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 1-9 of 9 results.

A199406 The number of inequivalent ways to color the edges of a cube using at most n colors.

Original entry on oeis.org

1, 144, 12111, 358120, 5131650, 45528756, 288936634, 1433251296, 5887880415, 20842168600, 65402344161, 185788177224, 485443851256, 1181242399260, 2703252560100, 5864398969216, 12138503871789, 24101498435616, 46112016365155, 85335258695400, 153249227870046
Offset: 1

Views

Author

Geoffrey Critzer, Nov 05 2011

Keywords

Comments

Two edge colorings are equivalent if one is the mirror image of the other or the cube can be picked up and rotated in any manner to obtain the other.
The group here has order 48 (compare A060530). - N. J. A. Sloane, Aug 14 2012
Also the number of unoriented colorings of the 12 edges of a regular octahedron with n or fewer colors. The Schläfli symbols of the cube and octahedron are {4,3} and {3,4} respectively. They are mutually dual. For an unoriented coloring, chiral pairs are counted as one. - Robert A. Russell, Oct 17 2020

Crossrefs

Cf. A060530 (oriented), A337406 (chiral), A331351 (achiral), A128766 (cube vertices, octahedron faces), A198833 (cube faces, octahedron vertices), A063842(n-1) (tetrahedron), A337963 (dodecahedron, icosahedron).
Row 3 of A337408 (orthotope edges, orthoplex ridges) and A337412 (orthoplex edges, orthotope ridges).

Programs

  • Mathematica
    Table[CycleIndex[KSubsetGroup[Automorphisms[CubicalGraph], Edges[CubicalGraph]],s] /. Table[s[i]->n, {i,1,6}], {n,1,15}]
    Table[(8n^2+12n^3+8n^4+4n^6+12n^7+3n^8+n^12)/48, {n,20}] (* Robert A. Russell, Oct 17 2020 *)

Formula

a(n) = n^12/48 + n^8/16 + n^7/4 + n^6/12 + n^4/6 + n^3/4 + n^2/6.
Cycle index = 1/48(s_1^12+3s_1^4s_2^4+12s_1^2s_2^5+4s_2^6+8s_3^4+12s_4^3+8s_6^2).
G.f.: -x*(76*x^10 +10016*x^9 +212772*x^8 +1380453*x^7 +3384939*x^6 +3388593*x^5 +1380279*x^4 +211623*x^3 +10317*x^2 +131*x +1)/(x -1)^13. [Colin Barker, Aug 13 2012]
From Robert A. Russell, Oct 17 2020: (Start)
a(n) = A060530(n) - A337406(n) = (A060530(n) + A331351(n)) / 2 = A337406(n) + A331351(n).
a(n) = 1*C(n,1) + 142*C(n,2) + 11682*C(n,3) + 310536*C(n,4) + 3460725*C(n,5) + 19870590*C(n,6) + 65886660*C(n,7) + 133585200*C(n,8) + 168399000*C(n,9) + 128898000*C(n,10) + 54885600*C(n,11) + 9979200*C(n,12), where the coefficient of C(n,k) is the number of unoriented colorings using exactly k colors. (End)

A327084 Array read by descending antidiagonals: A(n,k) is the number of unoriented 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, 10, 11, 1, 5, 20, 66, 34, 1, 6, 35, 276, 792, 156, 1, 7, 56, 900, 10688, 25506, 1044, 1, 8, 84, 2451, 90005, 1601952, 2302938, 12346, 1, 9, 120, 5831, 533358, 43571400, 892341888, 591901884, 274668
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 unoriented colorings are the same if congruent; chiral pairs are counted as one.
A(n,k) is also the number of unoriented 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 unoriented 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 ...
  1  4  10    20    35     56      84     120      165      220 ...
  1 11  66   276   900   2451    5831   12496    24651    45475 ...
  1 34 792 10688 90005 533358 2437848 9156288 29522961 84293770 ...
  ...
For A(2,3) = 10, the nine achiral colorings are AAA, AAB, AAC, ABB, ACC, BBB, BBC, BCC, and CCC. The chiral pair is ABC-ACB.
		

Crossrefs

Cf. A327083 (oriented), A327085 (chiral), A327086 (achiral), A327088 (exactly k colors), A325000 (vertices, facets), A337884 (faces, peaks), A337408 (orthotope edges, orthoplex ridges), A337412 (orthoplex edges, orthotope ridges).
Rows 1-4 are A000027, A000292, A063842(n-1), A063843.
Cf. A063841 (k-multigraphs on n nodes).

