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 1-7 of 7 results.

A331355 Number of unoriented colorings of the edges of a regular 4-dimensional orthoplex with n available colors.

Original entry on oeis.org

1, 49127, 740360358, 733776248840, 155261523065875, 12340612271439081, 498926608780739307, 12298018390569089088, 207726683413584244680, 2604177120221402303875, 25650403577338260144611, 207023317470352041578712
Offset: 1

Views

Author

Robert A. Russell, Jan 14 2020

Keywords

Comments

A regular 4-dimensional orthoplex (also hyperoctahedron or cross polytope) has 8 vertices and 24 edges. Its Schläfli symbol is {3,3,4}. Two unoriented colorings are the same if congruent; chiral pairs are counted as one. Also the number of unoriented colorings of the square faces of a tesseract {4,3,3} with n available colors.

Crossrefs

Cf. A331354 (oriented), A331356 (chiral), A331357 (achiral).
Other polychora: A063843 (5-cell), A331359 (8-cell), A338953 (24-cell), A338965 (120-cell, 600-cell).
Row 4 of A337412 (orthoplex edges, orthotope ridges) and A337888 (orthotope faces, orthoplex peaks).

Programs

  • Mathematica
    Table[(48 n^3 + 64 n^4 + 44 n^6 + 84 n^7 + 56 n^8 + 12 n^9 + 5 n^12 +
        36 n^13 + 18 n^14 + 12 n^15 + 4 n^18 + n^24)/384, {n, 1, 25}]

Formula

a(n) = (48*n^3 + 64*n^4 + 44*n^6 + 84*n^7 + 56*n^8 + 12*n^9 + 5*n^12 +
36*n^13 + 18*n^14 + 12*n^15 + 4*n^18 + n^24) / 384.
a(n) = C(n,1) + 49125*C(n, 2) + 740212980*C(n, 3) + 730815102166*C(n, 4) + 151600044933990*C(n, 5) + 11420034970306170*C(n, 6) + 415777158607920585*C(n, 7) + 8643499341510394200*C(n, 8) + 113988734942055623055*C(n, 9) + 1023002477284840979850*C(n, 10) + 6559265715033958749900*C(n, 11) + 31097943476763200314200*C(n, 12) + 111710751446923209781200*C(n, 13) + 309231173588248964052000*C(n, 14) + 666846649590586048584000*C(n, 15) + 1126625898539640346848000*C(n, 16) + 1492173541849975272288000*C(n, 17) + 1541987122059614438208000*C(n, 18) + 1229356526029003532160000*C(n, 19) + 741102367008078915840000*C(n, 20) + 326583680209195368960000*C(n, 21) + 99234043419574103040000*C(n, 22) + 18581137031073576960000*C(n, 23) + 1615751046180311040000*C(n, 24), where the coefficient of C(n,k) is the number of colorings using exactly k colors.
a(n) = A331354(n) - A331356(n) = (A331354(n) + A331357(n)) / 2 = A331356(n) + A331357(n).

A337884 Array read by descending antidiagonals: T(n,k) is the number of unoriented 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, 34, 1, 5, 35, 792, 2136, 1, 6, 70, 10688, 4977909, 7013320, 1, 7, 126, 90005, 1533771392, 9930666709494, 1788782616656, 1, 8, 210, 533358, 132597435125, 234249157811872000, 12979877431438089379035, 53304527811667897248, 1
Offset: 2

Views

Author

Robert A. Russell, Sep 28 2020

Keywords

Comments

Each chiral pair is counted as one when enumerating unoriented arrangements. 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 unoriented 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 ...
 1    5      15         35           70           126             210 ...
 1   34     792      10688        90005        533358         2437848 ...
 1 2136 4977909 1533771392 132597435125 5079767935320 110837593383153 ...
For T(3,4)=35, 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. The chiral pair is ABCD-ABDC.
		

Crossrefs

Cf. A337883 (oriented), A337885 (chiral), A337886 (achiral), A051168 (binary Lyndon words).
Other elements: A325000 (vertices), A327084 (edges).
Other polytopes: A337888 (orthotope), A337892 (orthoplex).
Rows 2-4 are A000027, A000332(n+3), A063843.

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[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) = A337883(n,k) - A337885(n,k) = (A337883(n,k) + A337886(n,k)) / 2 = A337885(n,k) + A337886(n,k).

A198833 The number of inequivalent ways to color the vertices of a regular octahedron using at most n colors.

Original entry on oeis.org

1, 10, 56, 220, 680, 1771, 4060, 8436, 16215, 29260, 50116, 82160, 129766, 198485, 295240, 428536, 608685, 848046, 1161280, 1565620, 2081156, 2731135, 3542276, 4545100, 5774275, 7268976, 9073260, 11236456, 13813570, 16865705, 20460496, 24672560, 29583961
Offset: 1

Views

Author

Geoffrey Critzer, Oct 30 2011

Keywords

Comments

The cycle index: 1/48 (s_1^6 + 3 s_1^4 s_2 + 9 s_1^2 s_2^2 +7 s_2^3 + 8 s_3^2 + 6 s_1^2 s_4 + 6 s_2 s_4 + 8 s_6) is returned in Mathematica by CycleIndex[ Automorphisms[ OctahedralGraph ], s].
One-sixth the area of the right triangles with sides 2b+2, b^2+2b, and b^2+2b+2 with b = A000217(n), the n-th triangular number. - J. M. Bergot, Aug 02 2013
Also the number of ways to color the faces of a cube with n colors, counting each pair of mirror images as one.

