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.

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A218004 Number of equivalence classes of compositions of n where two compositions a,b are considered equivalent if the summands of a can be permuted into the summands of b with an even number of transpositions.

Original entry on oeis.org

1, 1, 2, 4, 6, 9, 14, 19, 27, 37, 51, 67, 91, 118, 156, 202, 262, 334, 430, 543, 690, 867, 1090, 1358, 1696, 2099, 2600, 3201, 3939, 4820, 5899, 7181, 8738, 10590, 12821, 15467, 18644, 22396, 26878, 32166, 38450, 45842, 54599, 64870, 76990, 91181, 107861, 127343, 150182, 176788, 207883
Offset: 0

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Author

Geoffrey Critzer, Oct 17 2012

Keywords

Comments

a(n) = A000041(n) + A000009(n) - 1 where A000041 is the partition numbers and A000009 is the number of partitions into distinct parts.
From Gus Wiseman, Oct 14 2020: (Start)
Also the number of compositions of n that are either strictly increasing or weakly decreasing. For example, the a(1) = 1 through a(6) = 14 compositions are:
(1) (2) (3) (4) (5) (6)
(11) (12) (13) (14) (15)
(21) (22) (23) (24)
(111) (31) (32) (33)
(211) (41) (42)
(1111) (221) (51)
(311) (123)
(2111) (222)
(11111) (321)
(411)
(2211)
(3111)
(21111)
(111111)
A007997 counts only compositions of length 3.
A329398 appears to be the weakly increasing version.
A333147 is the strictly decreasing version.
A333255 union A114994 ranks these compositions using standard compositions (A066099).
A337482 counts the complement.
(End)

Examples

			a(4) = 6 because the 6 classes can be represented by: 4, 3+1, 1+3, 2+2, 2+1+1, 1+1+1+1.
		

Crossrefs

A000009 counts strictly increasing compositions, ranked by A333255.
A000041 counts weakly decreasing compositions, ranked by A114994.
A001523 counts unimodal compositions (strict: A072706).
A007318 and A097805 count compositions by length.
A032020 counts strict compositions, ranked by A233564.
A332834 counts compositions not increasing nor decreasing (strict: A333149).

Programs

  • Mathematica
    nn=50;p=CoefficientList[Series[Product[1/(1-x^i),{i,1,nn}],{x,0,nn}],x];d= CoefficientList[Series[Sum[Product[x^i/(1-x^i),{i,1,k}],{k,0,nn}],{x,0,nn}],x];p+d-1
    (* second program *)
    Table[Length[Select[Join@@Permutations/@IntegerPartitions[n],Less@@#||GreaterEqual@@#&]],{n,0,15}] (* Gus Wiseman, Oct 14 2020 *)

A058212 a(n) = 1 + floor(n*(n-3)/6).

Original entry on oeis.org

1, 0, 0, 1, 1, 2, 4, 5, 7, 10, 12, 15, 19, 22, 26, 31, 35, 40, 46, 51, 57, 64, 70, 77, 85, 92, 100, 109, 117, 126, 136, 145, 155, 166, 176, 187, 199, 210, 222, 235, 247, 260, 274, 287, 301, 316, 330, 345, 361, 376, 392, 409, 425, 442, 460, 477, 495, 514, 532, 551
Offset: 0

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Author

N. J. A. Sloane, Nov 30 2000

Keywords

Comments

For n >= 3, number of solutions to x+y+z == 0 (mod n) with 0 <= x < y < z < n. E.g., for n=3 there is a unique solution, x=0, y=1, z=2.

Crossrefs

Cf. A003035.
Apart from initial term, same as A007997.
The third diagonal of A061857.

