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

A325000 Array read by descending antidiagonals: T(n,k) is the number of unoriented colorings of the facets (or vertices) of a regular n-dimensional simplex using up to k colors.

Original entry on oeis.org

1, 3, 1, 6, 4, 1, 10, 10, 5, 1, 15, 20, 15, 6, 1, 21, 35, 35, 21, 7, 1, 28, 56, 70, 56, 28, 8, 1, 36, 84, 126, 126, 84, 36, 9, 1, 45, 120, 210, 252, 210, 120, 45, 10, 1, 55, 165, 330, 462, 462, 330, 165, 55, 11, 1, 66, 220, 495, 792, 924, 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. Two unoriented colorings are the same if congruent; chiral pairs are counted as one.
Note that antidiagonals are part of rows of the Pascal triangle.
T(n,k-n) is the number of chiral pairs of colorings of the facets (or vertices) of a regular n-dimensional simplex using k or fewer colors. - Robert A. Russell, Sep 28 2020

Examples

			The array begins with T(1,1):
  1  3  6  10  15   21   28    36    45    55    66     78     91    105 ...
  1  4 10  20  35   56   84   120   165   220   286    364    455    560 ...
  1  5 15  35  70  126  210   330   495   715  1001   1365   1820   2380 ...
  1  6 21  56 126  252  462   792  1287  2002  3003   4368   6188   8568 ...
  1  7 28  84 210  462  924  1716  3003  5005  8008  12376  18564  27132 ...
  1  8 36 120 330  792 1716  3432  6435 11440 19448  31824  50388  77520 ...
  1  9 45 165 495 1287 3003  6435 12870 24310 43758  75582 125970 203490 ...
  1 10 55 220 715 2002 5005 11440 24310 48620 92378 167960 293930 497420 ...
  ...
For T(1,2) = 3, the two achiral colorings use just one of the two colors for both vertices; the chiral pair uses two colors. For T(2,2)=4, the triangle may have 0, 1, 2, or 3 edges of one color.
		

Crossrefs

Cf. A324999 (oriented), A325001 (achiral).
Unoriented: A007318(n,k-1) (exactly k colors), A327084 (edges, ridges), A337884 (faces, peaks), A325005 (orthotope facets, orthoplex vertices), A325013 (orthoplex facets, orthotope vertices).
Chiral: A327085 (edges, ridges), A337885 (faces, peaks), A325006 (orthotope facets, orthoplex vertices), A325014 (orthoplex facets, orthotope vertices).
Cf. A104712 (same sequence for a triangle; same sequence apart from offset).
Rows 1-4 are A000217, A000292, A000332(n+3), A000389(n+4). - Robert A. Russell, Sep 28 2020

Programs

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

Formula

T(n,k) = binomial(n+k,n+1) = A007318(n+k,n+1).
T(n,k) = Sum_{j=1..n+1} A007318(n,j-1) * binomial(k,j).
T(n,k) = A324999(n,k) + T(n,k-n) = (A324999(n,k) - A325001(n,k)) / 2 = T(n,k-n) + A325001(n,k). - Robert A. Russell, Sep 28 2020
G.f. for row n: x / (1-x)^(n+2).
Linear recurrence for row n: T(n,k) = Sum_{j=1..n+2} -binomial(j-n-3,j) * T(n,k-j).
G.f. for column k: (1 - (1-x)^k) / (x * (1-x)^k) - k.
T(n,k-n) = A324999(n,k) - T(n,k) = (A324999(n,k) - A325001(n,k)) / 2 = T(n,k) - A325001(n,k). - Robert A. Russell, Oct 10 2020

A132366 Partial sum of centered tetrahedral numbers A005894.