Crossrefs

Cf. A047780 (oriented), A093566(n+1) (chiral), A337898 (achiral), A199406 (edges), A128766 (octahedron faces, cube vertices), A000332(n+3) (tetrahedron), A128766 (octahedron faces, cube vertices), A252705 (dodecahedron faces, icosahedron vertices), A252704 (icosahedron faces, dodecahedron vertices), A000217 (triangular numbers).
Row 3 of A325005 (orthotope facets, orthoplex vertices) and A337888 (orthotope faces, orthoplex peaks).

Programs

  • Magma
    [n*(n+1)*(n^2+n+2)*(n^2+n+4)/48: n in [1..35]]; // Vincenzo Librandi, Aug 04 2013
  • Mathematica
    Table[(n^6 + 3 n^5 + 9 n^4 + 13 n^3 + 14 n^2 + 8 n)/48, {n, 25}]
    CoefficientList[Series[-(1 + 3 x + 7 x^2 + 3 x^3 + x^4) / (x - 1)^7, {x, 0, 35}], x] (* Vincenzo Librandi, Aug 04 2013 *)
    LinearRecurrence[{7,-21,35,-35,21,-7,1},{1,10,56,220,680,1771,4060},40] (* Harvey P. Dale, Nov 06 2024 *)
  • PARI
    a(n)=n*(n+1)*(n^2+n+2)*(n^2+n+4)/48 \\ Charles R Greathouse IV, Aug 02 2013
    

Formula

a(n) = n*(n+1)*(n^2+n+2)*(n^2+n+4)/48.
G.f.: x*(1+3*x+7*x^2+3*x^3+x^4) / (1-x)^7. - R. J. Mathar, Oct 30 2011
a(n) = Sum_{i=1..A000217(n)} A000217(i). [Bruno Berselli, Sep 06 2013]
a(n) = 1*C(n,1) + 8*C(n,2) + 29*C(n,3) + 52*C(n,4) + 45*C(n,5) + 15*C(n,6), where the coefficient of C(n,k) is the number of unoriented colorings using exactly k colors.
a(n) = A047780(n) - A093566(n+1) = (A047780(n) + A337898(n)) / 2 = A093566(n+1) + A337898(n). - Robert A. Russell, Oct 19 2020

A337887 Array read by descending antidiagonals: T(n,k) is the number of oriented 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, 57, 90054, 1, 5, 240, 1471640157, 629648865588086369152, 1, 6, 800, 1466049174160, 76983765319971901895960429658208179, 76686070519895153193719509580895099970955878067526648007224125292544, 1
Offset: 2

Views

Author

Robert A. Russell, Sep 28 2020

Keywords

Comments

Each chiral pair is counted as two when enumerating oriented arrangements. 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 oriented colorings of peaks of an n-dimensional orthoplex. A peak is an (n-3)-dimensional simplex.

Examples

			Array begins with T(2,1):
 1     2          3             4               5                 6 ...
 1    10         57           240             800              2226 ...
 1 90054 1471640157 1466049174160 310441584462375 24679078461920106 ...
		

Crossrefs

Cf. A337888 (unoriented), A337889 (chiral), A337890 (achiral).
Other elements: A325012 (vertices), A337407 (edges).
Other polytopes: A337883 (simplex), A337891 (orthoplex).
Rows 2-4 are A000027, A047780, A331354.

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}]], (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,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) = A337888(n,k) + A337889(n,k) = 2*A337888(n,k) - A337890(n,k) = 2*A337889(n,k) + A337890(n,k).

A337889 Array read by descending antidiagonals: T(n,k) is the number of chiral pairs of colorings of the square faces of a regular n-dimensional orthotope (hypercube) using k or fewer colors.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 1, 40927, 0, 0, 20, 731279799, 314824333015938998688, 0, 0, 120, 732272925320, 38491882659300767730994725249684096, 38343035259947576596859560773963975000551460473665493534170658111488, 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. 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):
0     0         0            0               0                 0 ...
0     0         1           20             120               455 ...
0 40927 731279799 732272925320 155180061396500 12338466190481025 ...
		

Crossrefs

Cf. A337887 (oriented), A337888 (unoriented), A337890 (achiral).
Other elements: A325014 (vertices), A337409 (edges).
Other polytopes: A337885 (simplex), A337893 (orthoplex).
Rows 2-4 are A000004, A093566(n+1), A331356.

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+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,6}, {n,m,d+m-1}] // Flatten

Formula

T(n,k) = A337887(n,k) - A337888(n,k) = (A337887(n,k) - A337890(n,k)) / 2 = A337888(n,k) - A337890(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).

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 1-7 of 7 results.