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 1-10 of 10 results.

A006003 a(n) = n*(n^2 + 1)/2.

Original entry on oeis.org

0, 1, 5, 15, 34, 65, 111, 175, 260, 369, 505, 671, 870, 1105, 1379, 1695, 2056, 2465, 2925, 3439, 4010, 4641, 5335, 6095, 6924, 7825, 8801, 9855, 10990, 12209, 13515, 14911, 16400, 17985, 19669, 21455, 23346, 25345, 27455, 29679, 32020, 34481, 37065, 39775
Offset: 0

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Author

Keywords

Comments

Write the natural numbers in groups: 1; 2,3; 4,5,6; 7,8,9,10; ... and add the groups. In other words, "sum of the next n natural numbers". - Felice Russo
Number of rhombi in an n X n rhombus, if 'crossformed' rhombi are allowed. - Matti De Craene (Matti.DeCraene(AT)rug.ac.be), May 14 2000
Also the sum of the integers between T(n-1)+1 and T(n), the n-th triangular number (A000217). Sum of n-th row of A000027 regarded as a triangular array.
Unlike the cubes which have a similar definition, it is possible for 2 terms of this sequence to sum to a third. E.g., a(36) + a(37) = 23346 + 25345 = 48691 = a(46). Might be called 2nd-order triangular numbers, thus defining 3rd-order triangular numbers (A027441) as n(n^3+1)/2, etc. - Jon Perry, Jan 14 2004
Also as a(n)=(1/6)*(3*n^3+3*n), n > 0: structured trigonal diamond numbers (vertex structure 4) (cf. A000330 = alternate vertex; A000447 = structured diamonds; A100145 for more on structured numbers). - James A. Record (james.record(AT)gmail.com), Nov 07 2004
The sequence M(n) of magic constants for n X n magic squares (numbered 1 through n^2) from n=3 begins M(n) = 15, 34, 65, 111, 175, 260, ... - Lekraj Beedassy, Apr 16 2005 [comment corrected by Colin Hall, Sep 11 2009]
The sequence Q(n) of magic constants for the n-queens problem in chess begins 0, 0, 0, 0, 34, 65, 111, 175, 260, ... - Paul Muljadi, Aug 23 2005
Alternate terms of A057587. - Jeremy Gardiner, Apr 10 2005
Also partial differences of A063488(n) = (2*n-1)*(n^2-n+2)/2. a(n) = A063488(n) - A063488(n-1) for n>1. - Alexander Adamchuk, Jun 03 2006
In an n X n grid of numbers from 1 to n^2, select -- in any manner -- one number from each row and column. Sum the selected numbers. The sum is independent of the choices and is equal to the n-th term of this sequence. - F.-J. Papp (fjpapp(AT)umich.edu), Jun 06 2006
Nonnegative X values of solutions to the equation (X-Y)^3 - (X+Y) = 0. To find Y values: b(n) = (n^3-n)/2. - Mohamed Bouhamida, May 16 2006
For the equation: m*(X-Y)^k - (X+Y) = 0 with X >= Y, k >= 2 and m is an odd number the X values are given by the sequence defined by a(n) = (m*n^k+n)/2. The Y values are given by the sequence defined by b(n) = (m*n^k-n)/2. - Mohamed Bouhamida, May 16 2006
If X is an n-set and Y a fixed 3-subset of X then a(n-3) is equal to the number of 4-subsets of X intersecting Y. - Milan Janjic, Jul 30 2007
(m*(2n)^k+n, m*(2n)^k-n) solves the Diophantine equation: 2m*(X-Y)^k - (X+Y) = 0 with X >= Y, k >= 2 where m is a positive integer. - Mohamed Bouhamida, Oct 02 2007
Also c^(1/2) in a^(1/2) + b^(1/2) = c^(1/2) such that a^2 + b = c. - Cino Hilliard, Feb 09 2008
a(n) = n*A000217(n) - Sum_{i=0..n-1} A001477(i). - Bruno Berselli, Apr 25 2010
a(n) is the number of triples (w,x,y) having all terms in {0,...,n} such that at least one of these inequalities fails: x+y < w, y+w < x, w+x < y. - Clark Kimberling, Jun 14 2012
Sum of n-th row of the triangle in A209297. - Reinhard Zumkeller, Jan 19 2013
The sequence starting with "1" is the third partial sum of (1, 2, 3, 3, 3, ...). - Gary W. Adamson, Sep 11 2015
a(n) is the largest eigenvalue of the matrix returned by the MATLAB command magic(n) for n > 0. - Altug Alkan, Nov 10 2015
a(n) is the number of triples (x,y,z) having all terms in {1,...,n} such that all these triangle inequalities are satisfied: x+y > z, y+z > x, z+x > y. - Heinz Dabrock, Jun 03 2016
Shares its digital root with the stella octangula numbers (A007588). See A267017. - Peter M. Chema, Aug 28 2016
Can be proved to be the number of nonnegative solutions of a system of three linear Diophantine equations for n >= 0 even: 2*a_{11} + a_{12} + a_{13} = n, 2*a_{22} + a_{12} + a_{23} = n and 2*a_{33} + a_{13} + a_{23} = n. The number of solutions is f(n) = (1/16)*(n+2)*(n^2 + 4n + 8) and a(n) = n*(n^2 + 1)/2 is obtained by remapping n -> 2*n-2. - Kamil Bradler, Oct 11 2016
For n > 0, a(n) coincides with the trace of the matrix formed by writing the numbers 1...n^2 back and forth along the antidiagonals (proved, see A078475 for the examples of matrix). - Stefano Spezia, Aug 07 2018
The trace of an n X n square matrix where the elements are entered on the ascending antidiagonals. The determinant is A069480. - Robert G. Wilson v, Aug 07 2018
Bisections are A317297 and A005917. - Omar E. Pol, Sep 01 2018
Number of achiral colorings of the vertices (or faces) of a regular tetrahedron with n available colors. An achiral coloring is identical to its reflection. - Robert A. Russell, Jan 22 2020
a(n) is the n-th centered triangular pyramidal number. - Lechoslaw Ratajczak, Nov 02 2021
a(n) is the number of words of length n defined on 4 letters {b,c,d,e} that contain one or no b's, one c or two d's, and any number of e's. For example, a(3) = 15 since the words are (number of permutations in parentheses): bce (6), bdd (3), cee (3), and dde (3). - Enrique Navarrete, Jun 21 2025

