OFFSET
0,3
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).
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
MATHEMATICA
Table[Fibonacci[8*n]/21, {n, 15}] (* Michael De Vlieger, Apr 03 2015 *)
PROG
(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
(Magma) [Fibonacci(8*n)/21: n in [0..30]]; // G. C. Greubel, Dec 02 2017
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
STATUS
approved