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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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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
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OFFSET
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1,9
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COMMENTS
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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.
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LINKS
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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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MATHEMATICA
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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 *)
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PROG
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(PARI) a(n) = {cf = contfrac((3/2)^n); if (#cf < 3, return (0), return (cf[3])); } \\Michel Marcus, Aug 01 2013
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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STATUS
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approved
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