Examples

			G.f. = x + 5*x^2 + 15*x^3 + 34*x^4 + 65*x^5 + 111*x^6 + 175*x^7 + 260*x^8 + ...
For a(2)=5, the five tetrahedra have faces AAAA, AAAB, AABB, ABBB, and BBBB with colors A and B. - _Robert A. Russell_, Jan 31 2020
		

References

  • J.-M. De Koninck, Ces nombres qui nous fascinent, Entry 15, p. 5, Ellipses, Paris 2008.
  • F.-J. Papp, Colloquium Talk, Department of Mathematics, University of Michigan-Dearborn, March 6, 2005.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A000330, A000537, A066886, A057587, A027480, A002817 (partial sums).
Cf. A000578 (cubes).
(1/12)*t*(n^3-n)+n for t = 2, 4, 6, ... gives A004006, A006527, this sequence, A005900, A004068, A000578, A004126, A000447, A004188, A004466, A004467, A007588, A062025, A063521, A063522, A063523.
Antidiagonal sums of array in A000027. Row sums of the triangular view of A000027.
Cf. A063488 (sum of two consecutive terms), A005917 (bisection), A317297 (bisection).
Cf. A105374 / 8.
Tetrahedron colorings: A006008 (oriented), A000332(n+3) (unoriented), A000332 (chiral), A037270 (edges).
Other polyhedron colorings: A337898 (cube faces, octahedron vertices), A337897 (octahedron faces, cube vertices), A337962 (dodecahedron faces, icosahedron vertices), A337960 (icosahedron faces, dodecahedron vertices).
Row 3 of A325001 (simplex vertices and facets) and A337886 (simplex faces and peaks).