Original entry on oeis.org

1, 6, 21, 56, 125, 246, 441, 736, 1161, 1750, 2541, 3576, 4901, 6566, 8625, 11136, 14161, 17766, 22021, 27000, 32781, 39446, 47081, 55776, 65625, 76726, 89181, 103096, 118581, 135750, 154721, 175616, 198561, 223686, 251125, 281016, 313501, 348726, 386841
Offset: 0

Views

Author

Jonathan Vos Post, Nov 09 2007

Keywords

Comments

From Robert A. Russell, Oct 09 2020: (Start)
a(n-1) is the number of achiral colorings of the 5 tetrahedral facets (or vertices) of a regular 4-dimensional simplex using n or fewer colors. An achiral arrangement is identical to its reflection. The 4-dimensional simplex is also called a 5-cell or pentachoron. Its Schläfli symbol is {3,3,3}.
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 class of the permutation. The first formula for a(n-1) 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_1x_4^1
32 20 x_2^1x_3^1
2111 10 x_1^3x_2^1 (End)

Crossrefs

Cf. A337895 (oriented), A000389(n+4) (unoriented), A000389 (chiral), A331353 (5-cell edges, faces), A337955 (8-cell vertices, 16-cell facets), A337958 (16-cell vertices, 8-cell facets), A338951 (24-cell), A338967 (120-cell, 600-cell).
a(n-1) = A325001(4,n).

Programs

  • Mathematica
    Do[Print[n, " ", (n^4 + 4 n^3 + 11 n^2 + 14 n + 6)/6 ], {n, 0, 10000}]
    Accumulate[Table[(2n+1)(n^2+n+3)/3,{n,0,40}]] (* or *) LinearRecurrence[ {5,-10,10,-5,1},{1,6,21,56,125},40] (* Harvey P. Dale, Feb 26 2020 *)

Formula

a(n) = (n^4 + 4*n^3 + 11*n^2 + 14*n + 6)/6 = (n^2+2*n+6)*(n+1)^2/6.
G.f.: -(x+1)*(x^2+1) / (x-1)^5. - Colin Barker, May 04 2013
From Robert A. Russell, Oct 09 2020: (Start)
a(n-1) = n^2 * (5 + n^2) / 6.
a(n-1) = binomial(n+4,5) - binomial(n,5) = A000389(n+4) - A000389(n).
a(n-1) = 1*C(n,1) + 4*C(n,2) + 6*C(n,3) + 4*C(n,4), where the coefficient of C(n,k) is the number of achiral colorings using exactly k colors.
a(n-1) = 2*A000389(n+4) - A337895(n) = A337895(n) - 2*A000389(n) .
G.f. for a(n-1): x * (x+1) * (x^2+1) / (1-x)^5. (End)
From Amiram Eldar, Feb 14 2023: (Start)
Sum_{n>=0} 1/a(n) = Pi^2/5 + 3/25 - 3*Pi*coth(sqrt(5)*Pi)/(5*sqrt(5)).
Sum_{n>=0} (-1)^n/a(n) = Pi^2/10 - 3/25 + 3*Pi*cosech(sqrt(5)*Pi)/(5*sqrt(5)). (End)
a(n) = A006007(n) + A006007(n+1) = A002415(n) + A002415(n+2). - R. J. Mathar, Jun 05 2025

Extensions

Corrected offset, Mathematica program by Tomas J. Bulka (tbulka(AT)rodincoil.com), Sep 02 2009

A325007 Array read by descending antidiagonals: A(n,k) is the number of achiral colorings of the facets of a regular n-dimensional orthotope using up to k colors.

Original entry on oeis.org

1, 2, 1, 3, 6, 1, 4, 18, 10, 1, 5, 40, 55, 15, 1, 6, 75, 200, 126, 21, 1, 7, 126, 560, 700, 252, 28, 1, 8, 196, 1316, 2850, 1996, 462, 36, 1, 9, 288, 2730, 9261, 11376, 5004, 792, 45, 1, 10, 405, 5160, 25480, 50127, 38550, 11440, 1287, 55, 1, 11, 550, 9075, 61776, 181027, 225225, 116160, 24310, 2002, 66, 1
Offset: 1

Views

Author

Robert A. Russell, May 27 2019

Keywords

Comments

Also called hypercube, n-dimensional cube, and measure polytope. For n=1, the figure is a line segment with two vertices. For n=2 the figure is a square with four edges. For n=3 the figure is a cube with six square faces. For n=4, the figure is a tesseract with eight cubic facets. The Schläfli symbol, {4,3,...,3}, of the regular n-dimensional orthotope (n>1) consists of a four followed by n-2 threes. Each of its 2n facets is an (n-1)-dimensional orthotope. An achiral coloring is identical to its reflection.
Also the number of achiral colorings of the vertices of a regular n-dimensional orthoplex using up to k colors.

