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A071291 Second term of the continued fraction expansion of (3/2)^n; or 0 if no term is present. 1
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.

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.

LINKS

Table of n, a(n) for n=1..95.

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.

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 A068542 A036112

Adjacent sequences:  A071288 A071289 A071290 * A071292 A071293 A071294

KEYWORD

easy,nonn

AUTHOR

Paul D. Hanna, Jun 10 2002

STATUS

approved

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Last modified November 28 10:55 EST 2014. Contains 250323 sequences.