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A110919 Number of consecutive 1's in the continued fraction for floor(n*Phi)/n where Phi=(1+sqrt(5))/2. 0
1, 1, 1, 1, 3, 1, 3, 1, 3, 3, 3, 3, 5, 3, 3, 3, 3, 5, 3, 3, 3, 3, 5, 3, 3, 5, 3, 5, 3, 3, 5, 3, 5, 7, 3, 5, 3, 5, 5, 3, 5, 3, 5, 5, 3, 5, 7, 5, 5, 3, 5, 5, 5, 5, 3, 5, 5, 5, 5, 7, 5, 5, 5, 5, 5, 5, 5, 7, 5, 5, 5, 5, 7, 5, 5, 5, 5, 5, 5, 5, 7, 5, 5, 5, 5, 7, 5, 5, 9, 5, 5, 5, 5, 7, 5, 5, 5, 5, 7, 5, 5, 7, 5, 5, 5 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,5

COMMENTS

Terms are always odd.

LINKS

Table of n, a(n) for n=1..105.

FORMULA

sum(k=1, n, a(k)) seems to be asymptotic to c*n*log(n) with c around 1

EXAMPLE

The continued fraction for floor(128*Phi)/128 is [1, 1, 1, 1, 1, 1, 1, 2, 1, 2] with 7 consecutive 1's, thus a(128)=7

PROG

(PARI) a(n)=if(n<2, 1, s=1; while(component(contfrac(floor(n*(1+sqrt(5))/2)/n), s)==1, s++); s-1)

CROSSREFS

Sequence in context: A025810 A001319 A240833 * A109599 A066839 A176246

Adjacent sequences:  A110916 A110917 A110918 * A110920 A110921 A110922

KEYWORD

nonn

AUTHOR

Benoit Cloitre, Sep 22 2005

STATUS

approved

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Last modified September 2 14:44 EDT 2014. Contains 246361 sequences.