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A134566 a(n) = least m such that {-m*tau} > {n*tau}, where { } denotes fractional part and tau = (1 + sqrt(5))/2. 2

%I #6 Nov 29 2017 03:35:15

%S 2,1,5,2,1,2,1,13,2,1,5,2,1,2,1,5,2,1,2,1,34,2,1,5,2,1,2,1,13,2,1,5,2,

%T 1,2,1,5,2,1,2,1,13,2,1,5,2,1,2,1,5,2,1,2,1,89,2,1,5,2,1,2,1,13,2,1,5,

%U 2,1,2,1,5,2,1,2,1,34,2,1,5,2,1,2,1,13,2,1,5,2,1,2,1,5,2,1,2,1,13,2,1,5,2,1

%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 A001519, the odd-indexed Fibonacci numbers. The defining inequality {-m*tau} > {n*tau} is equivalent to {-m*tau} + {n*tau} < 1.

%C The terms belong to A001519, the odd-indexed Fibonacci numbers. The defining inequality {-m*tau} > {n*tau} is equivalent to {m*tau} + {n*tau} < 1. - _Clark Kimberling_, Nov 02 2007

%e a(3)=5 because {m*tau} < {3*tau} = 0.854... for m=1,2,3,4, whereas {-5*tau} = 0.909..., so that 5 is the least m for which {m*tau} > {3*tau}.

%e a(3)=5 because {-m*tau} < {3*tau} = 0.854... for m=1,2,3,4 whereas {-5*tau} = 0.9289..., so that 5 is the least m for which {-m*tau} > {2*tau}.

%Y Cf. A134567, A134570, A134571.

%K nonn

%O 1,1

%A _Clark Kimberling_, Nov 01 2007, Nov 02 2007

%E More terms from _Clark Kimberling_, Nov 02 2007

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