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 1-6 of 6 results.

A014125 Bisection of A001400.

Original entry on oeis.org

1, 3, 6, 11, 18, 27, 39, 54, 72, 94, 120, 150, 185, 225, 270, 321, 378, 441, 511, 588, 672, 764, 864, 972, 1089, 1215, 1350, 1495, 1650, 1815, 1991, 2178, 2376, 2586, 2808, 3042, 3289, 3549, 3822, 4109, 4410, 4725, 5055, 5400, 5760, 6136, 6528, 6936, 7361, 7803
Offset: 0

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Author

Keywords

Comments

Also Schoenheim bound L_1(n,5,4).
Degrees of polynomials defined by p(n) = (x^(n+1)*p(n-1)p(n-3) + p(n-2)^2)/p(n-4), p(-4)=p(-3)=p(-2)=p(-1)=1. - Michael Somos, Jul 21 2004
Degrees of polynomial tau-functions of q-discrete Painlevé I, which generate sequence A095708 when q=2 (up to an offset of 3). - Andrew Hone, Jul 29 2004
Because of the Laurent phenomenon for the general q-discrete Painlevé I tau-function recurrence p(n) = (a*x^(n+1)*p(n-1)*p(n-3) + b*p(n-2)^2)/p(n-4), p(n) for n > -1 will always be a polynomial in x and a Laurent polynomial in a,b and the initial data p(-4),p(-3),p(-2),p(-1). - Andrew Hone, Jul 29 2004
Create the sequence 0,0,0,0,0,6,18,36,66,108,... so that the sum of three consecutive terms b(n) + b(n+1) + b(n+2) = A007531(n), with b(0)=0; then a(n) = b(n+5)/6. - J. M. Bergot, Jul 30 2013
Number of partitions of n into three kinds of part 1 and one kind of part 3. - Joerg Arndt, Sep 28 2015
First differences are A001840(k) starting with k=2; second differences are A086161(k) starting with k=1. - Bob Selcoe, Sep 28 2015
Maximum Wiener index of all maximal planar graphs with n+2 vertices. The extremal graphs are cubes of paths. - Allan Bickle, Jul 09 2022
Maximum Wiener index of all maximal 3-degenerate graphs with n+2 vertices. (A maximal 3-degenerate graph can be constructed from a 3-clique by iteratively adding a new 3-leaf (vertex of degree 3) adjacent to three existing vertices.) The extremal graphs are cubes of paths, so the bound also applies to 3-trees. - Allan Bickle, Sep 18 2022

Examples

			Polynomials: p(0)=x+1, p(1)=x^3+x^2+1, p(2)=x^6+x^5+x^3+x^2+2x+1, ...
a(12)=185:  A000217(13)=91 + a(9)=94 == 91+55+28+10+1 = 185. - _Bob Selcoe_, Sep 27 2015
a(3)=11: the 11 partitions of 3 are {1a,1a,1a}, {1a,1a,1b}, {1a,1a,1c}, {1a,1b,1b}, {1a,1b,1c}, {1a,1c,1c}, {1b,1b,1b}, {1b,1b,1c}, {1b,1c,1c}, {1c,1c,1c}, {3}. - _Bob Selcoe_, Oct 04 2015
		

References

  • W. H. Mills and R. C. Mullin, Coverings and packings, pp. 371-399 of Jeffrey H. Dinitz and D. R. Stinson, editors, Contemporary Design Theory, Wiley, 1992. See Eq. 1.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • L. Smiley, Hidden Hexagons (preprint).

Crossrefs

A column of A036838.
Maximum Wiener index of all maximal k-degenerate graphs for k=1..6: A000292, A002623, A014125, A122046, A122047, A175724.

Programs

  • Magma
    [n^3/18+n^2/2+4*n/3+1+(((n+1) mod 3)-1)/9 : n in [0..50]]; // Wesley Ivan Hurt, Apr 14 2015
    
  • Magma
    I:=[1,3,6,11,18,27]; [n le 6 select I[n] else 3*Self(n-1) -3*Self(n-2) +2*Self(n-3)-3*Self(n-4)+3*Self(n-5)-Self(n-6): n in [1..50]]; // Vincenzo Librandi, Apr 15 2015
    
  • Maple
    L := proc(v,k,t,l) local i,t1; t1 := l; for i from v-t+1 to v do t1 := ceil(t1*i/(i-(v-k))); od: t1; end; # gives Schoenheim bound L_l(v,k,t). Current sequence is L_1(n,n-3,n-4,1).
  • Mathematica
    CoefficientList[Series[1/((1 - x)^3*(1 - x^3)), {x, 0, 50}], x] (* Wesley Ivan Hurt, Apr 14 2015 *)
  • PARI
    a(n)=if(n<-5,-a(-6-n),polcoeff(1/(1-x)^3/(1-x^3)+x^n*O(x),n)) /* Michael Somos, Jul 21 2004 */
    
