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 1-6 of 6 results.

A337886 Array read by descending antidiagonals: T(n,k) is the number of achiral colorings of the triangular faces of a regular n-dimensional simplex using k or fewer colors.

Original entry on oeis.org

1, 2, 1, 3, 5, 1, 4, 15, 28, 1, 5, 34, 387, 768, 1, 6, 65, 2784, 202203, 302032, 1, 7, 111, 13125, 11230976, 7109211078, 3098988832, 1, 8, 175, 46836, 254729375, 9393953524224, 50669807706182691, 1831011525739328, 1
Offset: 2

Views

Author

Robert A. Russell, Sep 28 2020

Keywords

Comments

An achiral arrangement is identical to its reflection. An n-simplex has n+1 vertices. For n=2, the figure is a triangle with one triangular face. For n=3, the figure is a tetrahedron with 4 triangular faces. For higher n, the number of triangular faces is C(n+1,3).
Also the number of achiral colorings of the peaks of a regular n-dimensional simplex. A peak of an n-simplex is an (n-3)-dimensional simplex.

Examples

			Table begins with T(2,1):
1   2      3        4         5          6           7            8 ...
1   5     15       34        65        111         175          260 ...
1  28    387     2784     13125      46836      137543       349952 ...
1 768 202203 11230976 254729375 3267720576 28271133933 183296831488 ...
For T(3,4)=34, the 34 achiral arrangements are AAAA, AAAB, AAAC, AAAD, AABB, AABC, AABD, AACC, AACD, AADD, ABBB, ABBC, ABBD, ABCC, ABDD, ACCC, ACCD, ACDD, ADDD, BBBB, BBBC, BBBD, BBCC, BBCD, BBDD, BCCC, BCCD, BCDD, BDDD, CCCC, CCCD, CCDD, CDDD, and DDDD.
		

Crossrefs

Cf. A337883 (oriented), A337884 (unoriented), A337885 (chiral), A051168 (binary Lyndon words).
Other elements: A325001 (vertices), A327086 (edges).
Other polytopes: A337890 (orthotope), A337894 (orthoplex).
Rows 2-4 are A000027, A006003, A331353.

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[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 vertices to a partition of n+1. It then determines the number of permutations for each partition and the cycle index for each partition using a formula for binary Lyndon words. If the value of m is increased, one can enumerate colorings of higher-dimensional elements beginning with T(m,1).
T(n,k) = A337884(n,k) - A337883(n,k) = A337883(n,k) - 2*A337885(n,k) = A337884(n,k) - A337885(n,k).

A337897 Number of achiral colorings of the 8 triangular faces of a regular octahedron or the 8 vertices of a cube using n or fewer colors.

Original entry on oeis.org

1, 21, 201, 1076, 4025, 11901, 29841, 66256, 134001, 251725, 445401, 750036, 1211561, 1888901, 2856225, 4205376, 6048481, 8520741, 11783401, 16026900, 21474201, 28384301, 37055921, 47831376, 61100625, 77305501, 96944121
Offset: 1

Views

Author

Robert A. Russell, Sep 28 2020

Keywords

Comments

An achiral coloring is identical to its reflection. 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 cube vertex (octahedron face) 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_2^4
Vertex rotation* 8 x_2^1x_6^1 Asterisk indicates that the
Edge rotation* 6 x_1^4x_2^2 operation is followed by an
Small face rotation* 3 x_4^2 inversion.
Large face rotation* 6 x_2^4

Crossrefs

Cf. A000543 (oriented), A128766 (unoriented), A337896 (chiral).
Other elements: A331351 (edges), A337898 (cube faces, octahedron vertices).
Other polyhedra: A006003 (tetrahedron), A337962 (dodecahedron faces, icosahedron vertices), A337960 (icosahedron faces, dodecahedron vertices).
Row 3 of A337894 (orthoplex faces, orthotope peaks) and A325015 (orthotope vertices, orthoplex facets).

Programs

  • Mathematica
    Table[n^2(7+2n^2+3n^4)/12, {n,30}]

Formula

a(n) = n^2 * (7 + 2*n^2 + 3*n^4) / 12.
a(n) = 1*C(n,1) + 19*C(n,2) + 141*C(n,3) + 394*C(n,4) + 450*C(n,5) + 180*C(n,6), where the coefficient of C(n,k) is the number of achiral colorings using exactly k colors.
a(n) = 2*A128766(n) - A000543(n) = A000543(n) - 2*A337896(n) = A128766(n) - A337896(n).
G.f.: x * (1+x) * (1 + 13*x + 62*x^2 + 13*x^3 + x^4) / (1-x)^7.

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).

A337891 Array read by descending antidiagonals: T(n,k) is the number of oriented colorings of the faces of a regular n-dimensional orthoplex (cross polytope) using k or fewer colors.

