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.

A080873 a(n)*a(n+3) - a(n+1)*a(n+2) = 5, given a(0)=a(1)=1, a(2)=2.

Original entry on oeis.org

1, 1, 2, 7, 19, 69, 188, 683, 1861, 6761, 18422, 66927, 182359, 662509, 1805168, 6558163, 17869321, 64919121, 176888042, 642633047, 1751011099, 6361411349, 17333222948, 62971480443, 171581218381, 623353393081, 1698478960862
Offset: 0

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Author

Paul D. Hanna, Feb 22 2003

Keywords

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[(-3x^3-8x^2+x+1)/(x^4-10x^2+1),{x,0,30}],x] (* or *) LinearRecurrence[{0,10,0,-1},{1,1,2,7},30] (* Harvey P. Dale, Feb 27 2023 *)

Formula

For n > 1: a(2*n-1) = 3*a(2*n-2) + 2*a(2*n-3) - a(2*n-4); a(2n) = 3*a(2*n-1) - a(2*n-2).
G.f.: (1 + x - 8*x^2 - 3*x^3) / (1 - 10*x^2 + x^4). - N. J. A. Sloane, Jul 19 2005
a(n+4) = 10*a(n+2) - a(n). [Richard Choulet, Dec 04 2008]
a(n) = (1/24*(3 + 3*3^(1/2)*2^(1/2) + 6*2^(1/2) + 2*3^(1/2))/(3^(1/2)*2^(1/2) + 2))*(sqrt(3) + sqrt(2))^n + (1/16*3^(1/2)*2^(1/2) + 1/8*2^(1/2) + 1/4 + 1/6*3^(1/2))*(sqrt(3) - sqrt(2))^n + (1/24*(3*3^(1/2)*2^(1/2) - 6*2^(1/2) + 3 - 2*3^(1/2))/(3^(1/2)*2^(1/2) + 2))*( - sqrt(3) - sqrt(2))^n + (1/16*3^(1/2)*2^(1/2) - 1/8*2^(1/2) + 1/4 - 1/6*3^(1/2))*( - sqrt(3) + sqrt(2))^n. [Richard Choulet, Dec 04 2008]