  • PARI
    my(x='x+O('x^50)); Vec(1/((1-x)^3*(1-x^3))) \\ Altug Alkan, Oct 16 2015
    
  • PARI
    a(n)=(n^3 + 9*n^2 + 24*n + 19)\/18 \\ Charles R Greathouse IV, Jun 29 2020
    
  • Sage
    [(binomial(n+4,3) - ((n+4)//3))/3 for n in (0..50)] # G. C. Greubel, Apr 28 2019

Formula

G.f.: 1/((1-x)^3*(1-x^3)).
a(n) = -a(-6-n) = 3*a(n-1) -3*a(n-2) +2*a(n-3) -3*a(n-4) +3*a(n-5) -a(n-6).
The simplest recurrence is fourth order: a(n) = a(n-1) + a(n-3) - a(n-4) + n + 1, which gives the g.f.: 1/((1-x)^3*(1-x^3)), with a(-4) = a(-3) = a(-2) = a(-1) = 0.
a(n) = n^3/18 + n^2/2 + 4*n/3 + 1 + 2/(9*sqrt(3))*sin(2*Pi*n/3). - Andrew Hone, Jul 29 2004
a(n) = n^3/18 + n^2/2 + 4*n/3 + 1 + (((n+1) mod 3) - 1)/9. - same formula, simplified by Gerald Hillier, Apr 14 2015
a(n) = (2*A000027(n+1) + 3*A000292(n+1) + A049347(n-1) + 1 + 3*A000217(n+1))/9. - R. J. Mathar, Nov 16 2007
From Johannes W. Meijer, May 20 2011: (Start)
a(n) = A144677(n) + A144677(n-1) + A144677(n-2).
a(n) = A190717(n-4) + 2*A190717(n-3) + 3*A190717(n-2) + 2*A190717(n-1) + A190717(n). (End)
3*a(n) = binomial(n+4,3) - floor((n+4)/3). - Bruno Berselli, Nov 08 2013
a(n) = A000217(n+1) + a(n-3) = Sum_{j>=0, n>=3*j} (n-3*j+1)*(n-3*j+2)/2. - Bob Selcoe, Sep 27 2015
a(n) = round(((2*n+5)^3 + 3*(2*n+5)^2 - 9*(2*n+5))/144). - Giacomo Guglieri, Jun 28 2020
a(n) = floor(((n+2)^3 + 3*(n+2)^2)/18). - Allan Bickle, Aug 01 2020
a(n) = Sum_{j=0..n} (n-j+1)*floor((j+3)/3). - G. C. Greubel, Oct 18 2021
E.g.f.: exp(x) + exp(x)*x*(34 + 12*x + x^2)/18 + 2*exp(-x/2)*sin(sqrt(3)*x/2)/(9*sqrt(3)). - Stefano Spezia, Apr 05 2023

Extensions

More terms from James Sellers, Dec 24 1999

A092498 Expansion of g.f. (1 + x + 2*x^2)/((1 - x)^3*(1 - x^3)).

Original entry on oeis.org

1, 4, 11, 23, 41, 67, 102, 147, 204, 274, 358, 458, 575, 710, 865, 1041, 1239, 1461, 1708, 1981, 2282, 2612, 2972, 3364, 3789, 4248, 4743, 5275, 5845, 6455, 7106, 7799, 8536, 9318, 10146, 11022, 11947, 12922, 13949, 15029, 16163, 17353, 18600, 19905, 21270
Offset: 0

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Author

N. J. A. Sloane, Apr 05 2004

Keywords

Comments

Arises from the Molien series for 4-dimensional group of structure S_3 X C_2 and order 12, which preserves the complete weight enumerators of even trace-Hermitian self-dual additive codes over GF(4). The Molien series is (1 + x^2 + 2*x^4)/((1 - x^2)^3 *(1 - x^6)).
From Philippe LALLOUET (philip.lallouet(AT)wanadoo.fr), Sep 24 2007: (Start)
Also arises when a pyramid is built row by row with squares of size 1.
At the first step, we put a single square on row 1. At the second, we put a square above the first one, on row 2, and a square on each of its sides on row 1. At each following step, we begin a new row with one square and add a square at each end of each of the previous rows. The term a(n) of the sequence is the total number of squares of any size which can be seen in the entire triangular array.
..........................
.....................|__|..
.....|__|.....|__||__|
||..||__||...||__||__||
The table below gives the number of squares by size, and the total number of squares (i.e., a(n)), for each row.
+-----------------------+
|size size size size |
n | 1 2 3 4 | a(n)
--+-----------------------+-----
1 | .1....................|....1
2 | .4....................|....4
3 | .9....2...............|...11
4 | 16....6....1..........|...23
5 | 25...12....4..........|...41
6 | 36...20....9....2.....|...67
(End)

