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A134568 a(n) = least m such that {-m*r} > {n*r}, where { } denotes fractional part and r = sqrt(2). 1

%I

%S 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 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 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 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 A079496? Are all a(n) denominators of intermediate convergents to sqrt(2)?

%e a(2)=5 because {-m*r} < {2*r} = 0.828... for m=1,2,3,4 whereas

%e {-5*r} = 0.9289..., so that 5 is the least m for which

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

%Y Cf. A134569.

%K nonn

%O 1,2

%A _Clark Kimberling_, Nov 02 2007

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