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A134569 a(n) = least m such that {-m*r} < {n*r}, where { } denotes fractional part and r = sqrt(2).

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%I A134569 #5 Nov 29 2017 03:35:24
%S A134569 2,1,2,1,12,2,1,2,1,7,2,1,2,1,2,1,12,2,1,2,1,7,2,1,2,1,2,1,70,2,1,2,1,
%T A134569 12,2,1,2,1,7,2,1,2,1,2,1,12,2,1,2,1,7,2,1,2,1,2,1,41,2,1,2,1,12,2,1,
%U A134569 2,1,7,2,1,2,1,2,1,12,2,1,2,1,7,2,1,2,1,2,1,12,2,1,2,1,7,2,1,2,1,2,1,70,2
%N A134569 a(n) = least m such that {-m*r} < {n*r}, where { } denotes fractional part and r = sqrt(2).
%C A134569 The defining inequality {-m*r} > {n*r} is equivalent to {m*r} + {n*r} < 1. Are all a(n) in A084068? Are all a(n) denominators of intermediate convergents to sqrt(2)?
%e A134569 a(3)=2 because {-m*r} < {3*r} = 0.2426... for m=1 whereas
%e A134569 {-2*r} = 0.1715..., so that 2 is the least m for which
%e A134569 {-m*r} < {3*r}.
%Y A134569 Cf. A134568.
%K A134569 nonn
%O A134569 1,1
%A A134569 _Clark Kimberling_, Nov 02 2007