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A134566
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a(n) = least m such that {-m*tau} > {n*tau}, where { } denotes fractional part and tau = (1 + sqrt(5))/2.
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2
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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, 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, 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
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OFFSET
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1,1
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COMMENTS
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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.
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
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LINKS
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EXAMPLE
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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}.
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}.
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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