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.

A035508 a(n) = Fibonacci(2*n+2) - 1.

Original entry on oeis.org

0, 2, 7, 20, 54, 143, 376, 986, 2583, 6764, 17710, 46367, 121392, 317810, 832039, 2178308, 5702886, 14930351, 39088168, 102334154, 267914295, 701408732, 1836311902, 4807526975, 12586269024, 32951280098, 86267571271, 225851433716
Offset: 0

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Comments

Except for 0, numbers whose dual Zeckendorf representation (A104326) has the same number of 0's as 1's. - Amiram Eldar, Mar 22 2021

Crossrefs

With different offset: 2nd row of Inverse Stolarsky array A035507.
Cf. A001906, A104326, A112844, A152891 (partial sums).

Programs

  • Magma
    [Fibonacci(2*n+2)-1: n in [0..30]]; // Vincenzo Librandi, Apr 18 2011
    
  • Maple
    g:=z/(1-3*z+z^2): gser:=series(g, z=0, 43): seq(abs(coeff(gser, z, n)-1), n=1..26); # Zerinvary Lajos, Mar 22 2009
    with(combinat):seq(fibonacci(4*n+2) mod fibonacci(2*n+2),n=0..25);
  • Mathematica
    Fibonacci[2*Range[0, 5!]] - 1 (* Vladimir Joseph Stephan Orlovsky, May 18 2010 *)
    LinearRecurrence[{4,-4,1},{0,2,7},40] (* Harvey P. Dale, Jan 15 2025 *)
  • Maxima
    makelist(fib(2*n+2)-1,n,0,30); /* Martin Ettl, Oct 21 2012 */
  • MuPAD
    numlib::fibonacci(2*n)-1 $ n = 1..38; // Zerinvary Lajos, May 08 2008
    
  • Sage
    [lucas_number1(n, 3, 1)-1 for n in range(1, 27)] # Zerinvary Lajos, Dec 07 2009
    

Formula

a(n) = A001906(n) - 1.
G.f.: x*(2 - x)/((1 - x)*(1 - 3*x + x^2)). a(n) = 4*a(n-1) - 4*a(n-2) + a(n-3). - R. J. Mathar, Dec 15 2008; adapted to the offset by Bruno Berselli, Apr 19 2011
a(n) = Fibonacci(4*n+2) mod Fibonacci(2*n+2). - Gary Detlefs, Nov 22 2010
a(n+1) = Sum_{k=0..n} Fibonacci(2*k+3). - Gary Detlefs, Dec 24 2010
a(n) = Sum_{i=1..n} A112844(i). - R. J. Mathar, Apr 19 2011
a(n) = floor(Fibonacci(2*n+2) - Fibonacci(n+1)^2/Fibonacci(2*n+2)). - Gary Detlefs, Dec 21 2012
From Peter Bala, Nov 14 2021: (Start)
a(n) = Fibonacci(2*n+4)*(Fibonacci(2*n+1) - 1)/(Fibonacci(2*n+3) - 1).
a(n)= -2 + Sum_{k = 1..2*n+3} (-1)^(k+1)*Fibonacci(k). (End)