for the n-D orthotope (hypercube) and the n-D orthoplex (hyperoctahedron, cross polytope) are {4,3,...,3,3} and {3,3,...,3,4} respectively, with n-2 3's in each case. The figures are mutually dual.
The algorithm used in the Mathematica program below assigns each permutation of the axes to a partition of n and then considers separate conjugacy classes for axis reversals. It uses the formulas in Balasubramanian's paper. 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(1,1):
  0
  0  0     3      3
  0 74 10482 303268 3440700 19842840 65867760 133580160 168399000
  ...
For T(2,3)=3, the chiral pairs are AABC-AACB, ABBC-ACBB, and ABCC-ACCB. For T(2,4)=3, the chiral pairs are ABCD-ADCB, ACBD-ADBC, and ABDC-ACDB.
		

Crossrefs

Cf. A338146 (oriented), A338147 (unoriented), A338149 (achiral), A337413 (k or fewer colors), A325010 (orthoplex vertices, orthotope facets).
Cf. A327089 (simplex), A338144 (orthotope edges, orthoplex ridges).

Programs

  • Mathematica
    m=1; (* dimension of color element, here an edge *)
    Fi1[p1_] := Module[{g, h}, Coefficient[Product[g = GCD[k1, p1]; h = GCD[2 k1, p1]; (1 + 2 x^(k1/g))^(r1[[k1]] g) If[Divisible[k1, h], 1, (1+2x^(2 k1/h))^(r2[[k1]] h/2)], {k1, Flatten[Position[cs, n1_ /; n1 > 0]]}], x, m+1]];
    FiSum[] := (Do[Fi2[k2] = Fi1[k2], {k2, Divisors[per]}]; DivisorSum[per, DivisorSum[d1 = #, MoebiusMu[d1/#] Fi2[#] &]/# &]);
    CCPol[r_List] := (r1 = r; r2 = cs - r1; per = LCM @@ Table[If[cs[[j2]] == r1[[j2]], If[0 == cs[[j2]], 1, j2], 2j2], {j2, n}]; If[EvenQ[Sum[If[EvenQ[j3], r1[[j3]], r2[[j3]]], {j3, n}]], 1, -1]Times @@ Binomial[cs, r1] 2^(n-Total[cs]) b^FiSum[]);
    PartPol[p_List] := (cs = Count[p, #]&/@ Range[n]; Total[CCPol[#]&/@ Tuples[Range[0, cs]]]);
    pc[p_List] := Module[{ci, mb}, mb = DeleteDuplicates[p]; ci = Count[p, #]&/@ mb; n!/(Times@@(ci!) Times@@(mb^ci))] (*partition count*)
    row[n_Integer] := row[n] = Factor[(Total[(PartPol[#] pc[#])&/@ IntegerPartitions[n]])/(n! 2^n)]
    array[n_, k_] := row[n] /. b -> k
    Join[{{0}},Table[LinearSolve[Table[Binomial[i,j],{i,2^(m+1)Binomial[n,m+1]},{j,2^(m+1)Binomial[n,m+1]}], Table[array[n,k],{k,2^(m+1)Binomial[n,m+1]}]], {n,m+1,m+4}]] // Flatten

Formula

For n>1, A337413(n,k) = Sum_{j=1..2*n*(n-1)} T(n,j) * binomial(k,j).
T(n,k) = A338146(n,k) - A338147(n,k) = (A338146(n,k) - A338149(n,k)) / 2 = A338147(n,k) - A338149(n,k).
T(2,k) = A338144(2,k) = A325018(2,k) = A325010(2,k); T(3,k) = A338144(3,k).

A338149 Triangle read by rows: T(n,k) is the number of achiral colorings of the edges of a regular n-D orthoplex (or ridges of a regular n-D orthotope) using exactly k colors. Row 1 has 1 column; row n>1 has 2*n*(n-1) columns.

