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.

A050534 Tritriangular numbers: a(n) = binomial(binomial(n,2),2) = n*(n+1)*(n-1)*(n-2)/8.

Original entry on oeis.org

0, 0, 0, 3, 15, 45, 105, 210, 378, 630, 990, 1485, 2145, 3003, 4095, 5460, 7140, 9180, 11628, 14535, 17955, 21945, 26565, 31878, 37950, 44850, 52650, 61425, 71253, 82215, 94395, 107880, 122760, 139128, 157080, 176715, 198135, 221445, 246753, 274170, 303810, 335790
Offset: 0

Views

Author

Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de), Dec 29 1999

Keywords

Comments

"There are n straight lines in a plane, no two of which are parallel and no three of which are concurrent. Their points of intersection being joined, show that the number of new lines drawn is (1/8)n(n-1)(n-2)(n-3)." (Schmall, 1915).
Several different versions of this sequence are possible, beginning with either one, two or three 0's.
If Y is a 3-subset of an n-set X then, for n>=6, a(n-4) is the number of (n-6)-subsets of X which have exactly one element in common with Y. - Milan Janjic, Dec 28 2007
Number of distinct ways to select 2 pairs of objects from a set of n+1 objects, when order doesn't matter. For example, with n = 3 (4 objects), the 3 possibilities are (12)(34), (13)(24), and (14)(23). - Brian Parsonnet, Jan 03 2012
Partial sums of A027480. - J. M. Bergot, Jul 09 2013
For the set {1,2,...,n}, the sum of the 2 smallest elements of all subsets with 3 elements is a(n) (see Bulut et al. link). - Serhat Bulut, Jan 20 2015
a(n) is also the number of subgroups of S_{n+1} (the symmetric group on n+1 elements) that are isomorphic to D_4 (the dihedral group of order 8). - Geoffrey Critzer, Sep 13 2015
a(n) is the coefficient of x1^(n-3)*x2^2 in exponential Bell polynomial B_{n+1}(x1,x2,...) (number of ways to select 2 pairs among n+1 objects, see above), hence its link with A000292 and A001296 (see formula). - Cyril Damamme, Feb 26 2018
Also the number of 4-cycles in the complete graph K_{n+1}. - Eric W. Weisstein, Mar 13 2018
Number of chiral pairs of colorings of the 4 edges or vertices of a square using n or fewer colors. Each member of a chiral pair is a reflection, but not a rotation, of the other. - Robert A. Russell, Oct 20 2020

Examples

			For a(3)=3, the chiral pairs of square colorings are AABC-AACB, ABBC-ACBB, and ABCC-ACCB. - _Robert A. Russell_, Oct 20 2020
		

References

  • Arthur T. Benjamin and Jennifer Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 154.
  • Louis Comtet, Advanced Combinatorics, Reidel, 1974, Problem 1, page 72.
  • Richard P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999; see Problem 5.5, case k=2.

Crossrefs

Cf. A000217, A000332, A033487, A107394, A034827, A210569, Second column of triangle A001498.
Cf. similar sequences listed in A241765.
Cf. (square colorings) A006528 (oriented), A002817 (unoriented), A002411 (achiral),
Row 2 of A325006 (orthoplex facets, orthotope vertices) and A337409 (orthotope edges, orthoplex ridges).
Row 4 of A293496 (cycles of n colors using k or fewer colors).

Programs

  • GAP
    List([0..40],n->3*Binomial(n+1,4)); # Muniru A Asiru, Mar 20 2018
  • Magma
    [3*Binomial(n+1, 4): n in [0..40]]; // Vincenzo Librandi, Feb 14 2015
    
  • Maple
    [seq(binomial(n+1,4)*3,n=0..40)]; # Zerinvary Lajos, Jul 18 2006
  • Mathematica
    Table[Binomial[Binomial[n, 2], 2], {n, 0, 50}] (* Stefan Steinerberger, Apr 08 2006 *)
    LinearRecurrence[{5, -10, 10, -5, 1}, {0, 0, 0, 3, 15}, 40] (* Harvey P. Dale, Dec 14 2011 *)
    (* Start from Eric W. Weisstein, Mar 13 2018 *)
    Binomial[Binomial[Range[0, 20], 2], 2]
    Nest[Binomial[#, 2] &, Range[0, 20], 2]
    Nest[PolygonalNumber[# - 1] &, Range[0, 20], 2]
    CoefficientList[Series[3 x^3/(1 - x)^5, {x, 0, 20}], x]
    (* End *)
  • PARI
    a(n)=n*(n+1)*(n-1)*(n-2)/8 \\ Charles R Greathouse IV, Nov 20 2012
    
