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 1-8 of 8 results.

A327086 Array read by descending antidiagonals: A(n,k) is the number of achiral colorings of the edges of a regular n-dimensional simplex using up to k colors.

Original entry on oeis.org

1, 2, 1, 3, 4, 1, 4, 9, 10, 1, 5, 16, 45, 28, 1, 6, 25, 136, 387, 128, 1, 7, 36, 325, 2784, 8352, 792, 1, 8, 49, 666, 13125, 186304, 382563, 7620, 1, 9, 64, 1225, 46836, 2117750, 36507008, 44526672, 124344
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. An achiral coloring is identical to its reflection.
A(n,k) is also the number of achiral colorings of (n-2)-dimensional regular simplices in an n-dimensional simplex using up to k colors. Thus, A(2,k) is also the number of achiral colorings of the vertices (0-dimensional simplices) of an equilateral triangle.

Examples

			Array begins with A(1,1):
  1  2   3    4     5     6      7      8      9      10      11      12 ...
  1  4   9   16    25    36     49     64     81     100     121     144 ...
  1 10  45  136   325   666   1225   2080   3321    5050    7381   10440 ...
  1 28 387 2784 13125 46836 137543 349952 797769 1667500 3248971 5973408 ...
  ...
For A(2,3) = 9, the colorings are AAA, AAB, AAC, ABB, ACC, BBB, BBC, BCC, and CCC.
		

Crossrefs

Cf. A327083 (oriented), A327084 (unoriented), A327085 (chiral), A327090 (exactly k colors), A325001 (vertices, facets), A337886 (faces, peaks), A337410 (orthotope edges, orthoplex ridges), A337414 (orthoplex edges, orthotope ridges).
Rows 1-4 are A000027, A000290, A037270, A331353.

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]]], 0, pc[#] j^Total[CycleX[#]][[2]]] & /@ IntegerPartitions[n+1]]/((n+1)!/2)]
    array[n_, k_] := row[n] /. j -> k
    Table[array[n,d-n+1], {d,1,10}, {n,1,d}] // Flatten
    (* Using Fripertinger's exponent per Andrew Howroyd's code in A063841: *)
    pc[p_] := Module[{ci, mb}, mb = DeleteDuplicates[p]; ci = Count[p, #] &/@ mb; Total[p]!/(Times @@ (ci!) Times @@ (mb^ci))]
    ex[v_] := Sum[GCD[v[[i]], v[[j]]], {i,2,Length[v]}, {j,i-1}] + Total[Quotient[v,2]]
    array[n_,k_] := Total[If[OddQ[Total[1-Mod[#,2]]], pc[#]k^ex[#], 0] &/@ IntegerPartitions[n+1]]/((n+1)!/2)
    Table[array[n,d-n+1], {d,10}, {n,d}] // 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.
A(n,k) = Sum_{j=1..(n+1)*n/2} A327090(n,j) * binomial(k,j).
A(n,k) = 2*A327084(n,k) - A327083(n,k) = A327083(n,k) - 2*A327085(n,k) = A327084(n,k) - A327085(n,k).

A337409 Array read by descending antidiagonals: T(n,k) is the number of chiral pairs of colorings of the edges of a regular n-dimensional orthotope (hypercube) using k or fewer colors.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 3, 74, 0, 0, 15, 10704, 11158298, 0, 0, 45, 345640, 4825452718593, 314824408633217132928, 0, 0, 105, 5062600, 48038354542204960, 38491882659952177472606694634030116, 136221825854745676076981182469325427379054390050209792, 0
Offset: 1

Views

Author

Robert A. Russell, Aug 26 2020

Keywords

Comments

Each member of a chiral pair is a reflection, but not a rotation, of the other. For n=1, the figure is a line segment with one edge. For n=2, the figure is a square with 4 edges. For n=3, the figure is a cube with 12 edges. The number of edges is n*2^(n-1).
Also the number of chiral pairs of colorings of the regular (n-2)-dimensional simplexes in a regular n-dimensional orthoplex.

Examples

			Table begins with T(1,1):
0  0     0      0       0        0         0          0          0 ...
0  0     3     15      45      105       210        378        630 ...
0 74 10704 345640 5062600 45246810 288005144 1430618784 5881281480 ...
For T(2,3)=3, the chiral arrangements are AABC-AACB, ABBC-ACBB, and ABCC-ACCB.
		

