0,1

Another version of the Fibonacci word.

(With offset 1) for k>0, a(ceiling(k*phi^2))=0 and a(floor(k*phi^2))=1 where phi=(1+sqrt(5))/2 is the Golden ratio - B. Cloitre (benoit7848c(AT)orange.fr), Apr 01 2006

(With offset 1) for n>1 a(A000045(n))=(1-(-1)^n)/2

Equals the Fibonacci word A005614 with an initial zero.

J.-P. Allouche and J. Shallit, Automatic Sequences, Cambridge Univ. Press, 2003.

Conjecture: a(n) is given recursively by a(1)=0 and, for n>1, by a(n)=1 if n=F(2k+1) and a(n)=a(n-F(2k+1)) otherwise, where F(2k+1) is the largest odd-index Fibonacci number smaller than or equal to n. (This has been confirmed for more than nine million terms.) The odd-index bisection of the Fibonacci numbers (A001519) is {1, 2, 5, 13, 34, 89, ...}. So by the conjecture, we would expect that a(30) = a(30-13) = a(17) = a(17-13) = a(4) = a(4-2) = a(2) = 1, which is in fact correct. - John W. Layman (layman(AT)math.vt.edu), Jun 29 2004

(With offset 1) a(n)=-1+floor(n*phi)-floor((n-1)*phi) where phi=(1+sqrt(5))/2 so a(n)=-1+A082389(n) - B. Cloitre (benoit7848c(AT)orange.fr), Apr 01 2006

Nest[ Function[l, {Flatten[(l /. {0 -> {0, 1}, 1 -> {0, 1, 1}})]}], {0}, 6] (from Robert G. Wilson v Feb 04 2005)

(PARI) a(n)=-1+floor(n*(1+sqrt(5))/2)-floor((n-1)*(1+sqrt(5))/2) [Cloitre]

Cf. A003849, A096268, A001519. See A005614, A114986 for other versions.

Sequence in context: A093879 A117872 A089809 this_sequence A123640 A022924 A023532

Adjacent sequences: A096267 A096268 A096269 this_sequence A096271 A096272 A096273

nonn,easy

njas, Jun 22 2004

More terms from John W. Layman (layman(AT)math.vt.edu), Jun 29 2004