Original entry on oeis.org

1, 1, 4, 3, 0, 1, 68, 1200, 7268, 20025, 27750, 18900, 5040, 0, 0, 0, 0, 1, 8198, 9055962, 1467050480, 74035775370, 1679679306420, 20864180531565, 159341117375160, 804216787965360, 2808560520334800, 6981656802951600
Offset: 1

Views

Author

Robert A. Russell, Oct 12 2020

Keywords

Comments

An achiral coloring is identical to its reflection. A ridge is an (n-2)-face of an n-D polytope. For n=1, the figure is a line segment with one edge. For n=2, the figure is a square with 4 edges (vertices). For n=3, the figure is an octahedron (cube) with 12 edges. For n>1, the number of edges (ridges) is 2*n*(n-1). The Schläfli symbols for the n-D orthotope (hypercube) and the n-D orthoplex (hyperoctahedron, cross polytope) are {4,3,...,3,3} and {3,3,...,3,4} respectively, with n-2 3's in each case. The figures are mutually dual.
The algorithm used in the Mathematica program below assigns each permutation of the axes to a partition of n and then considers separate conjugacy classes for axis reversals. It uses the formulas in Balasubramanian's paper. 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(1,1):
  1
  1    4       3          0
  1   68    1200       7268       20025         27750 18900 5040 0 0 0 0
  1 8198 9055962 1467050480 74035775370 1679679306420 ...
  ...
For T(2,2)=4, the achiral colorings are AAAB, AABB, ABAB, and ABBB. For T(2,3)=3, the 3 achiral colorings are ABAC, ABCB, and ACBC.
		

Crossrefs

Cf. A338146 (oriented), A338147 (unoriented), A338148 (chiral), A337414 (k or fewer colors), A325011 (orthoplex vertices, orthotope facets).
Cf. A327090 (simplex), A338145 (orthotope edges, orthoplex ridges).

Programs

  • Mathematica
    m=1; (* dimension of color element, here an edge *)
    Fi1[p1_] := Module[{g, h}, Coefficient[Product[g = GCD[k1, p1]; h = GCD[2 k1, p1]; (1 + 2 x^(k1/g))^(r1[[k1]] g) If[Divisible[k1, h], 1, (1+2x^(2 k1/h))^(r2[[k1]] h/2)], {k1, Flatten[Position[cs, n1_ /; n1 > 0]]}], x, m+1]];
    FiSum[] := (Do[Fi2[k2] = Fi1[k2], {k2, Divisors[per]}]; DivisorSum[per, DivisorSum[d1 = #, MoebiusMu[d1/#] Fi2[#] &]/# &]);
    CCPol[r_List] := (r1 = r; r2 = cs - r1; If[EvenQ[Sum[If[EvenQ[j3], r1[[j3]], r2[[j3]]], {j3, n}]], 0, (per = LCM @@ Table[If[cs[[j2]] == r1[[j2]], If[0 == cs[[j2]], 1, j2], 2j2], {j2, n}]; Times @@ Binomial[cs, r1] 2^(n-Total[cs]) b^FiSum[])]);
    PartPol[p_List] := (cs = Count[p, #]&/@ Range[n]; Total[CCPol[#]&/@ Tuples[Range[0, cs]]]);
    pc[p_List] := Module[{ci, mb}, mb = DeleteDuplicates[p]; ci = Count[p, #]&/@ mb; n!/(Times@@(ci!) Times@@(mb^ci))] (*partition count*)
    row[m]=b; row[n_Integer] := row[n] = Factor[(Total[(PartPol[#] pc[#])&/@ IntegerPartitions[n]])/(n! 2^(n-1))]
    array[n_, k_] := row[n] /. b -> k
    Join[{{1}},Table[LinearSolve[Table[Binomial[i,j],{i,2^(m+1)Binomial[n,m+1]},{j,2^(m+1)Binomial[n,m+1]}], Table[array[n,k],{k,2^(m+1)Binomial[n,m+1]}]], {n,m+1,m+4}]] // Flatten

Formula

For n>1, A337414(n,k) = Sum_{j=1..2*n*(n-1)} T(n,j) * binomial(k,j).
T(n,k) = 2*A338147(n,k) - A338146(n,k) = A338146(n,k) - 2*A338148(n,k) = A338147(n,k) - A338148(n,k).
T(2,k) = A338145(2,k) = A325019(2,k) = A325011(2,k); T(3,k) = A338145(3,k).
Showing 1-5 of 5 results.