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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 Adjacent sequences: A134565 A134566 A134567 * A134569 A134570 A134571 KEYWORD nonn AUTHOR Clark Kimberling, Nov 02 2007 STATUS approved

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Last modified December 6 15:38 EST 2022. Contains 358644 sequences. (Running on oeis4.)