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 A071291 Second term of the continued fraction expansion of (3/2)^n; or 0 if no term is present. 2
 0, 0, 1, 0, 1, 1, 1, 1, 3, 1, 101, 2, 1, 13, 8, 5, 1, 8, 5, 1, 7, 4, 2, 1, 1, 3, 1, 2, 3, 1, 1, 7, 4, 2, 1, 2, 1, 11, 7, 8, 12, 2, 1, 6, 4, 30, 19, 2, 129, 1, 8, 13, 2, 5, 1, 7, 5, 32, 21, 13, 1, 14, 1, 8, 1, 4, 2, 1, 1, 1, 1, 1, 1, 1, 2, 3, 1, 18, 2, 1, 20, 3, 1, 2, 1, 1, 12, 1, 1, 1, 2, 1, 1, 2, 1 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,9 COMMENTS What is the average of the reciprocal terms of this sequence? "Pisot and Vijayaraghavan proved that frac((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. C. Pisot, La repartition modulo 1 et les nombres algebriques, 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. LINKS G. C. Greubel, Table of n, a(n) for n = 1..1000 S. R. Finch, Powers of 3/2 Jeff Lagarias, 3x+1 Problem FORMULA a(n) = floor(1/frac(1/frac((3/2)^n))) when frac(1/frac((3/2)^n)) > 0; a(n) = 0 otherwise. EXAMPLE a(9) = 3 since floor(1/frac(1/frac(3^9/2^9))) = floor(1/frac(1/.443359375)) = 3. MATHEMATICA a[n_] := If[FractionalPart[1/FractionalPart[(3/2)^n]] > 0, Floor[1/FractionalPart[1/FractionalPart[(3/2)^n]]], 0]; Table[a[n], {n, 1, 100}] (* G. C. Greubel, Apr 18 2017 *) PROG (PARI) a(n) = {cf = contfrac((3/2)^n); if (#cf < 3, return (0), return (cf[3])); } \\Michel Marcus, Aug 01 2013 CROSSREFS Cf. A006543, A071353. Sequence in context: A012854 A241191 A221195 * A049330 A274040 A266363 Adjacent sequences:  A071288 A071289 A071290 * A071292 A071293 A071294 KEYWORD easy,nonn AUTHOR Paul D. Hanna, Jun 10 2002 STATUS approved

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