A134567 a(n) = least m such that {-m*tau} < {n*tau}, where { } denotes fractional part and tau = (1 + sqrt(5))/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

Offset: 1

Author

Clark Kimberling, Nov 01 2007

Keywords

• nonn

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.

Examples

			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

Cf. A134566, A134570, A134571.