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A134568 a(n) = least m such that {-m*r} > {n*r}, where { } denotes fractional part and r = sqrt(2).

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%I A134568 #5 Nov 29 2017 03:35:21
%S A134568 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,
%T A134568 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,
%U A134568 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
%N A134568 a(n) = least m such that {-m*r} > {n*r}, where { } denotes fractional part and r = sqrt(2).
%C A134568 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)?
%e A134568 a(2)=5 because {-m*r} < {2*r} = 0.828... for m=1,2,3,4 whereas
%e A134568 {-5*r} = 0.9289..., so that 5 is the least m for which
%e A134568 {-m*r} > {2*r}.
%Y A134568 Cf. A134569.
%K A134568 nonn
%O A134568 1,2
%A A134568 _Clark Kimberling_, Nov 02 2007