

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
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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. 192199.
C. Pisot, La repartition modulo 1 et les nombres algebriques, Ann. Scuola Norm. Sup. Pisa 7 (1938) 205248.
T. Vijayaraghavan, On the fractional parts of the powers of a number (I), J. London Math. Soc. 15 (1940) 159160.


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



