A071915 Number of 1's in the continued fraction expansion of (3/2)^n.

0, 0, 1, 0, 2, 3, 3, 6, 3, 5, 1, 2, 8, 2, 3, 5, 2, 3, 3, 6, 10, 8, 6, 4, 2, 3, 6, 5, 2, 9, 12, 7, 17, 10, 7, 9, 8, 10, 13, 13, 10, 12, 14, 9, 11, 10, 11, 6, 9, 5, 3, 13, 13, 19, 18, 13, 8, 12, 15, 14, 18, 7, 19, 19, 17, 15, 13, 14, 16, 13, 20, 16, 10, 20, 25, 17, 19, 14, 19, 14, 18, 22

Offset

1,5

Comments

It seems that lim n ->infinity a(n)/n = 0.2... << (log(4)-log(3))/log(2) = 0.415... the expected density of 1's (cf. measure theory of continued fraction).

Links

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

Example

The continued fraction of (3/2)^24 is [16834, 8, 1, 10, 2, 25, 1, 3, 1, 1, 57, 6] which contains 4 "1's", hence a(24)=4.

Mathematica

a[1] = 0; a[n_] := Count[ContinuedFraction[(3/2)^n], 1]; Array[a, 100] (* Amiram Eldar, Sep 05 2020 *)

Prog

(PARI) for(n=1, 200, s=contfrac(frac((3/2)^n)); print1(sum(i=1, length(s), if(1-component(s, i), 0, 1)), ", "))

Crossrefs

Cf. A071337, A071316, A071315, A071529.

Sequence in context: A104570 A103356 A345385 * A182578 A021432 A274824

Adjacent sequences: A071912 A071913 A071914 * A071916 A071917 A071918

Keyword

easy,nonn

Author

Benoit Cloitre, Jun 13 2002

Status

approved