Programs

  • GAP
    a_n:=List([0..nmax], n->n*(n^2 + 1)/2); # Stefano Spezia, Aug 12 2018
    
  • Haskell
    a006003 n = n * (n ^ 2 + 1) `div` 2
    a006003_list = scanl (+) 0 a005448_list
    -- Reinhard Zumkeller, Jun 20 2013
    
  • MATLAB
    % Also works with FreeMat.
    for(n=0:nmax); tm=n*(n^2 + 1)/2; fprintf('%d\t%0.f\n', n, tm); end
    % Stefano Spezia, Aug 12 2018
    
  • Magma
    [n*(n^2 + 1)/2 : n in [0..50]]; // Wesley Ivan Hurt, Sep 11 2015
    
  • Magma
    [Binomial(n,3)+Binomial(n-1,3)+Binomial(n-2,3): n in [2..60]]; // Vincenzo Librandi, Sep 12 2015
    
  • Mathematica
    Table[ n(n^2 + 1)/2, {n, 0, 45}]
    LinearRecurrence[{4,-6,4,-1}, {0,1,5,15},50] (* Harvey P. Dale, May 16 2012 *)
    CoefficientList[Series[x (1 + x + x^2)/(x - 1)^4, {x, 0, 45}], x] (* Vincenzo Librandi, Sep 12 2015 *)
    With[{n=50},Total/@TakeList[Range[(n(n^2+1))/2],Range[0,n]]] (* Requires Mathematica version 11 or later *) (* Harvey P. Dale, Nov 28 2017 *)
  • Maxima
    a(n):=n*(n^2 + 1)/2$ makelist(a(n), n, 0, nmax); /* Stefano Spezia, Aug 12 2018 */
    
  • PARI
    {a(n) = n * (n^2 + 1) / 2}; /* Michael Somos, Dec 24 2011 */
    
  • PARI
    concat(0, Vec(x*(1+x+x^2)/(x-1)^4 + O(x^20))) \\ Felix Fröhlich, Oct 11 2016
    
  • Python
    def A006003(n): return n*(n**2+1)>>1 # Chai Wah Wu, Mar 25 2024

Formula

a(n) = binomial(n+2, 3) + binomial(n+1, 3) + binomial(n, 3). [corrected by Michel Marcus, Jan 22 2020]
G.f.: x*(1+x+x^2)/(x-1)^4. - Floor van Lamoen, Feb 11 2002
Partial sums of A005448. - Jonathan Vos Post, Mar 16 2006
Binomial transform of [1, 4, 6, 3, 0, 0, 0, ...] = (1, 5, 15, 34, 65, ...). - Gary W. Adamson, Aug 10 2007
a(n) = -a(-n) for all n in Z. - Michael Somos, Dec 24 2011
a(n) = Sum_{k = 1..n} A(k-1, k-1-n) where A(i, j) = i^2 + i*j + j^2 + i + j + 1. - Michael Somos, Jan 02 2012
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4), with a(0)=0, a(1)=1, a(2)=5, a(3)=15. - Harvey P. Dale, May 16 2012
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + 3. - Ant King, Jun 13 2012
a(n) = A000217(n) + n*A000217(n-1). - Bruno Berselli, Jun 07 2013
a(n) = A057145(n+3,n). - Luciano Ancora, Apr 10 2015
E.g.f.: (1/2)*(2*x + 3*x^2 + x^3)*exp(x). - G. C. Greubel, Dec 18 2015; corrected by Ilya Gutkovskiy, Oct 12 2016
a(n) = T(n) + T(n-1) + T(n-2), where T means the tetrahedral numbers, A000292. - Heinz Dabrock, Jun 03 2016
From Ilya Gutkovskiy, Oct 11 2016: (Start)
Convolution of A001477 and A008486.
Convolution of A000217 and A158799.
Sum_{n>=1} 1/a(n) = H(-i) + H(i) = 1.343731971048019675756781..., where H(k) is the harmonic number, i is the imaginary unit. (End)
a(n) = A000578(n) - A135503(n). - Miquel Cerda, Dec 25 2016
Euler transform of length 3 sequence [5, 0, -1]. - Michael Somos, Dec 25 2016
a(n) = A037270(n)/n for n > 0. - Kritsada Moomuang, Dec 15 2018
a(n) = 3*A000292(n-1) + n. - Bruce J. Nicholson, Nov 23 2019
a(n) = A011863(n) - A011863(n-2). - Bruce J. Nicholson, Dec 22 2019
From Robert A. Russell, Jan 22 2020: (Start)
a(n) = C(n,1) + 3*C(n,2) + 3*C(n,3), where the coefficient of C(n,k) is the number of tetrahedron colorings using exactly k colors.
a(n) = C(n+3,4) - C(n,4).
a(n) = 2*A000332(n+3) - A006008(n) = A006008(n) - 2*A000332(n) = A000332(n+3) - A000332(n).
a(n) = A325001(3,n). (End)
From Amiram Eldar, Aug 21 2023: (Start)
Sum_{n>=1} 1/a(n) = 2 * (A248177 + A001620).
Product_{n>=2} (1 - 1/a(n)) = cosh(sqrt(7)*Pi/2)*cosech(Pi)/4.
Product_{n>=1} (1 + 1/a(n)) = cosh(sqrt(7)*Pi/2)*cosech(Pi). (End)

