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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
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, 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, 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 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,2
COMMENTS
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.
LINKS
EXAMPLE
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}.
CROSSREFS
Sequence in context: A049290 A297191 A147990 * A131932 A016462 A348693
KEYWORD
nonn
AUTHOR
Clark Kimberling, Nov 01 2007
STATUS
approved

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Last modified April 20 09:04 EDT 2024. Contains 371799 sequences. (Running on oeis4.)