Examples

			Array begins with A(1,1):
1  2   3     4      5      6       7        8         9        10 ...
1  6  18    40     75    126     196      288       405       550 ...
1 10  55   200    560   1316    2730     5160      9075     15070 ...
1 15 126   700   2850   9261   25480    61776    135675    275275 ...
1 21 252  1996  11376  50127  181027   559728   1529892   3784627 ...
1 28 462  5004  38550 225225 1053304  4119648  13942908  41918800 ...
1 36 792 11440 116160 881595 5263336 25794288 107427420 390891160 ...
For a(2,2)=6, all colorings are achiral: two with just one of the colors, two with one color on just one edge, one with opposite colors the same, and one with opposite colors different.
		

Crossrefs

Cf. A325004 (oriented), A325005 (unoriented), A325006 (chiral), A325011 (exactly k colors).
Other n-dimensional polytopes: A325001 (simplex), A325015 (orthoplex).
Rows 1-2 are A000027, A002411; column 2 is A186783(n+2).

Programs

  • Mathematica
    Table[Binomial[Binomial[d-n+2,2]+n-1,n]-Binomial[Binomial[d-n+1,2],n],{d,1,11},{n,1,d}] // Flatten
  • PARI
    a(n, k) = binomial(binomial(k+1, 2)+n-1, n) - binomial(binomial(k, 2), n)
    array(rows, cols) = for(x=1, rows, for(y=1, cols, print1(a(x, y), ", ")); print(""))
    /* Print initial 6 rows and 8 columns of array as follows: */
    array(6, 8) \\ Felix Fröhlich, May 30 2019

Formula

A(n,k) = binomial(binomial(k+1,2) + n-1, n) - binomial(binomial(k,2),n).
A(n,k) = Sum_{j=1..2*n} A325011(n,j) * binomial(k,j).
A(n,k) = 2*A325005(n,k) - A325004(n,k) = (A325004(n,k) - 2*A325006(n,k)) / 2 = A325005(n,k) + A325006(n,k).
G.f. for row n: Sum{j=1..2*n} A325011(n,j) * x^j / (1-x)^(j+1).
Linear recurrence for row n: T(n,k) = Sum_{j=0..2*n} binomial(-2-j,2*n-j) * T(n,k-1-j).
G.f. for column k: 1/(1-x)^binomial(k+1,2) - (1+x)^binomial(k,2).

A325015 Array read by descending antidiagonals: T(n,k) is the number of achiral colorings of the facets of a regular n-dimensional orthoplex using up to k colors.

Original entry on oeis.org

1, 2, 1, 3, 6, 1, 4, 18, 21, 1, 5, 40, 201, 308, 1, 6, 75, 1076, 34128, 180342, 1, 7, 126, 4025, 1056576, 2945136213, 366975285216, 1, 8, 196, 11901, 15303750, 2932338749408, 103863386269870076808, 10316179427644325573474464, 1
Offset: 1

Views

Author

Robert A. Russell, May 27 2019

Keywords

Comments

Also called cross polytope and hyperoctahedron. For n=1, the figure is a line segment with two vertices. For n=2 the figure is a square with four edges. For n=3 the figure is an octahedron with eight triangular faces. For n=4, the figure is a 16-cell with sixteen tetrahedral facets. The Schläfli symbol, {3,...,3,4}, of the regular n-dimensional orthoplex (n>1) consists of n-2 threes followed by a four. Each of its 2^n facets is an (n-1)-dimensional simplex. An achiral coloring is identical to its reflection.
Also the number of achiral colorings of the vertices of a regular n-dimensional orthotope (cube) using up to k colors.

Examples

			Array begins with T(1,1):
1   2     3       4        5         6         7          8 ...
1   6    18      40       75       126       196        288 ...
1  21   201    1076     4025     11901     29841      66256 ...
1 308 34128 1056576 15303750 136236276 865711763 4296782848 ...
...
For T(2,2)=6, two squares have all edges the same color, two have three edges the same color, one has opposite edges the same color, and one has opposite edges different colors.
		

