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.

A057088 Scaled Chebyshev U-polynomials evaluated at i*sqrt(5)/2. Generalized Fibonacci sequence.

Original entry on oeis.org

1, 5, 30, 175, 1025, 6000, 35125, 205625, 1203750, 7046875, 41253125, 241500000, 1413765625, 8276328125, 48450468750, 283633984375, 1660422265625, 9720281250000, 56903517578125, 333118994140625, 1950112558593750, 11416157763671875, 66831351611328125, 391237546875000000
Offset: 0

Views

Author

Wolfdieter Lang, Aug 11 2000

Keywords

Comments

a(n) gives the length of the word obtained after n steps with the substitution rule 0->11111, 1->111110, starting from 0. The number of 1's and 0's of this word is 5*a(n-1) and 5*a(n-2), resp.
a(n) / a(n-1) converges to (5 + (3 * sqrt(5))) / 2 as n approaches infinity. (5 + (3 * sqrt(5))) / 2 can also be written as phi^2 + (2 * phi), phi^3 + phi, phi + sqrt(5) + 2, (3 * phi) + 1, (3 * phi^2) - 2, phi^4 - 1 and (5 + (3 * (L(n) / F(n)))) / 2, where L(n) is the n-th Lucas number and F(n) is the n-th Fibonacci number as n approaches infinity. - Ross La Haye, Aug 18 2003, on another version
Pisano period lengths: 1, 3, 3, 6, 1, 3, 24, 12, 9, 3, 10, 6, 56, 24, 3, 24,288, 9, 18, 6, ... - R. J. Mathar, Aug 10 2012

Crossrefs

Programs

  • Magma
    I:=[1, 5]; [n le 2 select I[n] else 5*Self(n-1) + 5*Self(n-2): n in [0..30]]; // G. C. Greubel, Jan 16 2018
  • Maple
    a[0]:=0:a[1]:=1:for n from 2 to 50 do a[n]:=5*a[n-1]+5*a[n-2]od: seq(a[n], n=1..33); # Zerinvary Lajos, Dec 14 2008
  • Mathematica
    LinearRecurrence[{5,5}, {1,5}, 30] (* G. C. Greubel, Jan 16 2018 *)
  • PARI
    x='x+O('x^30); Vec(1/(1 - 5*x - 5*x^2)) \\ G. C. Greubel, Jan 16 2018
    
  • Sage
    [lucas_number1(n,5,-5) for n in range(1, 22)] # Zerinvary Lajos, Apr 24 2009
    

Formula

a(n) = 5*(a(n-1) + a(n-2)), a(-1)=0, a(0)=1.
a(n) = S(n, i*sqrt(5))*(-i*sqrt(5))^n with S(n, x) := U(n, x/2), Chebyshev's polynomials of the 2nd kind, A049310.
G.f.: 1/(1 - 5*x - 5*x^2).
a(n) = (1/3)*Sum_{k=0..n} binomial(n, k)*Fibonacci(k)*3^k. - Benoit Cloitre, Oct 25 2003
a(n) = ((5 + 3*sqrt(5))/2)^n(1/2 + sqrt(5)/6) + (1/2 - sqrt(5)/6)((5 - 3*sqrt(5))/2)^n. - Paul Barry, Sep 22 2004
(a(n)) appears to be given by the floretion - 0.75'i - 0.5'j + 'k - 0.75i' + 0.5j' + 0.5k' + 1.75'ii' - 1.25'jj' + 1.75'kk' - 'ij' - 0.5'ji' - 0.75'jk' - 0.75'kj' - 1.25e ("jes"). - Creighton Dement, Nov 28 2004
a(n) = Sum_{k=0..n} 4^k*A063967(n,k). - Philippe Deléham, Nov 03 2006
G.f.: G(0)/(2-5*x), where G(k)= 1 + 1/(1 - x*(9*k-5)/(x*(9*k+4) - 2/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 17 2013
From Ehren Metcalfe, Nov 18 2017: (Start)
With F(n) = A000045(n), L(n) = A000032(n), beta = (1-sqrt(5))/2:
a(2*n-1) = 5^n*F(4*n)/3 = (5^(n-1/2)*L(4*n) - 2*5^(n-1/2)*beta^(4*n))/3.
a(2*n) = 5^n*L(4*n+2)/3 = (5^(n+1/2)*F(4*n+2) + 2*5^n*beta^(4*n+2))/3.
a(n) = round 5^((n+1)/2)*F(2*(n+1))/3.
a(n) = round 5^(n/2)*L(2*(n+1))/3. (End)