Crossrefs

Cf. A014126.
Cf. A000969 (first differences). - R. J. Mathar, Jan 05 2009

Programs

  • Maple
    A092498:=n->(4*n^3+21*n^2+35*n+18-6*floor((n+2)/3))/18; seq(A092498(n), n=0..50); # Wesley Ivan Hurt, Apr 19 2014
  • Mathematica
    Table[(4*n^3 + 21*n^2 + 35*n + 18 - 6*Floor[(n + 2)/3])/18, {n, 0, 50}] (* Wesley Ivan Hurt, Apr 19 2014 *)
    CoefficientList[Series[(1 + x + 2 x^2)/((1 - x)^3 (1 -x^3)), {x, 0, 40 }], x] (* Vincenzo Librandi, Apr 20 2014 *)
    LinearRecurrence[{3,-3,2,-3,3,-1},{1,4,11,23,41,67},50] (* Harvey P. Dale, Jul 08 2017 *)

Formula

a(n) = (4*n^3+21*n^2+35*n+18-6*floor((n+2)/3))/18. - Luce ETIENNE, Apr 18 2014
a(n) = Sum_{j=0..floor(2*n/3)} ((4*n+5-6*j-(-1)^j)/4)*((4*n+3-6*j+(-1)^j)/4). - Luce ETIENNE, Oct 28 2014
E.g.f.: exp(-x/2)*(3*exp(3*x/2)*(2 + x)*(8 + 25*x + 4*x^2) + 6*cos(sqrt(3)*x/2) - 2*sqrt(3)*sin(sqrt(3)*x/2))/54. - Stefano Spezia, Apr 05 2023

Extensions

Edited by N. J. A. Sloane, May 15 2014

A256309 Number of partitions of 2n into exactly 5 parts.

Original entry on oeis.org

0, 0, 0, 1, 3, 7, 13, 23, 37, 57, 84, 119, 164, 221, 291, 377, 480, 603, 748, 918, 1115, 1342, 1602, 1898, 2233, 2611, 3034, 3507, 4033, 4616, 5260, 5969, 6747, 7599, 8529, 9542, 10642, 11835, 13125, 14518, 16019, 17633, 19366, 21224, 23212, 25337, 27604
Offset: 0

Views

Author

Colin Barker, Mar 22 2015

Keywords

Examples

			For n=4 the 3 partitions of 2*4 = 8 are [1,1,1,1,4], [1,1,1,2,3] and [1,1,2,2,2].
		

Crossrefs

Cf. Similar sequences: A000212 (3 parts), A001477 (2 parts), A014126 (4 parts), A256310 (6 parts).

Programs

  • Mathematica
    CoefficientList[Series[- x^3 (x^4 + x^2 + x + 1) / ((x - 1)^5 (x + 1) (x^2 + x + 1) (x^4 + x^3 + x^2 + x + 1)), {x, 0, 50}], x] (* Vincenzo Librandi, Mar 22 2015 *)
    LinearRecurrence[{2,0,-1,-1,1,0,-1,1,1,0,-2,1},{0,0,0,1,3,7,13,23,37,57,84,119},50] (* Harvey P. Dale, Mar 06 2023 *)
  • PARI
    concat(0, vector(40, n, k=0; forpart(p=2*n, k++, , [5,5]); k))
    
  • PARI
    concat([0,0,0], Vec(-x^3*(x^4+x^2+x+1)/((x-1)^5*(x+1)*(x^2+x+1)*(x^4+x^3+x^2+x+1)) + O(x^100)))

Formula

G.f.: -x^3*(x^4+x^2+x+1) / ((x-1)^5*(x+1)*(x^2+x+1)*(x^4+x^3+x^2+x+1)).

A256310 Number of partitions of 2n into exactly 6 parts.

Original entry on oeis.org

0, 0, 0, 1, 2, 5, 11, 20, 35, 58, 90, 136, 199, 282, 391, 532, 709, 931, 1206, 1540, 1945, 2432, 3009, 3692, 4494, 5427, 6510, 7760, 9192, 10829, 12692, 14800, 17180, 19858, 22856, 26207, 29941, 34085, 38677, 43752, 49342, 55491, 62239, 69624, 77695, 86499
Offset: 0

Views

Author

Colin Barker, Mar 22 2015

Keywords

Comments

The number of partitions of 2*(n-3) into at most 6 parts. - Colin Barker, Mar 31 2015

Examples

			For n=4 the 2 partitions of 2*4 = 8 are [1,1,1,1,1,3] and [1,1,1,1,2,2].
		