Extensions

Better description from Albert Rich (Albert_Rich(AT)msn.com), Mar 1997

A325001 Array read by descending antidiagonals: A(n,k) is the number of achiral colorings of the facets (or vertices) of a regular n-dimensional simplex using up to k colors.

Original entry on oeis.org

1, 2, 1, 3, 4, 1, 4, 9, 5, 1, 5, 16, 15, 6, 1, 6, 25, 34, 21, 7, 1, 7, 36, 65, 56, 28, 8, 1, 8, 49, 111, 125, 84, 36, 9, 1, 9, 64, 175, 246, 210, 120, 45, 10, 1, 10, 81, 260, 441, 461, 330, 165, 55, 11, 1, 11, 100, 369, 736, 917, 792, 495, 220, 66, 12, 1
Offset: 1

Views

Author

Robert A. Russell, Mar 23 2019

Keywords

Comments

For n=1, the figure is a line segment with two vertices. For n=2, the figure is a triangle with three edges. For n=3, the figure is a tetrahedron with four triangular faces. The Schläfli symbol, {3,...,3}, of the regular n-dimensional simplex consists of n-1 threes. Each of its n+1 facets is a regular (n-1)-dimensional simplex. An achiral coloring is the same as its reflection.

Examples

			The array begins with A(1,1):
  1  2  3   4   5    6    7     8     9    10    11     12     13 ...
  1  4  9  16  25   36   49    64    81   100   121    144    169 ...
  1  5 15  34  65  111  175   260   369   505   671    870   1105 ...
  1  6 21  56 125  246  441   736  1161  1750  2541   3576   4901 ...
  1  7 28  84 210  461  917  1688  2919  4795  7546  11452  16848 ...
  1  8 36 120 330  792 1715  3424  6399 11320 19118  31032  48672 ...
  1  9 45 165 495 1287 3003  6434 12861 24265 43593  75087 124683 ...
  1 10 55 220 715 2002 5005 11440 24309 48610 92323 167740 293215 ...
  ...
For A(2,2)=4, the triangle may have 0, 1, 2, or 3 edges of one color.
		

Crossrefs

Cf. A324999 (oriented), A325000 (unoriented), A325000(n,k-n) (chiral), A325003 (exactly k colors), A327086 (edges, ridges), A337886 (faces, peaks), A325007 (orthotope facets, orthoplex vertices), A325015 (orthoplex facets, orthotope vertices).
Rows 1-4 are A000027, A000290, A006003, A132366(n-1).
Column 2 is A162880.

Programs

  • Mathematica
    Table[Binomial[d+1,n+1] - Binomial[d+1-n,n+1], {d,1,15}, {n,1,d}] // Flatten

Formula

A(n,k) = binomial(n+k,n+1) - binomial(k,n+1).
A(n,k) = Sum_{j=1..n} A325003(n,j) * binomial(k,j).
A(n,k) = 2*A325000(n,k) - A324999(n,k) = A324999(n,k) - 2*A325000(n,k-n) = A325000(n,k) - A325000(n,k-n).
G.f. for row n: (x - x^(n+1)) / (1-x)^(n+2).
Linear recurrence for row n: A(n,k) = Sum_{j=1..n+1} -binomial(j-n-2,j) * A(n,k-j).
G.f. for column k: (1 - (1-x^2)^k) / (x*(1-x)^k).

