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A049629 a(n) = (F(6n+5) - F(6n+1))/4 = (F(6n+4) + F(6n+2))/4, where F=A000045 (the Fibonacci sequence). 17
1, 19, 341, 6119, 109801, 1970299, 35355581, 634430159, 11384387281, 204284540899, 3665737348901, 65778987739319, 1180356041958841, 21180629767519819, 380070979773397901, 6820097006153642399, 122381675130992165281, 2196050055351705332659, 39406519321199703822581 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,2

COMMENTS

x(n) := 2*a(n) and y(n) := A007805(n), n >= 0, give all the positive solutions of the Pell equation x^2 - 5*y^2 = -1.

The Gregory V. Richardson formula follows from this. - Wolfdieter Lang, Jun 20 2013

LINKS

Table of n, a(n) for n=0..18.

Tanya Khovanova, Recursive Sequences

H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277.

H. C. Williams and R. K. Guy, Some Monoapparitic Fourth Order Linear Divisibility Sequences Integers, Volume 12A (2012) The John Selfridge Memorial Volume.

Index entries for sequences related to Chebyshev polynomials.

Index entries for linear recurrences with constant coefficients, signature (18,-1).

FORMULA

a(n) ~ 1/4*(sqrt(5) + 2)^(2*n+1). - Joe Keane (jgk(AT)jgk.org), May 15 2002

For all members x of the sequence, 20*x^2 + 5 is a square. Lim. n -> Inf. a(n)/a(n-1) = 9 + 2*sqrt(20) = 9 + 4*sqrt(5). The 20 can be seen to derive from the equation "20*x^2 + 5 is a square". - Gregory V. Richardson, Oct 12 2002

a(n) = (((9 + 4*sqrt(5))^n - (9 - 4*sqrt(5))^n) + ((9 + 4*sqrt(5))^(n-1) - (9 - 4*sqrt(5))^(n-1)) / (8*sqrt(5)). - Gregory V. Richardson, Oct 12 2002

From R. J. Mathar, Nov 04 2008: (Start)

G.f.: (1+x)/(1 - 18x + x^2).

a(n) = A049660(n) + A049660(n+1). (End)

a(n) = 18*a(n-1) - a(n-2) for n>1; a(0)=1, a(1)=19. - Philippe Deléham, Nov 17 2008

a(n) = S(n,18) + S(n-1,18) with the Chebyshev S-polynomials (A049310). - Wolfdieter Lang, Jun 20 2013

From Peter Bala, Mar 23 2015: (Start)

a(n) = ( Fibonacci(6*n + 6 - 2*k) + Fibonacci(6*n + 2*k) )/( Fibonacci(6 - 2*k) + Fibonacci(2*k) ), for k an arbitrary integer.

a(n) = ( Fibonacci(6*n + 6 - 2*k - 1) - Fibonacci(6*n + 2*k + 1) )/( Fibonacci(6 - 2*k - 1) - Fibonacci(2*k + 1) ), for k an arbitrary integer, k != 1.

The aerated sequence (b(n))n>=1 = [1, 0, 19, 0, 341, 0, 6119, 0, ...] is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -16, Q = -1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047 for the connection with Chebyshev polynomials. (End)

a(n) = (A188378(n)^3 + (A188378(n)-2)^3) / 8. - Altug Alkan, Jan 24 2016

a(n) = sqrt(5 * Fibonacci(3 + 6*n)^2 - 4)/4. - Gerry Martens, Jul 25 2016

a(n) = Lucas(6*n + 3)/4. - Ehren Metcalfe, Feb 18 2017

EXAMPLE

Pell, n=1: (2*19)^2 - 5*17^2 = -1.

MATHEMATICA

q=10; s=0; lst={}; Do[s+=n; If[Sqrt[q*s+1]==Floor[Sqrt[q*s+1]], AppendTo[lst, Sqrt[q*s+1]]], {n, 0, 9!}]; lst (* Vladimir Joseph Stephan Orlovsky, Apr 02 2009 *)

f[n_] := Simplify[(2 + Sqrt@5)^(2 n - 1) + (2 - Sqrt@5)^(2 n - 1)]/4; Array[f, 16] (* Robert G. Wilson v, Oct 28 2010 *)

PROG

(Sage) [(lucas_number2(n, 18, 1)-lucas_number2(n-1, 18, 1))/16 for n in xrange(1, 13)] # Zerinvary Lajos, Nov 10 2009

CROSSREFS

Bisection of A001077 divided by 2.

Sequence in context: A093973 A202043 A142549 * A162805 A049664 A163110

Adjacent sequences:  A049626 A049627 A049628 * A049630 A049631 A049632

KEYWORD

nonn,easy

AUTHOR

Clark Kimberling

STATUS

approved

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Last modified July 24 21:38 EDT 2017. Contains 289777 sequences.