0,3

a(n)=L(n)-(1+(-1)^n)/2, where L(n) = Lucas numbers. - Mario Catalani (mario.catalani(AT)unito.it), Jan 17 2003

Floor{lim k->oo {Fibonacci(k)/Fibonacci(k-n)}} - Jon Perry (perry(AT)globalnet.co.uk), Jun 10 2003

For n>1 a(n) is the maximum element in the continued fraction for F(n)*Phi where F=A000045 and Phi=(1+sqrt(5))/2 - Benoit Cloitre (benoit7848c(AT)orange.fr), Jun 19 2005

An integer version of M.S. El Naschie's infinite-dimensional Markov process: d(n) = (1/d(0))^(n - 1); d(0)=(Sqrt[5] - 1)/2. - Roger L. Bagula and Gary W. Adamson (rlbagulatftn(AT)yahoo.com), Aug 08 2008

From 2: successive three evens and three odds. Recurrence a(n)=a(n-1)+2a(n-2)-a(n-3)-a(n-4) also valuable for successive differences ( like for instance a(n)=3a(n-1)-3a(n-2)+2a(n-3) ). See A062724 (2, 2, 3, 5) and A098600 (1, 2, 2). [From Paul Curtz (bpcrtz(AT)free.fr), Sep 20 2008]

Ayman A. El-Okaby, http://arxiv.org/pdf/0709.2394, Exceptional Lie Groups, E-infinity Theory and Higgs Boson. - Roger L. Bagula and Gary W. Adamson (rlbagulatftn(AT)yahoo.com), Aug 08 2008

T. D. Noe, Table of n, a(n) for n=0..300

G. Harman, One hundred years of normal numbers

a(n) = a(n-1) + 2*a(n-2) - a(n-3) - a(n-4). a(n) = a(n-1) + a(n-2) + (1-(-1)^n)/2.

G.f.: (1-x^2+x^3)/((1+x)(1-x)(1-x-x^2)). [From R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Sep 06 2008]

a[n_]:=Floor[GoldenRatio^n]; [From Vladimir Orlovsky (4vladimir(AT)gmail.com), Dec 12 2008]

(PARI) for (n=0, 20, print1(fibonacci(1000)/(1.0*fibonacci(1000-n))", "))

Cf. A057146, A062114, A052952, A000045, A020956.

Sequence in context: A018144 A115315 A004698 this_sequence A034297 A026636 A026658

Adjacent sequences: A014214 A014215 A014216 this_sequence A014218 A014219 A014220

nonn,easy,nice

Clark Kimberling (ck6(AT)evansville.edu)

Corrected by T. D. Noe (noe(AT)sspectra.com), Nov 09 2006

Edited by N. J. A. Sloane (njas(AT)research.att.com), Aug 29 2008 at the suggestion of R. J. Mathar