%I #5 Nov 29 2017 03:35:18
%S 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,
%T 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,
%U 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
%N a(n) = least m such that {m*tau} < {n*tau}, where { } denotes fractional part and tau = (1 + sqrt(5))/2.
%C The terms are members of A001906, the evenindexed Fibonacci numbers. The defining inequality {m*tau} < {n*tau} is equivalent to {m*tau} + {n*tau} > 1.
%e 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}.
%Y Cf. A134566, A134570, A134571.
%K nonn
%O 1,2
%A _Clark Kimberling_, Nov 01 2007
