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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
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,1
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COMMENTS
| 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 (ck6(AT)evansville.edu), Nov 02 2007
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EXAMPLE
| a(3)=5 because {m*tau}<{3*tau}=.854... for m=1,2,3,4, whereas {-5*tau}=.909..., so that 5 is the least m for which {m*tau}>{3*tau}.
a(3)=5 because {-m*tau}<{3*tau}=.854... for m=1,2,3,4 whereas {-5*tau}=.9289..., so that 5 is the least m for which {-m*tau}>{2*tau}.
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CROSSREFS
| Cf. A134567, A134570, A134571.
Sequence in context: A006556 A108790 A117941 * A128694 A088421 A146024
Adjacent sequences: A134563 A134564 A134565 * A134567 A134568 A134569
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KEYWORD
| nonn
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AUTHOR
| Clark Kimberling (ck6(AT)evansville.edu), Nov 01 2007, Nov 02 2007
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EXTENSIONS
| More terms from Clark Kimberling (ck6(AT)evansville.edu), Nov 02 2007
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