Crossrefs

Cf. A337407 (oriented), A337408 (unoriented), A337410 (achiral).
Rows 2-4 are A050534, A337406, A331360.
Cf. A327085 (simplex edges), A337413 (orthoplex edges), A325014 (orthotope vertices).

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+2x^(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; 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
    Table[array[n,d+m-n], {d,7}, {n,m,d+m-1}] // Flatten

Formula

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).
T(n,k) = A337407(n,k) - A337408(n,k) = (A337407(n,k) - A337410(n,k)) / 2 = A337408(n,k) - A337410(n,k).

A337414 Array read by descending antidiagonals: T(n,k) is the number of achiral colorings of the edges of a regular n-dimensional orthoplex (cross polytope) using k or fewer colors.

Original entry on oeis.org

1, 2, 1, 3, 6, 1, 4, 18, 70, 1, 5, 40, 1407, 8200, 1, 6, 75, 12480, 9080559, 12804908, 1, 7, 126, 69050, 1503323520, 4906480368591, 304899216832, 1, 8, 196, 281946, 81461669375, 48226825456539776, 187380251418565888983, 103685962258536432, 1
Offset: 1

Views

Author

Robert A. Russell, Aug 26 2020

Keywords

Comments

An achiral arrangement is identical to its reflection. For n=1, the figure is a line segment with one edge. For n=2, the figure is a square with 4 edges. For n=3, the figure is an octahedron with 12 edges. The number of edges is 2n*(n-1) for n>1.
Also the number of achiral colorings of the regular (n-2)-dimensional orthotopes (hypercubes) in a regular n-dimensional orthotope.

Examples

			Table begins with T(1,1):
1  2    3     4     5      6      7       8       9       10 ...
1  6   18    40    75    126    196     288     405      550 ...
1 70 1407 12480 69050 281946 931490 2632512 6598935 15041950 ...
For T(2,2)=6, the arrangements are AAAA, AAAB, AABB, ABAB, ABBB, and BBBB.
		

Crossrefs

Cf. A337411 (oriented), A337412 (unoriented), A337413 (chiral).
Rows 1-4 are A000027, A002411, A331351, A331357.
Cf. A327086 (simplex edges), A337410 (orthotope edges), A325007 (orthoplex vertices).

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
    Table[array[n,d+m-n], {d,8}, {n,m,d+m-1}] // Flatten

Formula

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).
T(n,k) = 2*A337412(n,k) - A337411(n,k) = A337411(n,k) - 2*A337413(n,k) = A337412(n,k) - A337413(n,k).

A331351 Number of achiral colorings of the edges of a cube or regular octahedron.

Original entry on oeis.org

1, 70, 1407, 12480, 69050, 281946, 931490, 2632512, 6598935, 15041950, 31740841, 62830560, 117855192, 211141490, 363551700, 604679936, 975561405, 1531968822, 2348375395, 3522668800, 5181705606, 7487800650, 10646250902
Offset: 1

Views

Author

Robert A. Russell, Jan 14 2020

Keywords

Comments

A cube has 8 vertices and 12 edges. A regular octahedron has 6 vertices and 12 edges. An achiral coloring is identical to its reflection.
From Robert A. Russell, Oct 08 2020: (Start)
The Schläfli symbols for the cube and regular octahedron are {4,3} and {3,4} respectively. They are mutually dual.
There are 24 elements in the automorphism group of the regular octahedron/cube that are not in the rotation group. They divide into five conjugacy classes. The first formula is obtained by averaging the edge cycle indices after replacing x_i^j with n^j according to the Pólya enumeration theorem.
Conjugacy Class Count Odd Cycle Indices
Inversion 1 x_2^6
Vertex rotation* 8 x_6^2 Asterisk indicates that the
Edge rotation* 6 x_1^2x_2^5 operation is followed by an
Small face rotation* 3 x_4^3 inversion.
Large face rotation* 6 x_1^4x_2^4 (End)

Crossrefs

Cf. A060530 (oriented), A199406 (unoriented), A337406 (chiral), A337897 (octahedron faces, cube vertices), A337898 (cube faces, octahedron vertices), A037270 (tetrahedron), A337953 (dodecahedron, icosahedron).
Row 3 of A337410 (orthotope edges, orthoplex ridges) and A337414 (orthoplex edges, orthotope ridges).