Crossrefs

Cf. A325012 (oriented), A325013 (unoriented), A325014 (chiral), A325019 (exactly k colors).
Other n-dimensional polytopes: A325001 (simplex), A325007 (orthotope).
Rows 1-2 are A000027, A002411.

Programs

  • Mathematica
    a48[n_] := a48[n] = DivisorSum[NestWhile[#/2&, n, EvenQ], MoebiusMu[#]2^(n/#)&]/(2n); (* A000048 *)
    a37[n_] := a37[n] = DivisorSum[n, MoebiusMu[n/#]2^#&]/n; (* A001037 *)
    CI0[{n_Integer}] := CI0[{n}] = CI[Transpose[If[EvenQ[n], p2 = IntegerExponent[n, 2]; sub = Divisors[n/2^p2]; {2^(p2+1) sub, a48 /@ (2^p2 sub) }, sub = Divisors[n]; {sub, a37 /@ sub}]]] 2^(n-1); (* even perm. *)
    CI1[{n_Integer}] := CI1[{n}] = CI[sub = Divisors[n]; Transpose[If[EvenQ[n], {sub, a37 /@ sub}, {2 sub, a48 /@ sub}]]] 2^(n-1); (* odd perm. *)
    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)
    cix[{a_, b_}, {c_, d_}] := {LCM[a, c], (a b c d)/LCM[a, c]};
    Unprotect[Times]; Times[CI[a_List], CI[b_List]] :=  (* combine *) CI[compress[Flatten[Outer[cix, a, b, 1], 1]]]; Protect[Times];
    CI0[p_List] := CI0[p] = Expand[CI0[Drop[p, -1]] CI0[{Last[p]}] + CI1[Drop[p, -1]] CI1[{Last[p]}]]
    CI1[p_List] := CI1[p] = Expand[CI0[Drop[p, -1]] CI1[{Last[p]}] + CI1[Drop[p, -1]] CI0[{Last[p]}]]
    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[(CI1[#] pc[#]) & /@ IntegerPartitions[n]])/(n! 2^(n - 1))] /. CI[l_List] :> j^(Total[l][[2]])
    array[n_, k_] := row[n] /. j -> k
    Table[array[n, d-n+1], {d, 1, 10}, {n, 1, d}] // Flatten

Formula

The algorithm used in the Mathematica program below assigns each permutation of the axes to a partition of n. It then determines the number of permutations for each partition and the cycle index for each partition.
T(n,k) = 2*A325013(n,k) - A325012(n,k) = A325012(n,k) - 2*A325014(n,k) = A325013(n,k) - A325014(n,k).
T(n,k) = Sum_{j=1..3*2^(n-2)} A325019(n,j) * binomial(k,j).

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).

A324999 Array read by descending antidiagonals: A(n,k) is the number of oriented colorings of the facets (or vertices) of a regular n-dimensional simplex using up to k colors.

Original entry on oeis.org

1, 4, 1, 9, 4, 1, 16, 11, 5, 1, 25, 24, 15, 6, 1, 36, 45, 36, 21, 7, 1, 49, 76, 75, 56, 28, 8, 1, 64, 119, 141, 127, 84, 36, 9, 1, 81, 176, 245, 258, 210, 120, 45, 10, 1, 100, 249, 400, 483, 463, 330, 165, 55, 11, 1, 121, 340, 621, 848, 931, 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. Two oriented colorings are the same if one is a rotation of the other; chiral pairs are counted as two.

Examples

			The array begins with A(1,1):
  1  4  9  16  25   36   49    64    81   100   121    144    169    196 ...
  1  4 11  24  45   76  119   176   249   340   451    584    741    924 ...
  1  5 15  36  75  141  245   400   621   925  1331   1860   2535   3381 ...
  1  6 21  56 127  258  483   848  1413  2254  3465   5160   7475  10570 ...
  1  7 28  84 210  463  931  1744  3087  5215  8470  13300  20280  30135 ...
  1  8 36 120 330  792 1717  3440  6471 11560 19778  32616  52104  80952 ...
  1  9 45 165 495 1287 3003  6436 12879 24355 43923  76077 127257 206493 ...
  1 10 55 220 715 2002 5005 11440 24311 48630 92433 168180 294645 499422 ...
  ...
For A(1,2) = 4, the two achiral colorings use just one of the two colors for both vertices; the chiral pair uses two colors. For A(2,2)=4, the triangle may have 0, 1, 2, or 3 edges of one color.
		