Programs

  • Haskell
    a058212 n = 1 + n * (n - 3) `div` 6  -- Reinhard Zumkeller, May 08 2012
    
  • Mathematica
    Table[Floor[(n(n-3))/6]+1,{n,0,70}] (* or *) LinearRecurrence[{2,-1,1,-2,1},{1,0,0,1,1},70] (* Harvey P. Dale, Jun 21 2021 *)
  • PARI
    a(n)=n*(n-3)\6 + 1 \\ Charles R Greathouse IV, Jun 11 2015
  • Sage
    [ceil(binomial(n,2)/3) for n in range(-1, 55)] # Zerinvary Lajos, Dec 03 2009
    

Formula

From Paul Barry, Mar 18 2004: (Start)
G.f.: (1 - 2x + x^2 + x^4)/((1 - x)^2(1 - x^3)).
a(n) = 4*cos(2*Pi*n/3)/9 + (3*n^2 - 9*n + 10)/18. (End)
E.g.f.: (exp(x)*(10 - 6*x + 3*x^2) + 8*exp(-x/2)*cos(sqrt(3)*x/2))/18. - Stefano Spezia, May 03 2023
Sum_{n>=3} 1/a(n) = 6 - (2*Pi/sqrt(3))*(1 - tanh(sqrt(5/3)*Pi/2)/sqrt(5)). - Amiram Eldar, May 06 2023

Extensions

Zerinvary Lajos, Dec 07 2009

A053618 a(n) = ceiling(binomial(n,4)/n).

Original entry on oeis.org

0, 0, 0, 1, 1, 3, 5, 9, 14, 21, 30, 42, 55, 72, 91, 114, 140, 170, 204, 243, 285, 333, 385, 443, 506, 575, 650, 732, 819, 914, 1015, 1124, 1240, 1364, 1496, 1637, 1785, 1943, 2109, 2285, 2470, 2665, 2870, 3086, 3311, 3548, 3795, 4054, 4324
Offset: 1

Views

Author

N. J. A. Sloane, Mar 25 2000

Keywords

Crossrefs

Programs

  • Magma
    [Ceiling(Binomial(n,4)/n): n in [1..60]]; // G. C. Greubel, May 16 2019
    
  • Mathematica
    CoefficientList[Series[x^4*(1-x+x^2)*(1-x+x^2+x^4)/((1-x)^3*(1-x^8)), {x,0,60}], x] (* G. C. Greubel, May 16 2019 *)
  • PARI
    concat([0,0,0], Vec(x^4*(x^2-x+1)*(x^4+x^2-x+1) / ((x-1)^4*(x+1)*(x^2+1)*(x^4+1)) + O(x^60))) \\ Colin Barker, Jan 20 2015
    
  • Sage
    [ceil(binomial(n,4)/n) for n in (1..60)] # G. C. Greubel, May 16 2019

Formula

a(n) = ( 2*n^3 - 12*n^2 + 22*n - 3 + 9*(-1)^n + 3*(1+(-1)^n)*(-1)^(n*(n-1)/2) - 6*(1 + (-1)^n)*(-1)^floor(n/4) )/48. - Luce ETIENNE, Jan 20 2015
G.f.: x^4*(1 - x + x^2)*(1 - x + x^2 + x^4)/((1-x)^3*(1-x^8)). - Colin Barker, Jan 20 2015

A032193 Number of necklaces with 8 black beads and n-8 white beads.

Original entry on oeis.org

1, 1, 5, 15, 43, 99, 217, 429, 810, 1430, 2438, 3978, 6310, 9690, 14550, 21318, 30667, 43263, 60115, 82225, 111041, 148005, 195143, 254475, 328756, 420732, 534076, 672452, 840652, 1043460, 1287036, 1577532, 1922741
Offset: 8

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Author

Keywords

Comments

The g.f. is Z(C_8,x)/x^8, the 8-variate cycle index polynomial for the cyclic group C_8, with substitution x[i]->1/(1-x^i), i=1,...,8. Therefore by Polya enumeration a(n+8) is the number of cyclically inequivalent 8-necklaces whose 8 beads are labeled with nonnegative integers such that the sum of labels is n, for n=0,1,2,... See A102190 for Z(C_8,x). See the comment in A032191 on the equivalence of this problem with the one given in the `Name' line. - Wolfdieter Lang, Feb 15 2005
From Petros Hadjicostas, Aug 31 2018: (Start)
The CIK[k] transform of sequence (c(n): n>=1) has generating function A_k(x) = (1/k)*Sum_{d|k} phi(d)*C(x^d)^{k/d}, where C(x) = Sum_{n>=1} c(n)*x^n is the g.f. of (c(n): n>=1).
When c(n) = 1 for all n >= 1, we get C(x) = x/(1-x) and A_k(x) = (x^k/k)*Sum_{d|k} phi(d)*(1-x^d)^{-k/d}, which is the g.f. of the number a_k(n) of necklaces of n beads of 2 colors with k of them black and n-k of them white.
Using Taylor expansions, we can easily prove that a_k(n) = (1/k)*Sum_{d|gcd(n,k)} phi(d)*binomial(n/d - 1, k/d - 1) = (1/n)*Sum_{d|gcd(n,k)} phi(d)*binomial(n/d, k/d), which is Robert A. Russell's formula in the Mathematica code below.
For this sequence k = 8, and thus we get the formulae below.
(End)

