login
This site is supported by donations to The OEIS Foundation.

 

Logo


Hints
(Greetings from The On-Line Encyclopedia of Integer Sequences!)
A060645 a(0) = 0, a(1) = 4, then a(n) = 18*a(n-1) - a(n-2). 10

%I

%S 0,4,72,1292,23184,416020,7465176,133957148,2403763488,43133785636,

%T 774004377960,13888945017644,249227005939632,4472197161895732,

%U 80250321908183544,1440033597185408060,25840354427429161536

%N a(0) = 0, a(1) = 4, then a(n) = 18*a(n-1) - a(n-2).

%C 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].

%C This sequence gives the values of y in solutions of the Diophantine equation x^2 - 5*y^2 = 1, the third simplest case of the Pell-Fermat type. The corresponding x values are in A023039.

%C n such that 5*n^2=floor(sqrt(5)*n*ceil(sqrt(5)*n)). - _Benoit Cloitre_, May 10 2003

%H Harry J. Smith, <a href="/A060645/b060645.txt">Table of n, a(n) for n = 0..200</a>

%H Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>

%H John Robertson, <a href="http://www.jpr2718.org/">Home page.</a>

%H <a href="/index/Rec">Index entries for linear recurrences with constant coefficients</a>, signature (18,-1).

%F G.f.: 4x/(1-18*x+x^2). - _Cino Hilliard_, Feb 02 2006

%F 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.

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

%F a(n)=4*A049660(n), a(n)=A000045(6*n)/2. - _Benoit Cloitre_, Feb 03 2006

%F 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

%F a(n)=-(1/10)*[9-4*sqrt(5)]^n*sqrt(5)+(1/10)*sqrt(5)*[9+4*sqrt(5)]^n, with n>=0. - _Paolo P. Lava_, Oct 02 2008

%F From _Johannes W. Meijer_, Jul 01 2010: (Start)

%F Limit(a(n+k)/a(k), k=infinity) = A023039(n) + A060645(n)*sqrt(5).

%F Limit(A023039(n)/a(n), n=infinity) = sqrt(5). (End)

%F a(n)= fibonacci(6*n)/2. - _Gary Detlefs_, Apr 02 2012

%F a(n) = 4*S(n-1, 18), with Chebyshev's S-polynomials. See A049310. S(-1, x)= 0. - _Wolfdieter Lang_, Aug 24 2014

%e 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.

%p 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:

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

%t Select[Select[Table[Fibonacci[n],{n,0,5!}],EvenQ]/2,EvenQ] (* _Vladimir Joseph Stephan Orlovsky_, May 10 2010 *)

%t LinearRecurrence[{18, -1} {0, 4}, 50] (* _Sture Sjöstedt_, Nov 29 2011 *)

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

%o (PARI) a(n)=fibonacci(6*n)/2 \\ _Benoit Cloitre_

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

%o (PARI) { for (n=0, 200, write("b060645.txt", n, " ", fibonacci(6*n)/2); ) } \\ _Harry J. Smith_, Jul 09 2009

%Y Cf. A023039.

%K nonn,easy

%O 0,2

%A _Lekraj Beedassy_, Apr 17 2001

%E More terms from _James A. Sellers_, Apr 19 2001

%E Entry revised by _N. J. A. Sloane_, Aug 13 2006

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.

License Agreements, Terms of Use, Privacy Policy. .

Last modified November 21 16:04 EST 2019. Contains 329371 sequences. (Running on oeis4.)