

A134566


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


2



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

1,1


COMMENTS

The terms are members of A001519, the oddindexed Fibonacci numbers. The defining inequality {m*tau} > {n*tau} is equivalent to {m*tau} + {n*tau} < 1.
The terms belong to A001519, the oddindexed Fibonacci numbers. The defining inequality {m*tau} > {n*tau} is equivalent to {m*tau} + {n*tau} < 1.  Clark Kimberling, Nov 02 2007


LINKS



EXAMPLE

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}.


CROSSREFS



KEYWORD

nonn


AUTHOR



EXTENSIONS



STATUS

approved