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[pc[#] j^Total[CycleX[#]][[2]] & /@ IntegerPartitions[n+1]]/(n+1)!]
    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 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[pc[#]k^ex[#] &/@ IntegerPartitions[n+1]]/(n+1)!
    Table[array[n,d-n+1], {d,10}, {n,d}] // Flatten
    (* Another program (translated from Andrew Howroyd's PARI code): *)
    permcount[v_] := Module[{m=1, s=0, k=0, t}, For[i=1, i <= Length[v], i++, t = v[[i]]; k = If[i>1 && t == v[[i-1]], k+1, 1]; m *= t*k; s += t]; s!/m];
    edges[v_] := Sum[GCD[v[[i]], v[[j]]], {i, 2, Length[v]}, {j, 1, i-1}] + Total[Quotient[v, 2]];
    T[n_, k_] := Module[{s = 0}, Do[s += permcount[p]*k^edges[p], {p, IntegerPartitions[n+1]}]; s/(n+1)!];
    Table[T[n-k+1, k], {n, 1, 9}, {k, n, 1, -1}] // Flatten (* Jean-François Alcover, Jan 08 2021 *)
  • PARI
    permcount(v) = {my(m=1, s=0, k=0, t); for(i=1, #v, t=v[i]; k=if(i>1&&t==v[i-1], k+1, 1); m*=t*k; s+=t); s!/m}
    edges(v) = {sum(i=2, #v, sum(j=1, i-1, gcd(v[i], v[j]))) + sum(i=1, #v, v[i]\2)}
    T(n, k) = {my(s=0); forpart(p=n+1, s+=permcount(p)*k^edges(p)); s/(n+1)!} \\ Andrew Howroyd, Sep 06 2019
    
  • Python
    from itertools import combinations
    from math import prod, gcd, factorial
    from fractions import Fraction
    from sympy.utilities.iterables import partitions
    def A327084_T(n,k): return int(sum(Fraction(k**(sum(p[r]*p[s]*gcd(r,s) for r,s in combinations(p.keys(),2))+sum((q>>1)*r+(q*r*(r-1)>>1) for q, r in p.items())),prod(q**r*factorial(r) for q, r in p.items())) for p in partitions(n+1))) # Chai Wah Wu, Jul 09 2024

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} A327088(n,j) * binomial(k,j).
A(n,k) = A327083(n,k) - A327085(n,k) = (A327083(n,k) + A327086(n,k)) / 2 = A327085(n,k) + A327086(n,k).
A(n,k) = A063841(n+1,k-1).

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).

A337411 Array read by descending antidiagonals: T(n,k) is the number of oriented 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, 24, 218, 1, 5, 70, 2285, 90054, 1, 6, 165, 703760, 1471640157, 573439556, 1, 7, 336, 10194250, 1466049174160, 6332134720430727, 50043770249328, 1, 8, 616, 90775566, 310441584462375, 629648890639384572032, 1839894096099964270283469, 59966884221697869216, 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 an octahedron with 12 edges. The number of edges is 2n*(n-1) for n>1.
Also the number of oriented 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 ...
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. A337412 (unoriented), A337413 (chiral), A337414 (achiral).
Rows 1-4 are A000027, A006528, A060530, A331354.
Cf. A327083 (simplex edges), A337407 (orthotope edges), A325004 (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}]], (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,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) = A337412(n,k) + A337413(n,k) = 2*A337412(n,k) - A337414(n,k) = 2*A337413(n,k) + A337414(n,k).

A337413 Array read by descending antidiagonals: T(n,k) is the number of chiral pairs of colorings of the edges of a regular n-dimensional orthoplex (cross polytope) using k or fewer colors.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 3, 74, 0, 0, 15, 10704, 40927, 0, 0, 45, 345640, 731279799, 280317324, 0, 0, 105, 5062600, 732272925320, 3163614120031068, 24869435516248, 0, 0, 210, 45246810, 155180061396500, 314800331906964016128, 919853357924272852197243, 29931599129719666392, 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 an octahedron with 12 edges. The number of edges is 2n*(n-1) for n>1.
Also the number of chiral pairs of colorings of the regular (n-2)-dimensional orthotopes (hypercubes) in a regular n-dimensional orthotope.

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. A337411 (oriented), A337412 (unoriented), A337414 (achiral).
Rows 2-4 are A050534, A337406, A331356.
Cf. A327085 (simplex edges), A337409 (orthotope edges), A325006 (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; 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,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) = A337411(n,k) - A337412(n,k) = (A337411(n,k) - A337414(n,k)) / 2 = A337412(n,k) - A337414(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).

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).

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).

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).
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