  • PARI
    x='x+O('x^100); concat([0, 0, 0], Vec(3*x^3/(1-x)^5)) \\ Altug Alkan, Nov 01 2015
    
  • Sage
    [(binomial(binomial(n,2),2)) for n in range(0, 39)] # Zerinvary Lajos, Nov 30 2009
    

Formula

a(n) = 3*binomial(n+1, 4) = 3*A000332(n+1).
From Vladeta Jovovic, May 03 2002: (Start)
Recurrence: a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5).
G.f.: 3*x^3 / (1-x)^5. (End)
a(n+1) = T(T(n)) - T(n); a(n+2) = T(T(n)+n) where T is A000217. - Jon Perry, Jun 11 2003
a(n+1) = T(n)^2 - T(T(n)) where T is A000217. - Jon Perry, Jul 23 2003
a(n) = T(T(n-1)-1) where T is A000217. - Jon E. Schoenfield, Dec 14 2014
a(n) = 3*C(n, 4) + 3*C(n, 3), for n>3.
From Alexander Adamchuk, Apr 11 2006: (Start)
a(n) = (1/2)*Sum_{k=1..n} k*(k-1)*(k-2).
a(n) = A033487(n-2)/2, n>1.
a(n) = C(n-1,2)*C(n+1,2)/2, n>2. (End)
a(n) = A052762(n+1)/8. - Zerinvary Lajos, Apr 26 2007
a(n) = (4x^4 - 4x^3 - x^2 + x)/2 where x = floor(n/2)*(-1)^n for n >= 0. - William A. Tedeschi, Aug 24 2010
E.g.f.: x^3*exp(x)*(4+x)/8. - Robert Israel, Nov 01 2015
a(n) = Sum_{k=1..n} Sum_{i=1..k} (n-i-1)*(n-k). - Wesley Ivan Hurt, Sep 12 2017
a(n) = A001296(n-1) - A000292(n-1). - Cyril Damamme, Feb 26 2018
Sum_{n>=3} 1/a(n) = 4/9. - Vaclav Kotesovec, May 01 2018
a(n) = A006528(n) - A002817(n) = (A006528(n) - A002411(n)) / 2 = A002817(n) - A002411(n). - Robert A. Russell, Oct 20 2020
Sum_{n>=3} (-1)^(n+1)/a(n) = 32*log(2)/3 - 64/9. - Amiram Eldar, Jan 09 2022
a(n) = Sum_{k=1..2} (-1)^(k+1)*binomial(n,2-k)*binomial(n,2+k). - Gerry Martens, Oct 09 2022

Extensions

Additional comments from Antreas P. Hatzipolakis, May 03 2002

A327085 Array read by descending antidiagonals: A(n,k) is the number of chiral pairs of colorings of the edges of a regular n-dimensional simplex using up to k colors.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 4, 21, 6, 0, 0, 10, 140, 405, 28, 0, 0, 20, 575, 7904, 17154, 252, 0, 0, 35, 1785, 76880, 1415648, 1920375, 4726, 0, 0, 56, 4606, 486522, 41453650, 855834880, 547375212, 150324, 0
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. The chiral colorings of its edges come in pairs, each the reflection of the other.
A(n,k) is also the number of chiral pairs of 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 chiral pairs of colorings of the vertices (0-dimensional simplices) of an equilateral triangle.

Examples

			Array begins with A(1,1):
  0 0   0    0     0      0       0       0        0        0         0 ...
  0 0   1    4    10     20      35      56       84      120       165 ...
  0 1  21  140   575   1785    4606   10416    21330    40425     71995 ...
  0 6 405 7904 76880 486522 2300305 8806336 28725192 82626270 214744629 ...
  ...
For A(2,3) = 1, the chiral pair is ABC-ACB.
		

Crossrefs

Cf. A327083 (oriented), A327084 (unoriented), A327086 (achiral), A327089 (exactly k colors), A325000(n,k-n) (vertices, facets), A337885 (faces, peaks), A337409 (orthotope edges, orthoplex ridges), A337413 (orthoplex edges, orthotope ridges).
Rows 1-4 are A000004, A000292(n-2), A337899, A331352.

