A033888 a(n) = Fibonacci(4*n).

0, 3, 21, 144, 987, 6765, 46368, 317811, 2178309, 14930352, 102334155, 701408733, 4807526976, 32951280099, 225851433717, 1548008755920, 10610209857723, 72723460248141, 498454011879264, 3416454622906707, 23416728348467685, 160500643816367088, 1100087778366101931, 7540113804746346429

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• easy

Comments

(x,y)=(a(n),a(n+1)) are solutions of (x+y)^2/(1+xy)=9, the other solutions are in A033890. - Floor van Lamoen, Dec 10 2001

Sequence A033888 provides half of the solutions to the equation 5*x^2 + 4 is a square. The other half are found in A033890. Lim_{n->infinity} a(n)/a(n-1) = phi^4 = (7+3*sqrt(5))/2. - Gregory V. Richardson, Oct 13 2002

Fibonacci numbers divisible by 3. - Reinhard Zumkeller, Aug 20 2011

Examples

			G.f. = 3*x + 21*x^2 + 144*x^3 + 987*x^4 + 6765*x^5 + 46368*x^6 + ...
		

Links

• Nathaniel Johnston, Table of n, a(n) for n = 0..300
• Piero Filipponi and Marco Bucci, On the Integrity of Certain Fibonacci Sums, The Fibonacci Quarterly, Vol. 32, No. 3 (1994), pp. 245-252.
• Tanya Khovanova, Recursive Sequences
• Index entries for linear recurrences with constant coefficients, signature (7,-1).

Crossrefs

Cf. A000032, A000045, A004187, A033890, A337928, A337929.

Fourth column of array A102310.

Programs

Magma

[ Fibonacci(4*n): n in [0..100]]; // Vincenzo Librandi, Apr 15 2011

Maple

A033888:=n->combinat[fibonacci](4*n): seq(A033888(n), n=0..30); # Wesley Ivan Hurt, Apr 26 2017

Mathematica

Table[Fibonacci[4*n], {n,0,14}] (* Vladimir Joseph Stephan Orlovsky, Jul 21 2008 *)
Table[Mod[Fibonacci[(4 n + 2)] , Fibonacci[(4 n + 1)]], {n, 1, 10}] (* Artur Jasinski, Nov 15 2011 (corrected by Iain Fox, Dec 18 2017) *)

MuPAD

numlib::fibonacci(n*4) $ n = 0..30; // Zerinvary Lajos, May 08 2008

PARI

a(n)=fibonacci(4*n) \\ Charles R Greathouse IV, Feb 03 2014

PARI

first(n) = Vec(3*x/(1 - 7*x + x^2) + O(x^n), -n) \\ Iain Fox, Dec 18 2017

PARI

a(n) = fibonacci(4*n + 2) % fibonacci(4*n + 1) \\ Iain Fox, Dec 18 2017

Sage

[lucas_number1(n,3,1)*lucas_number2(n,3,1) for n in range(0,21)] # Zerinvary Lajos, Jun 28 2008

Sage

[fibonacci(4*n) for n in range(0, 20)] # Zerinvary Lajos, May 15 2009

Formula

a(n) = 7*a(n-1) - a(n-2).

a(n) = ((7+3*sqrt(5))^(n-1) - (7-3*sqrt(5))^(n-1)) / ((2^(n-1))*sqrt(5)). - Gregory V. Richardson, Oct 13 2002

a(n) = Sum_{k=0..n} F(3*n-k)*binomial(n, k). - Benoit Cloitre, Jun 07 2004

a(n) = Lucas(2*n) * Lucas(n) * Fibonacci(n). - Ralf Stephan, Sep 25 2004

G.f.: 3*x/(1-7*x+x^2). - Philippe Deléham, Nov 17 2008

a(n) = 3*A004187(n). - R. J. Mathar, Sep 03 2010

a(n) = Fibonacci[(4*n + 2)] modulo Fibonacci[(4*n + 1)]. - Artur Jasinski, Nov 15 2011 (corrected by Iain Fox, Dec 18 2017)

a(n) = (A337929(n) + A337928(n)) / 2. - Flávio V. Fernandes, Feb 06 2021

E.g.f.: 2*exp(7*x/2)*sinh(3*sqrt(5)*x/2)/sqrt(5). - Stefano Spezia, Feb 07 2021

a(n) = Sum_{k>=0} Fibonacci(2*n*k)/Lucas(2*n)^k (Filipponi and Bucci, 1994). - Amiram Eldar, Jan 17 2022

From Peter Bala, May 22 2025: (Start)

Sum_{n >= 1} (-1)^(n+1)/(a(n) - 1/a(n)) = 1/3 (telescoping series: 3/(a(n) - 1/a(n)) = 1/A033890(n) + 1/A033890(n-1)).

Product_{n >= 1} (a(n) + 1)/(a(n) - 1) = sqrt(5) (telescoping product: ((a(n) + 1)/(a(n) - 1))^2 = (5 - 4/Fibonacci(2*n+1)^2)/(5 - 4/Fibonacci(2*n-1)^2) from which we get Product_{k = 1..n} ((a(k) + 1)/(a(k) - 1))^2 = 5 - 4/Fibonacci(2*n+1)^2). (End)