A049668 a(n) = Fibonacci(8*n)/21.

0, 1, 47, 2208, 103729, 4873055, 228929856, 10754830177, 505248088463, 23735905327584, 1115082302307985, 52385132303147711, 2460986135945634432, 115613963257141670593, 5431395286949712883439, 255159964523379363851040, 11987086937311880388115441

Offset: 0

Author

Clark Kimberling

Keywords

• nonn
• easy

Comments

This is the Lucas sequence U(47,1). Also partial sums of A333718. This sequence contains all nonnegative integers a(n) such that 2205*a(n)^2 + 4 = b(n)^2 = 2205*a(n-1)*a(n+1) + 2209, where b(n) = a(n+1) - a(n-1) = A087265(n). - Klaus Purath, Aug 14 2021

Links

• Colin Barker, Table of n, a(n) for n = 0..500
• R. Flórez, R. A. Higuita and A. Mukherjee, Alternating Sums in the Hosoya Polynomial Triangle, Article 14.9.5 Journal of Integer Sequences, Vol. 17 (2014).
• Tanya Khovanova, Recursive Sequences
• Index entries for linear recurrences with constant coefficients, signature (47,-1).

Crossrefs

A column of array A028412.

Cf. A000045.

Programs

Magma

[Fibonacci(8*n)/21: n in [0..30]]; // G. C. Greubel, Dec 02 2017

Mathematica

Table[Fibonacci[8*n]/21, {n, 15}] (* Michael De Vlieger, Apr 03 2015 *)

MuPAD

numlib::fibonacci(8*n)/21 $ n = 0..25; // Zerinvary Lajos, May 09 2008

PARI

concat(0, Vec(x/(1-47*x+x^2) + O(x^20))) \\ Colin Barker, Jun 03 2016

PARI

for(n=0,30, print1(fibonacci(8*n)/21, ", ")) \\ G. C. Greubel, Dec 02 2017

Formula

G.f.: x/(1-47*x+x^2), 47=L(8)=A000032(8) (Lucas).

a(n) = 47*a(n-1)-a(n-2) ; a(0)=0, a(1)=1. - Philippe Deléham, Nov 18 2008

From Peter Bala, Apr 03 2015: (Start)

For integer k, 1 + k*(14 - k)*Sum_{n >= 1} a(n)*x^(2*n) = ( 1 + k/3*Sum_{n >= 1} Fibonacci(4*n)*x^n )*( 1 + k/3*Sum_{n >= 1} Fibonacci(4*n)*(-x)^n ).

1 + 45*Sum_{n >= 1} a(n)*x^(2*n) = ( 1 + Sum_{n >= 1} Lucas(4*n)*x^n )*( 1 + Sum_{n >= 1} Lucas(4*n)*(-x)^n ).

1 - 36*Sum_{n >= 1} a(n)*x^(2*n) = ( 1 + 2*Sum_{n >= 1} Fibonacci(4*n+2)*x^n )*( 1 + 2*Sum_{n >= 1} Fibonacci(4*n+2)*(-x)^n ). (End)

a(n) = ((47 + 21*sqrt(5))^(1-n)*(-2^n + (2207 + 987*sqrt(5))^n )) /(2205 + 987*sqrt(5)). - Colin Barker, Jun 03 2016

a(n) = (a(n-1)*a(n-2) - 47)/a(n-3), n > 3; a(n) = (a(n-1)^2 - 1)/a(n-2), n > 2. - Klaus Purath, Aug 14 2021