Crossrefs

Programs

  • Mathematica
    k = 8; Table[Apply[Plus, Map[EulerPhi[ # ]Binomial[n/#, k/# ] &, Divisors[GCD[n, k]]]]/n, {n, k, 30}] (* Robert A. Russell, Sep 27 2004 *)
    CoefficientList[Series[1/8*(1/(1 - x)^8 + 1/(1 - x^2)^4 + 2/(1 - x^4)^2 + 4/(1 - x^8)^1),{x, 0, 30}], x] (* Stefano Spezia, Sep 01 2018 *)

Formula

"CIK[ 8 ]" (necklace, indistinct, unlabeled, 8 parts) transform of 1, 1, 1, 1...
G.f.: (x^8)*(1-3*x+5*x^2+3*x^3-4*x^4+4*x^5+6*x^6-4*x^7+7*x^8-x^9+x^10+x^11)/((1-x)^4*(1-x^2)^2*(1-x^4)*(1-x^8)).
G.f.: 1/8*x^8*(1/(1-x)^8+1/(1-x^2)^4+2/(1-x^4)^2+4/(1-x^8)^1). - Herbert Kociemba, Oct 22 2016
a(n) = (1/8)*Sum_{d|gcd(n,8)} phi(d)*binomial(n/d - 1, 8/d - 1) = (1/n)*Sum_{d|gcd(n,8)} phi(d)*binomial(n/d, 8/d). - Petros Hadjicostas, Aug 31 2018

A239438 Maximal number of points that can be placed on a triangular grid of side n so that there is no pair of adjacent points.

Original entry on oeis.org

1, 1, 3, 4, 6, 7, 10, 12, 15, 19, 22, 26, 31, 35, 40, 46, 51, 57, 64, 70, 77, 85, 92, 100, 109, 117, 126, 136, 145, 155, 166, 176, 187, 199, 210, 222, 235, 247, 260, 274, 287, 301, 316, 330, 345, 361, 376, 392, 409
Offset: 1

Views

Author

Heinrich Ludwig, Mar 18 2014

Keywords

Comments

In other words, the independence number of the (n-1)-triangular grid graph.
Apart from a(3) and a(5) same as A007997(n+4) and A058212(n+2). - Eric W. Weisstein, Jun 14 2017
Also the independence number of the n-triangular honeycomb king graph. - Eric W. Weisstein, Sep 06 2017

Examples

			On a triangular grid of side 5 at most a(5) = 6 points (X) can be placed so that there is no pair of adjacent points.
      X
     . .
    X . X
   . . . .
  X . X . X
		

Crossrefs

Programs

  • Mathematica
    Table[1/18 (Piecewise[{{28, n == 2 || n == 4}}, 10] + 3 n (3 + n) + 8 Cos[(2 n Pi)/3]), {n, 0, 20}] (* Eric W. Weisstein, Jun 14 2017 *)
  • PARI
    Vec(x*(x^9-2*x^8+2*x^7-3*x^6+3*x^5-2*x^4+2*x^3-2*x^2+x-1)/((x-1)^3*(x^2+x+1)) + O(x^100)) \\ Colin Barker, Feb 08 2015

Formula

a(n) = ceiling(n(n+1)/6) for n > 5, see Geramita, Gregory, & Roberts theorem 5.4. - Charles R Greathouse IV, Dec 04 2014
G.f.: x*(x^9-2*x^8+2*x^7-3*x^6+3*x^5-2*x^4+2*x^3-2*x^2+x-1) / ((x-1)^3*(x^2+x+1)). - Colin Barker, Feb 08 2015

Extensions

Extended by Charles R Greathouse IV, Dec 04 2014

A048259 Number of distinct solutions to x + y + z = 0 (mod n), where two solutions are equivalent if one can be obtained from the other by multiplying by units in Z/nZ and permuting x,y,z.

