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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.
LINKS
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(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
Sequence in context: A241191 A352232 A221195 * A049330 A274040 A367948
KEYWORD
easy,nonn
AUTHOR
Paul D. Hanna, Jun 10 2002
STATUS
approved

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Last modified March 28 17:42 EDT 2024. Contains 371254 sequences. (Running on oeis4.)