%I #5 Nov 29 2017 03:35:24
%S 2,1,2,1,12,2,1,2,1,7,2,1,2,1,2,1,12,2,1,2,1,7,2,1,2,1,2,1,70,2,1,2,1,
%T 12,2,1,2,1,7,2,1,2,1,2,1,12,2,1,2,1,7,2,1,2,1,2,1,41,2,1,2,1,12,2,1,
%U 2,1,7,2,1,2,1,2,1,12,2,1,2,1,7,2,1,2,1,2,1,12,2,1,2,1,7,2,1,2,1,2,1,70,2
%N a(n) = least m such that {m*r} < {n*r}, where { } denotes fractional part and r = sqrt(2).
%C The defining inequality {m*r} > {n*r} is equivalent to {m*r} + {n*r} < 1. Are all a(n) in A084068? Are all a(n) denominators of intermediate convergents to sqrt(2)?
%e a(3)=2 because {m*r} < {3*r} = 0.2426... for m=1 whereas
%e {2*r} = 0.1715..., so that 2 is the least m for which
%e {m*r} < {3*r}.
%Y Cf. A134568.
%K nonn
%O 1,1
%A _Clark Kimberling_, Nov 02 2007