Crossrefs

Cf. A325000 (unoriented), A325000(n,k-n) (chiral), A325001 (achiral), A325002 (exactly k colors), A327083 (edges, ridges), A337883 (faces, peaks), A325004 (orthotope facets, orthoplex vertices), A325012 (orthoplex facets, orthotope vertices).
Rows 1-4 are A000290, A006527, A006008, A337895.

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+1} A325002(n,j) * binomial(k,j).
A(n,k) = A325000(n,k) + A325000(n,k-n) = 2*A325000(n,k) - A325001(n,k) = 2*A325000(n,k-n) + A325001(n,k).
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+2} -binomial(j-n-3,j) * A(n,k-j).
G.f. for column k: (1 - 2*(1-x)^k + (1-x^2)^k) / (x*(1-x)^k) - 2*k.

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

Original entry on oeis.org

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

Views

Author

Robert A. Russell, Sep 28 2020

Keywords

Comments

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

Examples

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

Crossrefs

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

Programs

  • Mathematica
    m=2; (* dimension of color element, here a triangular face *)
    lw[n_,k_]:=lw[n, k]=DivisorSum[GCD[n,k],MoebiusMu[#]Binomial[n/#,k/#]&]/n (*A051168*)
    cxx[{a_, b_},{c_, d_}]:={LCM[a, c], GCD[a, c] b d}
    compress[x:{{, } ...}] := (s=Sort[x];For[i=Length[s],i>1,i-=1,If[s[[i,1]]==s[[i-1,1]], s[[i-1,2]]+=s[[i,2]]; s=Delete[s,i], Null]]; s)
    combine[a : {{, } ...}, b : {{, } ...}] := Outer[cxx, a, b, 1]
    CX[p_List, 0] := {{1, 1}} (* cycle index for partition p, m vertices *)
    CX[{n_Integer}, m_] := If[2m>n, CX[{n}, n-m], CX[{n},m] = Table[{n/k, lw[n/k, m/k]}, {k, Reverse[Divisors[GCD[n, m]]]}]]
    CX[p_List, m_Integer] := CX[p, m] = Module[{v = Total[p], q, r}, If[2 m > v, CX[p, v - m], q = Drop[p, -1]; r = Last[p]; compress[Flatten[Join[{{CX[q, m]}}, Table[combine[CX[q, m - j], CX[{r}, j]], {j, Min[m, r]}]], 2]]]]
    pc[p_] := Module[{ci, mb}, mb = DeleteDuplicates[p]; ci = Count[p, #] &/@ mb; Total[p]!/(Times @@ (ci!) Times @@ (mb^ci))] (* partition count *)
    row[n_Integer] := row[n] = Factor[Total[If[OddQ[Total[1-Mod[#, 2]]], pc[#] j^Total[CX[#, m+1]][[2]], 0] & /@ IntegerPartitions[n+1]]/((n+1)!/2)]
    array[n_, k_] := row[n] /. j -> k
    Table[array[n,d+m-n], {d,8}, {n,m,d+m-1}] // Flatten

Formula

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

A325002 Triangle read by rows: T(n,k) is the number of oriented colorings of the facets (or vertices) of a regular n-dimensional simplex using exactly k colors.

Original entry on oeis.org

1, 2, 1, 2, 2, 1, 3, 3, 2, 1, 4, 6, 4, 2, 1, 5, 10, 10, 5, 2, 1, 6, 15, 20, 15, 6, 2, 1, 7, 21, 35, 35, 21, 7, 2, 1, 8, 28, 56, 70, 56, 28, 8, 2, 1, 9, 36, 84, 126, 126, 84, 36, 9, 2, 1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 2, 1, 11, 55, 165, 330, 462, 462, 330, 165, 55, 11, 2
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 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. Two oriented colorings are the same if one is a rotation of the other; chiral pairs are counted as two. The only chiral pair occurs when k=n+1; for k <= n all the colorings are achiral.

