A134568 a(n) = least m such that {-m*r} > {n*r}, where { } denotes fractional part and r = sqrt(2).

1, 5, 1, 3, 1, 1, 5, 1, 3, 1, 1, 29, 1, 5, 1, 3, 1, 1, 5, 1, 3, 1, 1, 17, 1, 5, 1, 3, 1, 1, 5, 1, 3, 1, 1, 5, 1, 3, 1, 1, 29, 1, 5, 1, 3, 1, 1, 5, 1, 3, 1, 1, 17, 1, 5, 1, 3, 1, 1, 5, 1, 3, 1, 1, 5, 1, 3, 1, 1, 169, 1, 5, 1, 3, 1, 1, 5, 1, 3, 1, 1, 29, 1, 5, 1, 3, 1, 1, 5, 1, 3, 1, 1, 17, 1, 5, 1, 3, 1, 1, 5

Offset

1,2

Comments

The defining inequality {-m*r} < {n*r} is equivalent to {m*r} + {n*r} > 1. Are all a(n) in A079496? Are all a(n) denominators of intermediate convergents to sqrt(2)?

Links

Table of n, a(n) for n=1..101.

Example

a(2)=5 because {-m*r} < {2*r} = 0.828... for m=1,2,3,4 whereas
{-5*r} = 0.9289..., so that 5 is the least m for which
{-m*r} > {2*r}.

Crossrefs

Cf. A134569.

Sequence in context: A011305 A348317 A254378 * A198798 A229181 A121267

Adjacent sequences: A134565 A134566 A134567 * A134569 A134570 A134571

Keyword

nonn

Author

Clark Kimberling, Nov 02 2007

Status

approved