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 A071353 First term of the continued fraction expansion of (3/2)^n. 2
 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 (list; graph; refs; listen; history; text; internal format)
 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 A290645 A152878 Adjacent sequences:  A071350 A071351 A071352 * A071354 A071355 A071356 KEYWORD easy,nonn AUTHOR Paul D. Hanna, Jun 10 2002 STATUS approved

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Last modified June 20 06:56 EDT 2021. Contains 345157 sequences. (Running on oeis4.)