Crossrefs

Cf. Similar sequences: A000212 (3 parts), A001477 (2 parts), A014126 (4 parts), A256309 (5 parts).

Programs

  • Mathematica
    CoefficientList[Series[x^3 (x^4 + x^3 + x^2 + 1) / ((x - 1)^6 (x + 1) (x^2 + x + 1)^2 (x^4 + x^3 + x^2 + x + 1)), {x, 0, 50}], x] (* Vincenzo Librandi, Mar 22 2015 *)
  • PARI
    concat(0, vector(40, n, k=0; forpart(p=2*n, k++, , [6,6]); k))
    
  • PARI
    concat([0,0,0], Vec(x^3*(x^4+x^3+x^2+1)/((x-1)^6*(x+1)*(x^2+x+1)^2*(x^4+x^3+x^2+x+1)) + O(x^100)))

Formula

G.f.: x^3*(x^4+x^3+x^2+1) / ((x-1)^6*(x+1)*(x^2+x+1)^2*(x^4+x^3+x^2+x+1)).
a(n) = 2*a(n-1) - 3*a(n-4) + a(n-5) + a(n-6) + a(n-9) + a(n-10) - 3*a(n-11) + 2*a(n-14) - a(n-15). - Wesley Ivan Hurt, Jul 28 2022

A191593 Number of partitions of 12*n into parts < 5.

Original entry on oeis.org

1, 34, 169, 478, 1033, 1906, 3169, 4894, 7153, 10018, 13561, 17854, 22969, 28978, 35953, 43966, 53089, 63394, 74953, 87838, 102121, 117874, 135169, 154078, 174673, 197026, 221209, 247294, 275353, 305458, 337681
Offset: 0

Views

Author

Adi Dani, Jun 07 2011

Keywords

Comments

Number of ways of placing of 12*n indistinguishable objects into indistinguishable boxes with condition that in each box can be at most 4 objects.

Examples

			a(1)=34 all partitions of 1*12=12 into parts < 5 are:
[1,1,1,1,1,1,1,1,1,1,1,1],
[1,1,1,1,1,1,1,1,1,1,2],
[1,1,1,1,1,1,1,1,1,3],
[1,1,1,1,1,1,1,1,2,2],
[1,1,1,1,1,1,1,1,4],
[1,1,1,1,1,1,1,2,3],
[1,1,1,1,1,1,2,2,2],
[1,1,1,1,1,1,2,4],
[1,1,1,1,1,1,3,3],
[1,1,1,1,1,2,2,3],
[1,1,1,1,2,2,2,2],
[1,1,1,1,1,3,4],
[1,1,1,1,2,2,4],
[1,1,1,1,2,3,3],
[1,1,1,2,2,2,3],
[1,1,2,2,2,2,2],
[1,1,1,1,4,4],
[1,1,1,2,3,4],
[1,1,1,3,3,3],
[1,1,2,2,2,4],
[1,1,2,2,3,3],
[1,2,2,2,2,3],
[2,2,2,2,2,2],
[1,1,2,4,4],
[1,1,3,3,4],
[1,2,2,3,4],
[1,2,3,3,3],
[2,2,2,2,4],
[2,2,2,3,3],
[1,3,4,4],
[2,2,4,4],
[2,3,3,4],
[3,3,3,3],
[4,4,4].
		

Programs

  • Magma
    [12*n^3+15*n^2+6*n+1: n in [0..30]]; // Vincenzo Librandi, Jun 16 2011
  • Mathematica
    Table[12n^3 + 15n^2 + 6n + 1, {n, 0, 30}]

Formula

a(n) = 12*n^3 + 15*n^2 + 6*n + 1.
From R. J. Mathar, Jun 08 2011: (Start)
a(n) = A001400(12n) = A014126(6n).
G.f.: (1 + 30*x + 39*x^2 + 2*x^3) / (x-1)^4. (End)

A282044 Reduced Kronecker coefficients for the case a=2, b=3, i=4.

Original entry on oeis.org

0, 0, 0, 0, 1, 2, 5, 9, 16, 24, 37, 51, 71, 93
Offset: 0

Views

Author

N. J. A. Sloane, Feb 21 2017

Keywords

Comments

Table 3 of Colmenarejo (2016) shows this sequence as the missing member of the family A266769, A000601, A006918, A014126, A282044, A175287.
It would be nice to have a g.f.

Crossrefs

Formula

Conjectured g.f.: x^4*(1+x^2)/(1-2*x+x^3+3*x^5-4*x^6). - Jean-François Alcover, Feb 18 2019.
Showing 1-6 of 6 results.