Original entry on oeis.org

1, 1, 2, 3, 4, 3, 7, 4, 8, 6, 8, 4, 15, 5, 10, 11, 14, 5, 17, 6, 18, 14, 12, 6, 31, 9, 14, 13, 22, 7, 33, 8, 24, 16, 16, 16, 39, 9, 18, 19, 38, 9, 41, 10, 28, 28, 20, 10, 57, 15, 30, 21, 32, 11, 43, 20, 46, 24, 24, 12, 77, 13, 26, 35, 42, 23, 53, 14, 38, 26, 52, 14, 83
Offset: 0

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Author

Keywords

Examples

			For n=6 the 7 solutions are (x,y,z) = (0,0,0), (5,1,0), (4,2,0), (4,1,1), (3,3,0), (3,2,1), (2,2,2).
		

Crossrefs

Programs

  • PARI
    iscanon(n,v)={for(d=1, n-1, if(gcd(n,d)==1 && lex(v,vecsort(v*d%n))>0, return(0))); 1}
    a(n)={if(n==0, 1, sum(x=0, n-1, sum(y=x, n-1, my(z=(-x-y)%n); y<=z && iscanon(n,[x,y,z]) )))} \\ Andrew Howroyd, Jun 11 2021

Extensions

a(42) onward corrected by Sean A. Irvine, Jun 10 2021

A053731 a(n) = ceiling(binomial(n,8)/n).

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 1, 1, 5, 15, 42, 99, 215, 429, 805, 1430, 2431, 3978, 6299, 9690, 14535, 21318, 30645, 43263, 60088, 82225, 111004, 148005, 195098, 254475, 328697, 420732, 534006, 672452, 840565, 1043460, 1286934, 1577532, 1922618
Offset: 1

Views

Author

N. J. A. Sloane, Mar 25 2000

Keywords

Crossrefs

Cf. Sequences of the form ceiling(binomial(n,k)/n): A000012 (k=1), A004526 (k=2), A007997 (k=3), A008646 (k=5), A032192 (k=7), A053618 (k=4), A053643 (k=6), this sequence (k=8), A053733 (k=9).

Programs

  • Magma
    [Ceiling(Binomial(n,8)/n): n in [1..45]]; // G. C. Greubel, Sep 06 2019
    
  • Maple
    seq(ceil(binomial(n,8)/n), n=1..45); # G. C. Greubel, Sep 06 2019
  • Mathematica
    Table[Ceiling[Binomial[n, 8]/n], {n, 45}] (* G. C. Greubel, Sep 06 2019 *)
  • PARI
    vector(45, n, ceil(binomial(n,8)/n)) \\ G. C. Greubel, Sep 06 2019
    
  • Sage
    [ceil(binomial(n,8)/n) for n in (1..45)] # G. C. Greubel, Sep 06 2019

A032194 Number of necklaces with 9 black beads and n-9 white beads.

Original entry on oeis.org

1, 1, 5, 19, 55, 143, 335, 715, 1430, 2704, 4862, 8398, 14000, 22610, 35530, 54484, 81719, 120175, 173593, 246675, 345345, 476913, 650325, 876525, 1168710, 1542684, 2017356, 2615104, 3362260, 4289780, 5433736, 6835972
Offset: 9

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Author

Keywords

Comments

The g.f. is Z(C_9,x)/x^9, the 9-variate cycle index polynomial for the cyclic group C_9, with substitution x[i]->1/(1-x^i), i=1,...,9. Therefore by Polya enumeration a(n+9) is the number of cyclically inequivalent 9-necklaces whose 9 beads are labeled with nonnegative integers such that the sum of labels is n, for n=0,1,2,... See A102190 for Z(C_9,x). See the comment in A032191 on the equivalence of this problem with the one given in the `Name' line. - Wolfdieter Lang, Feb 15 2005

