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A134569 a(n) = least m such that {-m*r} < {n*r}, where { } denotes fractional part and r = sqrt(2). 1

%I #5 Nov 29 2017 03:35:24

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

%C 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 a(3)=2 because {-m*r} < {3*r} = 0.2426... for m=1 whereas

%e {-2*r} = 0.1715..., so that 2 is the least m for which

%e {-m*r} < {3*r}.

%Y Cf. A134568.

%K nonn

%O 1,1

%A _Clark Kimberling_, Nov 02 2007

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