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Search: id:A110919
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| A110919 |
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Number of consecutive 1's in the continued fraction for floor(n*Phi)/n where Phi=(1+sqrt(5))/2. |
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+0
1
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| 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; listen)
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OFFSET
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1,5
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COMMENT
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Terms are always odd.
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FORMULA
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sum(k=1, n, a(k)) seems to be asymptotic to c*n*log(n) with c around 1
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EXAMPLE
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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
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PROGRAM
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(PARI) a(n)=if(n<2, 1, s=1; while(component(contfrac(floor(n*(1+sqrt(5))/2)/n), s)==1, s++); s-1)
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CROSSREFS
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Adjacent sequences: A110916 A110917 A110918 this_sequence A110920 A110921 A110922
Sequence in context: A087822 A025810 A001319 this_sequence A109599 A066839 A046933
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KEYWORD
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nonn
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AUTHOR
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Benoit Cloitre (benoit7848c(AT)orange.fr), Sep 22 2005
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