

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
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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++); s1)


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



