A071353 First term of the continued fraction expansion of (3/2)^n.

2, 4, 2, 16, 1, 2, 11, 1, 2, 1, 2, 1, 1, 1, 1, 1, 3, 1, 1, 3, 1, 1, 1, 8, 5, 1, 7, 1, 25, 16, 1, 1, 1, 1, 1, 2, 1, 1, 1, 3, 2, 1, 1, 1, 3, 1, 1, 2, 7, 4, 3, 2, 4, 1, 3, 1, 3, 1, 1, 1, 2, 10, 1, 2, 4, 1, 4, 2, 1, 3, 2, 14, 9, 6, 1, 11, 1, 1, 2, 1, 1, 2, 6, 1, 12, 1, 1, 2, 1, 2, 19, 12, 8, 1, 89, 59, 1, 3

Offset

1,1

Comments

If uniformly distributed, then the average of the reciprocal terms of this sequence is 1/2.

"Pisot and Vijayaraghavan proved that (3/2)^n has infinitely many accumulation points, i.e. infinitely many convergent subsequences with distinct limits. The sequence is believed to be uniformly distributed, but no one has even proved that it is dense in [0,1]." - S. R. Finch.

References

S. R. Finch, Mathematical Constants, Cambridge, 2003, pp. 192-199.

Links

G. C. Greubel, Table of n, a(n) for n = 1..1000

Steven R. Finch, Powers of 3/2 Modulo One [From Steven Finch, Apr 20 2019]

Steven R. Finch, Non-Ideal Waring's Problem [From Steven Finch, Apr 20 2019]

Jeff Lagarias, 3x+1 Problem

C. Pisot, La répartition modulo 1 et les nombres algébriques, Ann. Scuola Norm. Sup. Pisa, 7 (1938), 205-248.

T. Vijayaraghavan, On the fractional parts of the powers of a number (I), J. London Math. Soc. 15 (1940) 159-160.

Formula

a(n) = floor(1/frac((3/2)^n)).

Example

a(7) = 11 since floor(1/frac(3^7/2^7)) = floor(1/.0859375) = 11.

Mathematica

Table[Floor[1/FractionalPart[(3/2)^n]], {n, 1, 100}] (* G. C. Greubel, Apr 18 2017 *)

Prog

(PARI) a(n) = contfrac((3/2)^n)[2] \\ Michel Marcus, Aug 01 2013

Crossrefs

Cf. A055500, A071291.

Sequence in context: A113539 A215055 A152877 * A134763 A370135 A290645

Adjacent sequences: A071350 A071351 A071352 * A071354 A071355 A071356

Keyword

easy,nonn

Author

Paul D. Hanna, Jun 10 2002

Status

approved