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Search: id:A071291
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| A071291 |
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Second term of the continued fraction expansion of (3/2)^n; or 0 if no term is present. |
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+0
2
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| 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; listen)
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OFFSET
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1,9
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COMMENT
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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
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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(1/frac((3/2)^n))) when frac(1/frac((3/2)^n)) > 0; a(n) = 0 otherwise.
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EXAMPLE
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a(9) = 3 since floor(1/frac(1/frac(3^9/2^9))) = floor(1/frac(1/.443359375)) = 3
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CROSSREFS
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Cf. A006543, A071353.
Adjacent sequences: A071288 A071289 A071290 this_sequence A071292 A071293 A071294
Sequence in context: A095988 A082525 A016482 this_sequence A049330 A068542 A036112
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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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