A134566 a(n) = least m such that {-m*tau} > {n*tau}, where { } denotes fractional part and tau = (1 + sqrt(5))/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

Offset: 1

Author

Clark Kimberling, Nov 01 2007, Nov 02 2007

Keywords

• nonn

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, Nov 02 2007

Examples

			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

Cf. A134567, A134570, A134571.

Extensions

More terms from Clark Kimberling, Nov 02 2007