Programs

  • Mathematica
    Table[(8n^2 + 6n^3 + n^6 + 6n^7 + 3n^8)/24, {n, 1, 30}]
    LinearRecurrence[{9, -36, 84, -126, 126, -84, 36, -9, 1}, {1, 70, 1407, 12480, 69050, 281946, 931490, 2632512, 6598935}, 25]

Formula

a(n) = (8*n^2 + 6*n^3 + n^6 + 6*n^7 + 3*n^8) / 24.
a(n) = 1*C(n,1) + 68*C(n,2) + 1200*C(n,3) + 7268*C(n,4) + 20025*C(n,5) + 27750*C(n,6) + 18900*C(n,7) + 5040*C(n,8), where the coefficient of C(n,k) is the number of colorings using exactly k colors.
a(n) = 2*A199406(n) - A060530(n) = A060530(n) - 2*A337406(n) = A199406(n) - A337406(n). - Robert A. Russell, Oct 08 2020
G.f.: (x + 61*x^2 + 813*x^3 + 2253*x^4 + 1628*x^5 + 282*x^6 + 2*x^7) / (1-x)^9.
E.g.f.: (1/24)*exp(x)*x*(24 + 816*x + 4800*x^2 + 7268*x^3 + 4005*x^4 + 925*x^5 + 90*x^6 + 3*x^7). - Stefano Spezia, Jan 17 2020

A337407 Array read by descending antidiagonals: T(n,k) is the number of oriented colorings of the edges of a regular n-dimensional orthotope (hypercube) using k or fewer colors.

Original entry on oeis.org

1, 2, 1, 3, 6, 1, 4, 24, 218, 1, 5, 70, 22815, 22409620, 1, 6, 165, 703760, 9651199594275, 629648865090036960064, 1, 7, 336, 10194250, 96076801068337216, 76983765319971869475595432431084156, 272443651709491352597039736725488834366101875164020736, 1
Offset: 1

Views

Author

Robert A. Russell, Aug 26 2020

Keywords

Comments

Each chiral pair is counted as two when enumerating oriented arrangements. For n=1, the figure is a line segment with one edge. For n=2, the figure is a square with 4 edges. For n=3, the figure is a cube with 12 edges. The number of edges is n*2^(n-1).
Also the number of oriented colorings of the regular (n-2)-dimensional simplexes in a regular n-dimensional orthoplex.

Examples

			Table begins with T(1,1):
1   2     3      4        5        6         7          8           9 ...
1   6    24     70      165      336       616       1044        1665 ...
1 218 22815 703760 10194250 90775566 576941778 2863870080 11769161895 ...
For T(2,2)=6, the arrangements are AAAA, AAAB, AABB, ABAB, ABBB, and BBBB.
		

Crossrefs

Cf. A337408 (unoriented), A337409 (chiral), A337410 (achiral).
Rows 1-4 are A000027, A006528, A060530, A331358.
Cf. A327083 (simplex edges), A337411 (orthoplex edges), A325012 (orthotope vertices).

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[array[n,d+m-n], {d,7}, {n,m,d+m-1}] // Flatten

Formula

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).
T(n,k) = A337408(n,k) + A337409(n,k) = 2*A337408(n,k) - A337410(n,k) = 2*A337409(n,k) + A337410(n,k).

A337408 Array read by descending antidiagonals: T(n,k) is the number of unoriented colorings of the edges of a regular n-dimensional orthotope (hypercube) using k or fewer colors.

Original entry on oeis.org

1, 2, 1, 3, 6, 1, 4, 21, 144, 1, 5, 55, 12111, 11251322, 1, 6, 120, 358120, 4825746875682, 314824456456819827136, 1, 7, 231, 5131650, 48038446526132256, 38491882660019692002988737797054040, 136221825854745676520058554256163406987047485113810944, 1
Offset: 1

Views

Author

Robert A. Russell, Aug 26 2020

Keywords

Comments

Each chiral pair is counted as one when enumerating unoriented arrangements. For n=1, the figure is a line segment with one edge. For n=2, the figure is a square with 4 edges. For n=3, the figure is a cube with 12 edges. The number of edges is n*2^(n-1).
Also the number of unoriented colorings of the regular (n-2)-dimensional simplexes in a regular n-dimensional orthoplex.

