0,3

Here's how to form the lazy Fibonacci representation of n>=1. First, define g(n) to be the Fibonacci number F(k-1), where k is the number satisfying F(k)-1 <= n <= F(k+1)-2. Then let g(1)=g(n), g(2)=g(n-g(1)), g(3)=g(n-g(1)-g(2)) and so on, until reaching h for which g(h) is 1 or 2. The desired representation is n = g(1)+g(2)+...+g(h).

P. Erdos and I. Joo, "On the Expansion of 1 = Sum{q^(-n_i)}," Period. Math. Hung. 23 (1991), no. 1, 25-28. (This paper introduces lazy Fibonacci representations.)

Vienna University of Technology, The Joint Distribution of Greedy and Lazy Fibonacci Expansions.

1, 1, then F(3) 2's, then F(4) 3's, then F(5) 4's, ..., then F(k+1) k's, ...

a(0)=a(1)=1 then a(n)=a(floor(n/tau))+1 where tau=(1+sqrt(5))/2 - Benoit Cloitre (benoit7848c(AT)orange.fr), Dec 17 2006

a(n) = least k such that f^(k)(n)=0 where f^(k+1)(x)=f(f^(k)(x)) and f(x)=floor(x/Phi) where Phi=(1+sqrt(5))/2 (see pari-gp program) - Benoit Cloitre (benoit7848c(AT)orange.fr), May 24 2007

The lazy Fibonacci representation of 14 is 8+3+2+1, which in binary notation is 10111, which consists of 5 digits.

(PARI) a(n)=if(n<2, 1, a(floor(n*(-1+sqrt(5))/2))+1) - Benoit Cloitre (benoit7848c(AT)orange.fr), Dec 17 2006

(PARI) a(n)=if(n<0, 0, c=1; s=n; while(floor(s*2/(1+sqrt(5)))>0, c++; s=floor(s*2/(1+sqrt(5)))); c) - Benoit Cloitre (benoit7848c(AT)orange.fr), May 24 2007

Cf. A000045, A072649, A095792.

Sequence in context: A085727 A143442 A137300 this_sequence A036042 A162988 A143824

Adjacent sequences: A095788 A095789 A095790 this_sequence A095792 A095793 A095794

nonn

Clark Kimberling (ck6(AT)evansville.edu), Jun 05 2004