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 A134567 a(n) = least m such that {-m*tau} < {n*tau}, where { } denotes fractional part and tau = (1 + sqrt(5))/2. 2

%I #5 Nov 29 2017 03:35:18

%S 1,3,1,1,8,1,3,1,1,3,1,1,21,1,3,1,1,8,1,3,1,1,3,1,1,8,1,3,1,1,3,1,1,

%T 55,1,3,1,1,8,1,3,1,1,3,1,1,21,1,3,1,1,8,1,3,1,1,3,1,1,8,1,3,1,1,3,1,

%U 1,21,1,3,1,1,8,1,3,1,1,3,1,1,8,1,3,1,1,3,1,1,144,1,3,1,1,8,1,3,1,1,3,1,1,21

%N a(n) = least m such that {-m*tau} < {n*tau}, where { } denotes fractional part and tau = (1 + sqrt(5))/2.

%C The terms are members of A001906, the even-indexed Fibonacci numbers. The defining inequality {-m*tau} < {n*tau} is equivalent to {m*tau} + {n*tau} > 1.

%e a(2)=3 because {-m*tau} > {2*tau} = 0.236... for m=1,2, whereas {-3*tau} = 0.145..., so that 3 is the least m for which {-m*tau} < {3*tau}.

%Y Cf. A134566, A134570, A134571.

%K nonn

%O 1,2

%A _Clark Kimberling_, Nov 01 2007

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