Examples

			Triangle begins with T(1,1):
1  2
1  2   2
1  3   3   2
1  4   6   4    2
1  5  10  10    5    2
1  6  15  20   15    6    2
1  7  21  35   35   21    7    2
1  8  28  56   70   56   28    8    2
1  9  36  84  126  126   84   36    9    2
1 10  45 120  210  252  210  120   45   10    2
1 11  55 165  330  462  462  330  165   55   11    2
1 12  66 220  495  792  924  792  495  220   66   12   2
1 13  78 286  715 1287 1716 1716 1287  715  286   78  13   2
1 14  91 364 1001 2002 3003 3432 3003 2002 1001  364  91  14  2
1 15 105 455 1365 3003 5005 6435 6435 5005 3003 1365 455 105 15 2
For T(3,2)=3, the tetrahedron may have one, two, or three faces of one color.
		

Crossrefs

Cf. A007318(n,k-1) (unoriented), A325003 (achiral), A325001 (up to k colors).
Other n-dimensional polytopes: A325008 (orthotope), A325016 (orthoplex).

Programs

  • Mathematica
    Table[Binomial[n,k-1] + Boole[k==n+1], {n,1,15}, {k,1,n+1}] \\ Flatten

Formula

T(n,k) = binomial(n,k-1) + [k==n+1] = A007318(n,k-1) + [k==n+1].
T(n,k) = 2*A007318(n,k-1) - A325003(n,k) = [k==n+1] + A325003(n,k).
G.f. for row n: x * (1+x)^n + x^(n+1).
G.f. for column k>1: x^(k-1)/(1-x)^k + x^(k-1).

A325003 Triangle read by rows: T(n,k) is the number of achiral colorings of the facets (or vertices) of a regular n-dimensional simplex using exactly k colors.

Original entry on oeis.org

1, 0, 1, 2, 0, 1, 3, 3, 0, 1, 4, 6, 4, 0, 1, 5, 10, 10, 5, 0, 1, 6, 15, 20, 15, 6, 0, 1, 7, 21, 35, 35, 21, 7, 0, 1, 8, 28, 56, 70, 56, 28, 8, 0, 1, 9, 36, 84, 126, 126, 84, 36, 9, 0, 1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 0, 1, 11, 55, 165, 330, 462, 462, 330, 165, 55, 11, 0
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. For k <= n all the colorings are achiral.
The final zero in each row indicates no achiral colorings when each facet has a different color.

Examples

			Triangle begins with T(1,1):
1  0
1  2   0
1  3   3   0
1  4   6   4    0
1  5  10  10    5    0
1  6  15  20   15    6    0
1  7  21  35   35   21    7    0
1  8  28  56   70   56   28    8    0
1  9  36  84  126  126   84   36    9    0
1 10  45 120  210  252  210  120   45   10    0
1 11  55 165  330  462  462  330  165   55   11    0
1 12  66 220  495  792  924  792  495  220   66   12   0
1 13  78 286  715 1287 1716 1716 1287  715  286   78  13   0
1 14  91 364 1001 2002 3003 3432 3003 2002 1001  364  91  14  0
1 15 105 455 1365 3003 5005 6435 6435 5005 3003 1365 455 105 15 0
For T(3,2)=3, the tetrahedron may have one, two, or three faces of one color.
		

Crossrefs

Cf. A325002 (oriented), A007318(n,k-1) (unoriented), A325001 (up to k colors).
Other n-dimensional polytopes: A325011 (orthotope), A325019 (orthoplex).
Cf. A198321.

Programs

  • Mathematica
    Table[Binomial[n, k-1] - Boole[k==n+1], {n,1,15}, {k,1,n+1}] \\ Flatten

Formula

T(n,k) = binomial(n,k-1) - [k==n+1] = A007318(n,k-1) - [k==n+1].
T(n,k) = A325002(n,k) - 2*[k==n+1] = 2*A007318(n,k-1) - A325002(n,k).
G.f. for row n: x * (1+x)^n - x^(n+1).
G.f. for column k>1: x^(k-1)/(1-x)^k - x^(k-1).
Showing 1-10 of 10 results.