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A134568 a(n) = least m such that {-m*r} > {n*r}, where { } denotes fractional part and r = sqrt(2). 1
1, 5, 1, 3, 1, 1, 5, 1, 3, 1, 1, 29, 1, 5, 1, 3, 1, 1, 5, 1, 3, 1, 1, 17, 1, 5, 1, 3, 1, 1, 5, 1, 3, 1, 1, 5, 1, 3, 1, 1, 29, 1, 5, 1, 3, 1, 1, 5, 1, 3, 1, 1, 17, 1, 5, 1, 3, 1, 1, 5, 1, 3, 1, 1, 5, 1, 3, 1, 1, 169, 1, 5, 1, 3, 1, 1, 5, 1, 3, 1, 1, 29, 1, 5, 1, 3, 1, 1, 5, 1, 3, 1, 1, 17, 1, 5, 1, 3, 1, 1, 5 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,2
COMMENTS
The defining inequality {-m*r} < {n*r} is equivalent to {m*r} + {n*r} > 1. Are all a(n) in A079496? Are all a(n) denominators of intermediate convergents to sqrt(2)?
LINKS
EXAMPLE
a(2)=5 because {-m*r} < {2*r} = 0.828... for m=1,2,3,4 whereas
{-5*r} = 0.9289..., so that 5 is the least m for which
{-m*r} > {2*r}.
CROSSREFS
Cf. A134569.
Sequence in context: A011305 A348317 A254378 * A198798 A229181 A121267
KEYWORD
nonn
AUTHOR
Clark Kimberling, Nov 02 2007
STATUS
approved

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Last modified May 26 16:43 EDT 2024. Contains 372840 sequences. (Running on oeis4.)