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.

A113225 a(2n) = A011900(n), a(2n+1) = A001109(n+1).

Original entry on oeis.org

1, 1, 3, 6, 15, 35, 85, 204, 493, 1189, 2871, 6930, 16731, 40391, 97513, 235416, 568345, 1372105, 3312555, 7997214, 19306983, 46611179, 112529341, 271669860, 655869061, 1583407981, 3822685023, 9228778026, 22280241075, 53789260175
Offset: 0

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Author

Creighton Dement, Oct 18 2005

Keywords

Comments

a(n+1) - a(n) = A097075(n+1), a(n) + a(n+1) = A024537(n+1), a(n+2) - a(n+1) - a(n) = A105635(n+1).
For n >= 1, a(n) is also the edge cover number and edge cut count of the n-Pell graph. - Eric W. Weisstein, Aug 01 2023
Also the independence number, Lovasz number, and Shannon capacity of the n-Pell graph. - Eric W. Weisstein, Aug 01 2023
Floretion Algebra Multiplication Program, FAMP Code: -2jbasejseq[B*C], B = - .5'i + .5'j - .5i' + .5j' - 'kk' - .5'ik' - .5'jk' - .5'ki' - .5'kj'; C = + .5'i + .5i' + .5'ii' + .5e

References

  • C. Dement, Floretion Integer Sequences (work in progress).

Crossrefs

Programs

  • Maple
    seq(iquo(fibonacci(n,2),1)-iquo(fibonacci(n,2),2),n=1..30); # Zerinvary Lajos, Apr 20 2008
    with(combinat):seq(ceil(fibonacci(n,2)/2), n=1..30); # Zerinvary Lajos, Jan 12 2009
  • Mathematica
    Ceiling[Fibonacci[Range[20], 2]/2]
    Table[(1 + (-1)^n + 2 Fibonacci[n + 1, 2])/4, {n, 0, 20}] // Expand
    CoefficientList[Series[-(-1 + x + x^2)/(1 - 2 x - 2 x^2 + 2 x^3 + x^4), {x, 0, 20}], x]
    LinearRecurrence[{2, 2, -2, -1}, {1, 1, 3, 6}, 20]
  • PARI
    {a(n)=local(y); if(n<0, 0, n++; y=x/(x^2+x-1)+x*O(x^n); polcoeff( y/(y^2-1), n))} /* Michael Somos, Sep 09 2006 */

Formula

G.f.: y/(y^2-1) where y=x/(x^2+x-1) if offset=1. - Michael Somos, Sep 09 2006
G.f.: (-1+x+x^2)/((1-x)*(x+1)*(x^2+2*x-1)).
Diagonal sums of A119468. - Paul Barry, May 21 2006
a(n) = (1 + (-1)^n + 2 A000129(n+1))/4. - Eric W. Weisstein, Aug 01 2023
a(n) = 2*a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4). - Eric W. Weisstein, Aug 01 2023