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

Author

Paul D. Hanna, Jun 10 2002

Keywords

• easy
• nonn

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.

Examples

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

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.

Crossrefs

Cf. A055500, A071291.

Programs

Mathematica

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

PARI

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

Formula

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