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

Original entry on oeis.org

1, 2, 1, 3, 4, 1, 4, 9, 10, 1, 5, 16, 45, 28, 1, 6, 25, 136, 387, 128, 1, 7, 36, 325, 2784, 8352, 792, 1, 8, 49, 666, 13125, 186304, 382563, 7620, 1, 9, 64, 1225, 46836, 2117750, 36507008, 44526672, 124344
Offset: 1

Views

Author

Robert A. Russell, Aug 19 2019

Keywords

Comments

An n-dimensional simplex has n+1 vertices and (n+1)*n/2 edges. For n=1, the figure is a line segment with one edge. For n-2, the figure is a triangle with three edges. For n=3, the figure is a tetrahedron with six edges. The Schläfli symbol, {3,...,3}, of the regular n-dimensional simplex consists of n-1 threes. An achiral coloring is identical to its reflection.
A(n,k) is also the number of achiral colorings of (n-2)-dimensional regular simplices in an n-dimensional simplex using up to k colors. Thus, A(2,k) is also the number of achiral colorings of the vertices (0-dimensional simplices) of an equilateral triangle.

Examples

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

Crossrefs

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

Programs

  • Mathematica
    CycleX[{2}] = {{1,1}}; (* cycle index for permutation with given cycle structure *)
    CycleX[{n_Integer}] := CycleX[n] = If[EvenQ[n], {{n/2,1}, {n,(n-2)/2}}, {{n,(n-1)/2}}]
    compress[x : {{, } ...}] := (s = Sort[x]; For[i = Length[s], i > 1, i -= 1, If[s[[i,1]] == s[[i-1,1]], s[[i-1,2]] += s[[i,2]]; s = Delete[s,i], Null]]; s)
    CycleX[p_List] := CycleX[p] = compress[Join[CycleX[Drop[p, -1]], If[Last[p] > 1, CycleX[{Last[p]}], ## &[]], If[# == Last[p], {#, Last[p]}, {LCM[#, Last[p]], GCD[#, Last[p]]}] & /@ Drop[p, -1]]]
    pc[p_List] := Module[{ci, mb}, mb = DeleteDuplicates[p]; ci = Count[p, #] & /@ mb; Total[p]!/(Times @@ (ci!) Times @@ (mb^ci))] (* partition count *)
    row[n_Integer] := row[n] = Factor[Total[If[EvenQ[Total[1-Mod[#,2]]], 0, pc[#] j^Total[CycleX[#]][[2]]] & /@ IntegerPartitions[n+1]]/((n+1)!/2)]
    array[n_, k_] := row[n] /. j -> k
    Table[array[n,d-n+1], {d,1,10}, {n,1,d}] // Flatten
    (* Using Fripertinger's exponent per Andrew Howroyd's code in A063841: *)
    pc[p_] := Module[{ci, mb}, mb = DeleteDuplicates[p]; ci = Count[p, #] &/@ mb; Total[p]!/(Times @@ (ci!) Times @@ (mb^ci))]
    ex[v_] := Sum[GCD[v[[i]], v[[j]]], {i,2,Length[v]}, {j,i-1}] + Total[Quotient[v,2]]
    array[n_,k_] := Total[If[OddQ[Total[1-Mod[#,2]]], pc[#]k^ex[#], 0] &/@ IntegerPartitions[n+1]]/((n+1)!/2)
    Table[array[n,d-n+1], {d,10}, {n,d}] // Flatten

Formula

The algorithm used in the Mathematica program below assigns each permutation of the vertices to a partition of n+1. It then determines the number of permutations for each partition and the cycle index for each partition.
A(n,k) = Sum_{j=1..(n+1)*n/2} A327090(n,j) * binomial(k,j).
A(n,k) = 2*A327084(n,k) - A327083(n,k) = A327083(n,k) - 2*A327085(n,k) = A327084(n,k) - A327085(n,k).

A331353 Number of achiral colorings of the edges (or triangular faces) of a regular 4-dimensional simplex with n available colors.

