A110919 Number of consecutive 1's in the continued fraction for floor(n*phi)/n where phi = (1+sqrt(5))/2 (A001622).

1, 1, 1, 1, 3, 1, 3, 1, 3, 3, 3, 3, 5, 3, 3, 3, 3, 5, 3, 3, 3, 3, 5, 3, 3, 5, 3, 5, 3, 3, 5, 3, 5, 7, 3, 5, 3, 5, 5, 3, 5, 3, 5, 5, 3, 5, 7, 5, 5, 3, 5, 5, 5, 5, 3, 5, 5, 5, 5, 7, 5, 5, 5, 5, 5, 5, 5, 7, 5, 5, 5, 5, 7, 5, 5, 5, 5, 5, 5, 5, 7, 5, 5, 5, 5, 7, 5, 5, 9, 5, 5, 5, 5, 7, 5, 5, 5, 5, 7, 5, 5, 7, 5, 5, 5

Offset

1,5

Comments

Terms are always odd.

Links

Amiram Eldar, Table of n, a(n) for n = 1..10000

Amiram Eldar, Plot of Sum_{k=1..n} a(n)/(n*log(n)) for n = 100..100000

Formula

Sum_{k=1..n} a(k) seems to be asymptotic to c*n*log(n) with c around 1.

Example

The continued fraction for floor(128*phi)/128 is [1, 1, 1, 1, 1, 1, 1, 2, 1, 2] with 7 consecutive 1's, thus a(128) = 7.

Mathematica

a[1] = 1; a[n_] := -1 + FirstPosition[ContinuedFraction[Floor[n*GoldenRatio]/n], _?(# > 1 &)][[1]]; Array[a, 100] (* Amiram Eldar, May 08 2022 *)

Prog

(PARI) a(n)=if(n<2, 1, s=1; while(component(contfrac(floor(n*(1+sqrt(5))/2)/n), s)==1, s++); s-1)

Crossrefs

Cf. A001622.

Sequence in context: A025810 A001319 A240833 * A281273 A109599 A333752

Adjacent sequences: A110916 A110917 A110918 * A110920 A110921 A110922

Keyword

nonn

Author

Benoit Cloitre, Sep 22 2005

Status

approved