%I #6 Nov 29 2017 03:35:15
%S 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,
%T 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,
%U 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
%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 A001519, the oddindexed Fibonacci numbers. The defining inequality {m*tau} > {n*tau} is equivalent to {m*tau} + {n*tau} < 1.
%C 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
%e 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}.
%e 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}.
%Y Cf. A134567, A134570, A134571.
%K nonn
%O 1,1
%A _Clark Kimberling_, Nov 01 2007, Nov 02 2007
%E More terms from _Clark Kimberling_, Nov 02 2007