Original entry on oeis.org

1, 28, 387, 2784, 13125, 46836, 137543, 349952, 797769, 1667500, 3248971, 5973408, 10459917, 17571204, 28479375, 44742656, 68393873, 102041532, 148984339, 213340000, 300189141, 415735188, 567481047, 764423424
Offset: 1

Views

Author

Robert A. Russell, Jan 14 2020

Keywords

Comments

A 4-dimensional simplex has 5 vertices and 10 edges. Its Schläfli symbol is {3,3,3}. An achiral coloring is identical to its reflection,
There are 60 elements in the automorphism group of the 4-dimensional simplex that are not in its rotation group. Each is an odd permutation of the vertices and can be associated with a partition of 5 based on the conjugacy group of the permutation. The first formula is obtained by averaging their cycle indices after replacing x_i^j with n^j according to the Pólya enumeration theorem.
Partition Count Odd Cycle Indices
41 30 x_2^1x_4^2
32 20 x_1^1x_3^1x_6^1
2111 10 x_1^4x_2^3

Crossrefs

Cf. A331350 (oriented), A063843 (unoriented), A331352 (chiral).
Other polychora: A331361 (8-cell), A331357 (16-cell), A338955 (24-cell), A338967 (120-cell, 600-cell).
Row 4 of A327086 (simplex edges and ridges) and A337886 (simplex faces and peaks).

Programs

  • Mathematica
    Table[(5 n^3 + n^7)/6, {n, 1, 25}]
  • PARI
    Vec(x*(1 + 20*x + 191*x^2 + 416*x^3 + 191*x^4 + 20*x^5 + x^6) / (1 - x)^8 + O(x^25)) \\ Colin Barker, Jan 15 2020

Formula

a(n) = (5*n^3 + n^7) / 6.
a(n) = C(n,1) + 26*C(n,2) + 306*C(n,3) + 1400*C(n,4) + 2800*C(n,5) + 2520*C(n,6) + 840*C(n,7), where the coefficient of C(n,k) is the number of colorings using exactly k colors.
a(n) = 2*A063843(n) - A331350(n) = A331350(n) - 2*A331352(n) = A063843(n) - A331352(n).
From Colin Barker, Jan 15 2020: (Start)
G.f.: x*(1 + 20*x + 191*x^2 + 416*x^3 + 191*x^4 + 20*x^5 + x^6) / (1 - x)^8.
a(n) = 8*a(n-1) - 28*a(n-2) + 56*a(n-3) - 70*a(n-4) + 56*a(n-5) - 28*a(n-6) + 8*a(n-7) - a(n-8) for n>8.
(End)

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).

A337883 Array read by descending antidiagonals: T(n,k) is the number of oriented 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, 40, 1, 5, 36, 1197, 3504, 1, 6, 75, 18592, 9753615, 13724608, 1, 7, 141, 166885, 3056311808, 19854224207910, 3574466244480, 1, 8, 245, 1019880, 264940140875, 468488921670219776, 25959704193068472575379, 106607224611810055168, 1
Offset: 2

Views

Author

Robert A. Russell, Sep 28 2020

Keywords

Comments

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

Examples

			The table begins with T(2,1):
 1    2       3          4            5              6               7 ...
 1    5      15         36           75            141             245 ...
 1   40    1197      18592       166885        1019880         4738153 ...
 1 3504 9753615 3056311808 264940140875 10156268150064 221646915632373 ...
For T(3,4)=36, 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. A337884 (unoriented), A337885 (chiral), A337886 (achiral), A051168 (binary Lyndon words).
Other elements: A324999 (vertices), A327083 (edges).
Other polytopes: A337887 (orthotope), A337891 (orthoplex).
Rows 2-4 are A000027, A006008, A331350.

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[EvenQ[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) + A337885(n,k) = 2*A337884(n,k) - A337886(n,k) = 2*A337885(n,k) + A337886(n,k).

A337885 Array read by descending antidiagonals: T(n,k) is the number of chiral pairs of colorings of the triangular faces of a regular n-dimensional simplex using k or fewer colors.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 6, 0, 0, 1, 405, 1368, 0, 0, 5, 7904, 4775706, 6711288, 0, 0, 15, 76880, 1522540416, 9923557498416, 1785683627824, 0, 0, 35, 486522, 132342705750, 234239763858347776, 12979826761630383196344, 53302696800142157920, 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. 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 chiral pairs of 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):
 0    0       0          0            0             0               0 ...
 0    0       0          1            5            15              35 ...
 0    6     405       7904        76880        486522         2300305 ...
 0 1368 4775706 1522540416 132342705750 5076500214744 110809322249220 ...
For T(3,4)=1, the chiral pair is ABCD-ABDC.
		