Crossrefs

Programs

  • Mathematica
    k = 9; Table[Apply[Plus, Map[EulerPhi[ # ]Binomial[n/#, k/# ] &, Divisors[GCD[n, k]]]]/n, {n, k, 30}] (* Robert A. Russell, Sep 27 2004 *)

Formula

"CIK[ 9 ]" (necklace, indistinct, unlabeled, 9 parts) transform of 1, 1, 1, 1...
G.f.: (x^9)*(1-5*x+14*x^2-18*x^3+21*x^4-21*x^5+25*x^6 -21*x^7 +21*x^8 -18*x^9 +14*x^10 -5*x^11 +x^12) / ((1-x)^6*(1-x^3)^2*(1-x^9)).
G.f.: (1/9)*x^9*(1/(1-x)^9+2/(1-x^3)^3+6/(1-x^9)^1). - Herbert Kociemba, Oct 22 2016

A053733 a(n) = ceiling(binomial(n,9)/n).

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 5, 19, 55, 143, 334, 715, 1430, 2702, 4862, 8398, 13997, 22610, 35530, 54480, 81719, 120175, 173587, 246675, 345345, 476905, 650325, 876525, 1168700, 1542684, 2017356, 2615092, 3362260, 4289780, 5433722
Offset: 1

Views

Author

N. J. A. Sloane, Mar 25 2000

Keywords

Crossrefs

Cf. Sequences of the form ceiling(binomial(n,k)/n): A000012 (k=1), A004526 (k=2), A007997 (k=3), A008646 (k=5), A032192 (k=7), A053618 (k=4), A053643 (k=6), A053731 (k=8), this sequence (k=9).

Programs

  • Magma
    [Ceiling(Binomial(n,9)/n): n in [1..40]]; // G. C. Greubel, Sep 06 2019
    
  • Maple
    seq(ceil(binomial(n,9)/n), n=1..40); # G. C. Greubel, Sep 06 2019
  • Mathematica
    Table[Ceiling[Binomial[n, 9]/n], {n, 40}] (* G. C. Greubel, Sep 06 2019 *)
  • PARI
    vector(40, n, ceil(binomial(n,9)/n)) \\ G. C. Greubel, Sep 06 2019
    
  • Sage
    [ceil(binomial(n,9)/n) for n in (1..40)] # G. C. Greubel, Sep 06 2019

A347971 Triangle read by rows: T(n, k) is the number of k-dimensional subspaces in (F_4)^n, counted up to coordinate permutation (n >= 0, 0 <= k <= n).

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 7, 7, 1, 1, 12, 31, 12, 1, 1, 19, 111, 111, 19, 1, 1, 29, 361, 964, 361, 29, 1, 1, 41, 1068, 8042, 8042, 1068, 41, 1, 1, 56, 2954, 64674, 205065, 64674, 2954, 56, 1, 1, 75, 7681, 492387, 5402621, 5402621, 492387, 7681, 75, 1, 1, 97, 18880, 3507681, 137287827
Offset: 0

Views

Author

Álvar Ibeas, Sep 21 2021

Keywords

Comments

Columns can be computed by a method analogous to that of Fripertinger for isometry classes of linear codes, disallowing scalar transformation of individual coordinates.
Regarding the formula for column k = 1, note that A241926(q - 1, n) counts, up to coordinate permutation, one-dimensional subspaces of (F_q)^n generated by a vector with no zero component.

Examples

			Triangle begins:
  k:  0    1    2    3    4    5    6
      -------------------------------
n=0:  1
n=1:  1    1
n=2:  1    3    1
n=3:  1    7    7    1
n=4:  1   12   31   12    1
n=5:  1   19  111  111   19    1
n=6:  1   29  361  964  361   29    1
There are 5 = A022168(2, 1) one-dimensional subspaces in (F_4)^2, namely, those generated by vectors (0, 1), (1, 0), (1, 1), (1, x), and (1, x + 1), where F_4 = F_2[x] / (x^2 + x + 1). The coordinate swap identifies the first two on the one hand and the last two on the other, while <(1, 1)> is invariant. Hence, T(2, 1) = 3.
		

Crossrefs

Formula

T(n, 1) = T(n - 1, 1) + A007997(n + 5).
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