Examples

			Table begins with T(1,1):
1   2     3      4       5        6         7          8          9 ...
1   6    21     55     120      231       406        666       1035 ...
1 144 12111 358120 5131650 45528756 288936634 1433251296 5887880415 ...
For T(2,2)=6, the arrangements are AAAA, AAAB, AABB, ABAB, ABBB, and BBBB.
		

Crossrefs

Cf. A337407 (oriented), A337409 (chiral), A337410 (achiral).
Rows 1-4 are A000027, A002817, A199406, A331359.
Cf. A327084 (simplex edges), A337412 (orthoplex edges), A325013 (orthotope vertices).

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; 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[n_Integer] := row[n] = Factor[(Total[(PartPol[#] pc[#])&/@ IntegerPartitions[n]])/(n! 2^n)]
    array[n_, k_] := row[n] /. b -> k
    Table[array[n,d+m-n], {d,7}, {n,m,d+m-1}] // Flatten

Formula

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).
T(n,k) = A337407(n,k) - A337409(n,k) = (A337407(n,k) - A337410(n,k)) / 2 = A337409(n,k) + A337410(n,k).

A337890 Array read by descending antidiagonals: T(n,k) is the number of achiral colorings of the square faces of a regular n-dimensional orthotope (hypercube) using k or fewer colors.

Original entry on oeis.org

1, 2, 1, 3, 10, 1, 4, 55, 8200, 1, 5, 200, 9080559, 199556208371776, 1, 6, 560, 1503323520, 1370366433970979158839987, 388032967149969852957120195660938882809069568, 1
Offset: 2

Views

Author

Robert A. Russell, Sep 28 2020

Keywords

Comments

An achiral arrangement is identical to its reflection. Each face is a square bounded by four edges. For n=2, the figure is a square with one face. For n=3, the figure is a cube with 6 faces. For n=4, the figure is a tesseract with 24 faces. The number of faces is 2^(n-2)*C(n,2).
Also the number of chiral pairs of colorings of peaks of an n-dimensional orthoplex. A peak is an (n-3)-dimensional simplex.
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

			Array begins with T(2,1):
1    2       3          4           5             6              7 ...
1   10      55        200         560          1316           2730 ...
1 8200 9080559 1503323520 81461669375 2146080958056 34228350856910 ...
		

Crossrefs

Cf. A337887 (oriented), A337888 (unoriented), A337889 (chiral).
Other elements: A325015 (vertices), A337410 (edges).
Other polytopes: A337886 (simplex), A337894 (orthoplex).
Rows 2-4 are A000027, A337897, A331357.

Programs

  • Mathematica
    m=2; (* dimension of color element, here a square face *)
    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}]],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[n_Integer] := row[n] = Factor[(Total[(PartPol[#] pc[#])&/@ IntegerPartitions[n]])/(n! 2^(n-1))]
    array[n_, k_] := row[n] /. b -> k
    Table[array[n,d+m-n], {d,6}, {n,m,d+m-1}] // Flatten

Formula

T(n,k) = 2*A337888(n,k) - A337887(n,k) = A337887(n,k) - 2*A337889(n,k) = A337888(n,k) - A337889(n,k).

A338145 Triangle read by rows: T(n,k) is the number of achiral 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, 3, 0, 1, 68, 1200, 7268, 20025, 27750, 18900, 5040, 0, 0, 0, 0, 1, 93022, 293878020, 90807857080, 7503022894800, 258528829444320, 4681671089961600, 50981530073846400, 363246007692204000
Offset: 1

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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 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    3    0
  1 68 1200 7268 20025 27750 18900 5040 0 0 0 0
  ...
For T(2,2)=4, the achiral colorings are AAAB, AABB, ABAB, and ABBB. For T(2,3)=3, the achiral colorings are ABAC, ABCB, and ACBC.
		

Crossrefs

Cf. A338142 (oriented), A338143 (unoriented), A338144 (chiral), A337410 (k or fewer colors), A325019 (orthotope vertices, orthoplex facets).
Cf. A327090 (simplex), A338149 (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}]], 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[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

A337410(n,k) = Sum_{j=1..n*2^(n-1)} T(n,j) * binomial(k,j).
T(n,k) = 2*A338143(n,k) - A338142(n,k) = A338142(n,k) - 2*A338144(n,k) = A338143(n,k) - A338144(n,k).
T(2,k) = A338149(2,k) = A325019(2,k) = A325011(2,k); T(3,k) = A338149(3,k).
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