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 1-3 of 3 results.

A097063 Expansion of (1-2*x+3*x^2)/((1+x)*(1-x)^3).

Original entry on oeis.org

1, 0, 3, 4, 9, 12, 19, 24, 33, 40, 51, 60, 73, 84, 99, 112, 129, 144, 163, 180, 201, 220, 243, 264, 289, 312, 339, 364, 393, 420, 451, 480, 513, 544, 579, 612, 649, 684, 723, 760, 801, 840, 883, 924, 969, 1012, 1059, 1104, 1153, 1200, 1251, 1300, 1353, 1404
Offset: 0

Views

Author

Paul Barry, Jul 22 2004

Keywords

Comments

Partial sums of A097062. Pairwise sums are A002061. Binomial transform is essentially A007466.

Links

Crossrefs

A diagonal of A326296.

Programs

Formula

G.f. : (1-2*x+3*x^2)/((1-x^2)(1-x)^2).
a(n) = 2*a(n-1) - 2*a(n-3) + a(n-4).
a(n) = Sum_{k=0..n} (k^2-k+1)*(-1)^(n-k).
a(2n) = A058331(n); a(2n+1) = A046092(n). - R. J. Mathar, Oct 27 2008
a(n) = binomial(n+1, 2) - ceiling((n+1)/2) + 2((n+1) mod 2). - Wesley Ivan Hurt, Mar 08 2014
a(n) = 2*floor(n/2) + ceiling((n-1)^2/2). - M. Ryan Julian Jr., Sep 10 2019
a(n) = A326296(n + 1, n) for n > 0. - Andrew Howroyd, Sep 23 2019

A326657 a(n) = 4*floor(n/2) + ceiling((n-1)^2/2).

Original entry on oeis.org

1, 0, 5, 6, 13, 16, 25, 30, 41, 48, 61, 70, 85, 96, 113, 126, 145, 160, 181, 198, 221, 240, 265, 286, 313, 336, 365, 390, 421, 448, 481, 510, 545, 576, 613, 646, 685, 720, 761, 798, 841, 880, 925, 966, 1013, 1056, 1105, 1150, 1201, 1248, 1301, 1350, 1405, 1456, 1513, 1566, 1625, 1680
Offset: 0

Views

Author

M. Ryan Julian Jr., Sep 12 2019

Keywords

Comments

a(n) gives the maximum number of inversions in a permutation on n + 2 symbols consisting of a single n-cycle and 2 fixed points.
Sequence is a diagonal of A326296.

Links

Crossrefs

Diagonal of A326296.

Programs

  • PARI
    a(n) = 4*floor(n/2) + ceil((n-1)^2/2) \\ Andrew Howroyd, Sep 23 2019
  • PARI
    Vec((1 - 2*x + 5*x^2 - 2*x^3) / ((1 - x)^3*(1 + x)) + O(x^40)) \\ Andrew Howroyd, Sep 23 2019

Formula

a(n) = 4*floor(n/2) + ceiling((n-1)^2/2).
a(n) = A326296(2 + n, n) for n > 0.
From Colin Barker, Sep 15 2019: (Start)
G.f.: (1 - 2*x + 5*x^2 - 2*x^3) / ((1 - x)^3*(1 + x)).
a(n) = 2*a(n-1) - 2*a(n-3) + a(n-4) for n >= 4.
a(n) = (-1 + 5*(-1)^n + 4*n + 2*n^2) / 4.
(End)
E.g.f.: (1/4)*(5*exp(-x) + exp(x)*(-1 + 6*x + 2*x^2)). - Stefano Spezia, Sep 16 2019 after Colin Barker

A326658 a(n) = 6*floor(n/2) + ceiling((n-1)^2/2).

Original entry on oeis.org

1, 0, 7, 8, 17, 20, 31, 36, 49, 56, 71, 80, 97, 108, 127, 140, 161, 176, 199, 216, 241, 260, 287, 308, 337, 360, 391, 416, 449, 476, 511, 540, 577, 608, 647, 680, 721, 756, 799, 836, 881, 920, 967, 1008, 1057, 1100, 1151, 1196, 1249, 1296, 1351, 1400, 1457, 1508, 1567, 1620, 1681
Offset: 0

Views

Author

M. Ryan Julian Jr., Sep 12 2019

Keywords

Comments

a(n) gives the maximum number of inversions in a permutation on n + 3 symbols consisting of a single n-cycle and 3 fixed points.
Sequence is a diagonal of A326296.

Links

Crossrefs

Diagonal of A326296.

Programs

  • Mathematica
    Table[6*Floor[n/2] + Ceiling[(n - 1)^2/2], {n, 80}] (* Wesley Ivan Hurt, Sep 13 2019 *)
  • PARI
    a(n) = 6*floor(n/2) + ceil((n-1)^2/2) \\ Andrew Howroyd, Sep 23 2019
  • PARI
    Vec((1 - 2*x + 7*x^2 - 4*x^3) / ((1 - x)^3*(1 + x)) + O(x^40)) \\ Andrew Howroyd, Sep 23 2019

Formula

a(n) = 6*floor(n/2) + ceiling((n-1)^2/2).
a(n) = A326296(3 + n, n) for n > 0.
From Colin Barker, Sep 13 2019: (Start)
G.f.: (1 - 2*x + 7*x^2 - 4*x^3) / ((1 - x)^3*(1 + x)).
a(n) = 2*a(n-1) - 2*a(n-3) + a(n-4) for n >= 4.
a(n) = (-3 + 7*(-1)^n + 8*n + 2*n^2) / 4.
(End)
Showing 1-3 of 3 results.

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