0,2

a(n) = 18*a(n-1) - a(n-2), with a(1) = denominator of continued fraction [2;4] and a(2) = denominator of [2;4,4,4].

a(n) solves for y in the Diophantine equation x^2 - 5*y^2 = 1, the third simplest case of the Pell-Fermat type. The corresponding x solutions are provided by A023039.

n such that 5*n^2=floor(sqrt(5)*n*ceil(sqrt(5)*n)) Benoit Cloitre (benoit7848c(AT)orange.fr), May 10 2003

Tanya Khovanova, Recursive Sequences

Author?, Title?

John Robertson, Home page.

G.f.: 4x/(1-18*x+x^2). - Cino Hilliard (hillcino368(AT)gmail.com), Feb 02 2006

a(n) may be computed either as i) the denominator of the (2n-1)-th convergent of the continued fraction [2;4, 4, 4, ...] = sqrt(5), or ii) as the coefficient of sqrt(5) in {9+sqrt(5)}^n.

n such that Mod(sigma(5*n^2+1), 2 ) = 1 - Mohammed Bouayoun (bouyao(AT)wanadoo.fr), Mar 26 2004

a(n)=4*A049660(n), a(n)=A000045(6*n)/2. - Benoit Cloitre (benoit7848c(AT)orange.fr), Feb 03 2006

a(n) = 17*(a(n-1)+a(n-2))-a(n-3), a(n) = 19*(a(n-1)-a(n-2))+a(n-3). - Mohamed Bouhamida (bhmd95(AT)yahoo.fr), Sep 20 2006

Given a(1) = 4, a(2) = 72 we have for instance a(4) = 18*a(3) - a(2) = 18*{18*a(2) -a(1)} - a(2), i.e. a(4) = 323*a(2) - 18*a(1) = 323*72 - 18*4 = 23184.

A060645 := proc(n) option remember: if n=1 then RETURN(4) fi: if n=2 then RETURN(72) fi: 18*A060645(n -1)- A060645(n-2): end: for n from 1 to 30 do printf(`%d, `, A060645(n)) od:

CoefficientList[ Series[4x/(1 - 18x + x^2), {x, 0, 16}], x] (* Robert G. Wilson v *)

(PARI) g(n, k) = for(y=0, n, x=k*y^2+1; if(issquare(x), print1(y", ")))

(PARI) a(n)=fibonacci(6*n)/2 (Cloitre)

(PARI) for (i=1, 10000, if(Mod(sigma(5*i^2+1), 2)==1, print1(i, ", ")))

Cf. A023039.

Sequence in context: A066992 A100521 A111868 this_sequence A003718 A012947 A013066

Adjacent sequences: A060642 A060643 A060644 this_sequence A060646 A060647 A060648

nonn

Lekraj Beedassy (blekraj(AT)yahoo.com), Apr 17 2001

More terms from James A. Sellers (sellersj(AT)math.psu.edu), Apr 19 2001

Entry revised by njas, Aug 13 2006