Original entry on oeis.org

1, 2, 1, 3, 23, 1, 4, 333, 22409620, 1, 5, 2916, 9651199594275, 629648865588086369152, 1, 6, 16725, 96076801068337216, 76983765319971901895960429658208179, 63433230786931550329738915431918588874940416, 1
Offset: 2

Views

Author

Robert A. Russell, Sep 28 2020

Keywords

Comments

Each chiral pair is counted as two when enumerating oriented arrangements. For n=2, the figure is a square with one square face. For n=3, the figure is an octahedron with 8 triangular faces. For higher n, the number of triangular faces is 8*C(n,3).
Also the number of oriented colorings of the peaks of an n-dimensional orthotope (hypercube). A peak is an (n-3)-dimensional orthotope.
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 ...
1       23           333              2916                 16725 ...
1 22409620 9651199594275 96076801068337216 121265960728368199375 ...
		

Crossrefs

Cf. A337892 (unoriented), A337893 (chiral), A337894 (achiral).
Other elements: A325004 (vertices), A337411 (edges).
Other polytopes: A337883 (simplex), A337887 (orthotope).
Rows 2-4 are A000027, A000543, A331358

Programs

  • Mathematica
    m=2; (* dimension of color element, here a 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, 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}]], (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[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,6}, {n,m,d+m-1}] // Flatten

Formula

T(n,k) = A337892(n,k) + A337893(n,k) = 2*A337892(n,k) - A337894(n,k) = 2*A337893(n,k) + A337894(n,k).

A337893 Array read by descending antidiagonals: T(n,k) is the number of chiral pairs of colorings of the faces of a regular n-dimensional orthoplex (cross polytope) using k or fewer colors.

Original entry on oeis.org

0, 0, 0, 0, 1, 0, 0, 66, 11158298, 0, 0, 920, 4825452718593, 314824333015938998688, 0, 0, 6350, 48038354542204960, 38491882659300767730994725249684096, 31716615393292685397985382790580028572676096, 0
Offset: 2

Views

Author

Robert A. Russell, Sep 28 2020

Keywords

Comments

Each member of a chiral pair is a reflection, but not a rotation, of the other. For n=2, the figure is a square with one square face. For n=3, the figure is an octahedron with 8 triangular faces. For higher n, the number of triangular faces is 8*C(n,3).
Also the number of chiral pairs of colorings of the peaks of an n-dimensional orthotope (hypercube). A peak is an (n-3)-dimensional orthotope.

Examples

			Table begins with T(2,1):
 0        0             0                 0                    0 ...
 0        1            66               920                 6350 ...
 0 11158298 4825452718593 48038354542204960 60632976384183154375 ...
		

Crossrefs

Cf. A337891 (oriented), A337892 (unoriented), A337894 (achiral).
Other elements: A325006 (vertices), A337413 (edges).
Other polytopes: A337885 (simplex), A337889 (orthotope).
Rows 2-4 are A000004, A337896, A331360.

Programs

  • Mathematica
    m=2; (* dimension of color element, here a 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, 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
    Table[array[n,d+m-n], {d,6}, {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) = A337891(n,k) - A337892(n,k) = (A337891(n,k) - A337894(n,k)) / 2 = A337892(n,k) - A337894(n,k).

A337892 Array read by descending antidiagonals: T(n,k) is the number of unoriented colorings of the faces of a regular n-dimensional orthoplex (cross polytope) using k or fewer colors.

Original entry on oeis.org

1, 2, 1, 3, 22, 1, 4, 267, 11251322, 1, 5, 1996, 4825746875682, 314824532572147370464, 1, 6, 10375, 48038446526132256, 38491882660671134164965704408524083, 31716615393638864931753532641338560302264320, 1
Offset: 2

Views

Author

Robert A. Russell, Sep 28 2020

Keywords

Comments

Each chiral pair is counted as one when enumerating unoriented arrangements. For n=2, the figure is a square with one square face. For n=3, the figure is an octahedron with 8 triangular faces. For higher n, the number of triangular faces is 8*C(n,3).
Also the number of unoriented colorings of the peaks of an n-dimensional orthotope (hypercube). A peak is an (n-3)-dimensional orthotope.

Examples

			Array begins with T(2,1):
 1        2             3                 4                    5 ...
 1       22           267              1996                10375 ...
 1 11251322 4825746875682 48038446526132256 60632984344185045000 ...
		

Crossrefs

Cf. A337891 (oriented), A337893 (chiral), A337894 (achiral).
Other elements: A325005 (vertices), A337412 (edges).
Other polytopes: A337884 (simplex), A337888 (orthotope).
Rows 2-4 are A000027, A128766, A331359

Programs

  • Mathematica
    m=2; (* dimension of color element, here a 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, 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
    Table[array[n,d+m-n], {d,6}, {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) = A337891(n,k) - A337893(n,k) = (A337891(n,k) + A337894(n,k)) / 2 = A337893(n,k) + A337894(n,k).
Showing 1-6 of 6 results.