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]]], 1, -1] 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'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[EvenQ[Total[1-Mod[#,2]]],1,-1] pc[#]k^ex[#] &/@ IntegerPartitions[n+1]]/(n+1)!
    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} A327089(n,j) * binomial(k,j).
A(n,k) = A327083(n,k) - A327084(n,k) = (A327083(n,k) - A327086(n,k)) / 2 = A327084(n,k) - A327086(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).

A337406 Number of chiral pairs of colorings of the edges of a cube (or regular octahedron) using n or fewer colors.

Original entry on oeis.org

0, 74, 10704, 345640, 5062600, 45246810, 288005144, 1430618784, 5881281480, 20827126650, 65370603320, 185725346664, 485325996064, 1181031257770, 2702889008400, 5863794289280, 12137528310384, 24099966466794
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. Both the cube and the octahedron have 12 edges.

Crossrefs

Cf. A060530 (oriented), A199406 (unoriented), A331351 (achiral).
Row 3 of A337409 (orthotope edge colorings) and A337413 (orthoplex edge colorings).

Programs

  • Mathematica
    Table[(n-1)n^2(n+1)(n^8+n^6-2n^4+8)/48, {n,20}]
    LinearRecurrence[{13,-78,286,-715,1287,-1716,1716,-1287,715,-286,78,-13,1},{0,74,10704,345640,5062600,45246810,288005144,1430618784,5881281480,20827126650,65370603320,185725346664,485325996064},20] (* Harvey P. Dale, Jul 11 2025 *)

Formula

a(n) = (n-1) * n^2 * (n+1) * (n^8 + n^6 - 2n^4 + 8) / 48.
a(n) = 74*C(n,2) + 10482*C(n,3) + 303268*C(n,4) + 3440700*C(n,5) + 19842840*C(n,6) + 65867760*C(n,7) + 133580160*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 colorings using exactly k colors.
a(n) = (A060530(n) - A331351(n)) / 2 = A060530(n) - A199406(n) = A199406(n) - A331351(n).
G.f.: 2 * (37*x^2 + 4871*x^3 + 106130*x^4 + 691514*x^5 + 1692248*x^6 + 1692248*x^7 + 691514*x^8 + 106130*x^9 + 4871*x^10 + 37*x^11) / (1-x)^13.

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).

A337410 Array read by descending antidiagonals: T(n,k) is the number of achiral 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, 18, 70, 1, 5, 40, 1407, 93024, 1, 6, 75, 12480, 294157089, 47823602694208, 1, 7, 126, 69050, 91983927296, 67514530382043163023924, 443077371786837979607993095063601152, 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 a cube with 12 edges. The number of edges is n*2^(n-1).
Also the number of achiral 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       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. A337407 (oriented), A337408 (unoriented), A337409 (chiral).
Rows 1-4 are A000027, A002411, A331351, A331361.
Cf. A327086 (simplex edges), A337414 (orthoplex edges), A325015 (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}]],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,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) = 2*A337408(n,k) - A337407(n,k) = A337407(n,k) - 2*A337409(n,k) = A337408(n,k) - A337409(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).

A338144 Triangle read by rows: T(n,k) is the number of chiral pairs of 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

0, 0, 0, 3, 3, 0, 74, 10482, 303268, 3440700, 19842840, 65867760, 133580160, 168399000, 128898000, 54885600, 9979200, 0, 11158298, 4825419243699, 48019052798280376, 60392832865887732525, 20362602448352682660450
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 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):
  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. A338142 (oriented), A338143 (unoriented), A338145 (achiral), A337409 (k or fewer colors), A325018 (orthotope vertices, orthoplex facets).
Cf. A327089 (simplex), A338148 (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+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[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

A337409(n,k) = Sum_{j=1..n*2^(n-1)} T(n,j) * binomial(k,j).
T(n,k) = A338142(n,k) - A338143(n,k) = (A338142(n,k) - A338145(n,k)) / 2 = A338143(n,k) - A338145(n,k).
T(2,k) = A338148(2,k) = A325018(2,k) = A325010(2,k); T(3,k) = A338148(3,k).
Showing 1-9 of 9 results.