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Search: id:A071353
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| A071353 |
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First term of the continued fraction expansion of (3/2)^n. |
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+0
2
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| 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; listen)
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OFFSET
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1,1
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COMMENT
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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]." - Steve Finch.
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REFERENCES
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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.
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LINKS
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S. R. Finch, Powers of 3/2
Jeff Lagarias, 3x+1 Problem
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FORMULA
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a(n) = floor(1/frac((3/2)^n))
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EXAMPLE
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a(7) = 11 since floor(1/frac(3^7/2^7)) = floor(1/.0859375) = 11
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CROSSREFS
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Cf. A055500, A071291.
Sequence in context: A006018 A062867 A113539 this_sequence A134763 A100944 A059890
Adjacent sequences: A071350 A071351 A071352 this_sequence A071354 A071355 A071356
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KEYWORD
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easy,nonn
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AUTHOR
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Paul D. Hanna (pauldhanna(AT)juno.com), Jun 10 2002
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