Crossrefs

Cf. A337883 (oriented), A337884 (unoriented), A337886 (achiral), A051168 (binary Lyndon words).
Other elements: A325000(n,k-n) (vertices), A327085 (edges).
Other polytopes: A337889 (orthotope), A337893 (orthoplex).
Rows 2-4 are A000004, A000332, A331352.

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[EvenQ[Total[1-Mod[#, 2]]],1,-1] 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) - A337884(n,k) = (A337883(n,k) - A337886(n,k)) / 2 = A337884(n,k) - A337886(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).

A337894 Array read by descending antidiagonals: T(n,k) is the number of achiral 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, 21, 1, 4, 201, 93024, 1, 5, 1076, 294157089, 199556208371776, 1, 6, 4025, 91983927296, 1370366433970979158839987, 346179533768149850758531729588224, 1
Offset: 2

Views

Author

Robert A. Russell, Sep 28 2020

Keywords

Comments

An achiral arrangement is identical to its reflection. 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 achiral 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):
1     2         3           4             5               6 ...
1    21       201        1076          4025           11901 ...
1 93024 294157089 91983927296 7960001890625 304914963625056 ...
		

Crossrefs

Cf. A337891 (oriented), A337892 (unoriented), A337893 (chiral).
Other elements: A325007 (vertices), A337414 (edges).
Other polytopes: A337886 (simplex), A337890 (orthotope).
Rows 2-4 are A000027, A337897, A331361.

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}]],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,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*A337892(n,k) - A337891(n,k) = A337891(n,k) - 2*A337893(n,k) = A337892(n,k) - A337893(n,k).

A338116 Triangle read by rows: T(n,k) is the number of achiral colorings of the faces (and peaks) of a regular n-dimensional simplex using exactly k colors. Row n has C(n+1,3) columns.

Original entry on oeis.org

1, 1, 3, 3, 0, 1, 26, 306, 1400, 2800, 2520, 840, 0, 0, 0, 1, 766, 199902, 10426768, 200588850, 1903776420, 10360383600, 35133957600, 77643846000, 113816253600, 109880971200, 67199932800, 23610787200, 3632428800, 0, 0, 0, 0, 0, 0
Offset: 2

Views

Author

Robert A. Russell, Oct 10 2020

Keywords

Comments

An n-dimensional simplex has n+1 vertices, C(n+1,3) faces, and C(n+1,3) peaks, which are (n-3)-dimensional simplexes. For n=2, the figure is a triangle with one face. For n=3, the figure is a tetrahedron with four triangular faces and four peaks (vertices). For n=4, the figure is a 4-simplex with ten triangular faces and ten peaks (edges). The Schläfli symbol {3,...,3}, of the regular n-dimensional simplex consists of n-1 3's. An achiral coloring is identical to its reflection.
The algorithm used in the Mathematica program below assigns each permutation of the vertices to a cycle-structure partition of n+1. It then determines the number of permutations for each partition and the cycle index for each partition. 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(2,1):
  1
  1   3      3        0
  1  26    306     1400      2800       2520         840           0   0   0
  1 766 199902 10426768 200588850 1903776420 10360383600 35133957600 ...
  ...
For T(3,3)=3, one of the three colors appears on two faces (vertices) of the tetrahedron.
		

Crossrefs

Cf. A338113 (oriented), A338114 (unoriented), A338115 (chiral), A337886 (k or fewer colors), A325003 (vertices and facets), A327090 (edges and ridges).

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[LinearSolve[Table[Binomial[i,j],{i,Binomial[n+1,m+1]},{j,Binomial[n+1,m+1]}], Table[array[n,k],{k,Binomial[n+1,m+1]}]], {n,m,m+4}] // Flatten

Formula

A337886(n,k) = Sum_{j=1..C(n+1,3)} T(n,j) * binomial(k,j).
T(n,k) = 2*A338114(n,k) - A338113(n,k) = A338113(n,k) - 2*A338115(n,k) = A338114(n,k) - A338115(n,k).
T(3,k) = A325003(3,k); T(4,k) = A327090(4,k).
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