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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; 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)?

EXAMPLE

a(2)=5 because {-m*r}<{2*r}=.828... for m=1,2,3,4 whereas

{-5*r}=.9289..., so that 5 is the least m for which

{-m*r}>{2*r}.

CROSSREFS

Cf. A134569.

Sequence in context: A201526 A091384 A011305 * A198798 A121267 A111740

Adjacent sequences:  A134565 A134566 A134567 * A134569 A134570 A134571

KEYWORD

nonn

AUTHOR

Clark Kimberling (ck6(AT)evansville.edu), Nov 02 2007

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Last modified February 17 04:45 EST 2012. Contains 205984 sequences.