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.

A072265 Variant of Lucas numbers: a(n) = a(n-1) + 4*a(n-2) starting with a(0)=2 and a(1)=1.

Original entry on oeis.org

2, 1, 9, 13, 49, 101, 297, 701, 1889, 4693, 12249, 31021, 80017, 204101, 524169, 1340573, 3437249, 8799541, 22548537, 57746701, 147940849, 378927653, 970691049, 2486401661, 6369165857, 16314772501, 41791435929, 107050525933, 274216269649, 702418373381
Offset: 0

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Author

Miklos Kristof, Jul 08 2002

Keywords

Comments

Pisano period lengths: 1, 1, 8, 1, 6, 8, 48, 2, 24, 6,120, 8, 12, 48, 24, 4, 8, 24, 18, 6, ... . - R. J. Mathar, Aug 10 2012
The Lucas sequence V(1,-4). - Peter Bala, Jun 23 2015

References

  • Thomas Koshy, "Fibonacci and Lucas Numbers with Applications", Wiley, 2001, p. 471.

Crossrefs

Cf. A006131.

Programs

  • GAP
    a:=[2,1];; for n in [3..30] do a[n]:=a[n-1]+4*a[n-2]; od; a; # G. C. Greubel, Jan 15 2020
  • Magma
    I:=[2,1]; [n le 2 select I[n] else Self(n-1) + 4*Self(n-2): n in [1..30]]; // G. C. Greubel, Jan 15 2020
    
  • Maple
    a:= n-> (Matrix([[1,2]]). Matrix([[1,1], [4,0]])^n)[1,2]:
    seq(a(n), n=0..32);  # Alois P. Heinz, Aug 15 2008
    a := n -> 2*(2*I)^n*ChebyshevT(n, -I/4):
    seq(simplify(a(n)), n = 0..29);  # Peter Luschny, Dec 03 2023
  • Mathematica
    CoefficientList[Series[(2-x)/(1-x-4*x^2), {x,0,30}], z] (* Vladimir Joseph Stephan Orlovsky, Jun 11 2011 *)
    Table[2^n*LucasL[n, 1/2], {n,0,30}] (* G. C. Greubel, Jan 15 2020 *)
  • PARI
    polsym(x^2-x-4, 44)
    
  • Sage
    [lucas_number2(n,1,-4) for n in range(0, 27)] # Zerinvary Lajos, Apr 30 2009
    

Formula

G.f.: (2-x)/(1-x-4*x^2). - Gary W. Adamson, Jul 02 2003
a(n) = ((1+sqrt(17))/2)^n + ((1-sqrt(17))/2)^n = 4*A006131(n-1) + A006131(n+1) = A075117(4, n).
a(n) = [x^n] ( (1 + x + sqrt(1 + 2*x + 17*x^2))/2 )^n for n >= 1. - Peter Bala, Jun 23 2015
a(n) = 2^n * Lucas(n, 1/2). - G. C. Greubel, Jan 15 2020

Extensions

Edited and extended by Henry Bottomley, Sep 03 2002