A134569 a(n) = least m such that {-m*r} < {n*r}, where { } denotes fractional part and r = sqrt(2).

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, 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, 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

Offset

1,1

Comments

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)?

Links

Table of n, a(n) for n=1..100.

Example

a(3)=2 because {-m*r} < {3*r} = 0.2426... for m=1 whereas
{-2*r} = 0.1715..., so that 2 is the least m for which
{-m*r} < {3*r}.

Crossrefs

Cf. A134568.

Sequence in context: A179508 A134304 A211096 * A295853 A287541 A288196

Adjacent sequences: A134566 A134567 A134568 * A134570 A134571 A134572

Keyword

nonn

Author

Clark Kimberling, Nov 02 2007

Status

approved