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.

A271389 a(n) = 2*a(n-1) + a(n-2) + n^2 for n > 1, with a(0) = 0, a(1) = 1.

Original entry on oeis.org

0, 1, 6, 22, 66, 179, 460, 1148, 2820, 6869, 16658, 40306, 97414, 235303, 568216, 1371960, 3312392, 7997033, 19306782, 46610958, 112529098, 271669595, 655868772, 1583407668, 3822684684, 9228777661, 22280240682, 53789259754, 129858760974, 313506782543, 756872326960
Offset: 0

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Author

Ilya Gutkovskiy, Apr 06 2016

Keywords

Crossrefs

Programs

  • Mathematica
    RecurrenceTable[{a[0] == 0, a[1] == 1, a[n] == 2 a[n - 1] + a[n - 2] + n^2}, a, {n, 31}]
    LinearRecurrence[{5, -8, 4, 1, -1}, {0, 1, 6, 22, 66}, 32]
  • PARI
    x='x+O('x^99); concat(0, Vec(x*(1+x)/((1-x)^3*(1-2*x-x^2)))) \\ Altug Alkan, Apr 06 2016

Formula

G.f.: x*(1 + x)/((1 - x)^3*(1 - 2*x - x^2)).
E.g.f.: (1/4)*exp(x)*(-2*(x*(x + 5) + 5) + 7*sqrt(2)*sinh(sqrt(2)*x) + 10*cosh(sqrt(2)*x)).
a(n) = 5*a(n-1) - 8*a(n-2) + 4*a(n-3) + a(n-4) - a(n-5).
a(n) = (1/8)*(-4*(n*(n + 4) + 5) + (10 - 7*sqrt(2))*(1 - sqrt(2))^n + (10 + 7*sqrt(2))*(1 + sqrt(2))^n).
Lim_{n->infinity} a(n + 1)/a(n) = 1 + sqrt(2) = A014176.
a(n) = (A000129(n+3) - A002522(n+2))/2. - R. J. Mathar, Jun 07 2016