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A159885 For n>=1, let f(2n+1)=(3n+2)/A006519(3n+2) and let f^k be the k-th iteration of f. Then a(n) is the least k such that A000120(f^k(2n+1))<=A000120(n) 9
2, 1, 2, 6, 1, 1, 2, 3, 3, 1, 1, 4, 1, 1, 2, 8, 2, 3, 3, 39, 1, 1, 1, 4, 3, 1, 1, 2, 1, 1, 2, 8, 5, 2, 2, 41, 3, 2, 3, 5, 5, 1, 1, 1, 1, 1, 1, 42, 2, 1, 4, 6, 1, 1, 1, 2, 2, 1, 1, 2, 1, 1, 2, 44, 5, 5, 5, 31, 5, 2, 2, 41, 7, 1, 3, 3, 3, 2, 3, 34, 3, 5, 13, 12, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 42, 8, 1, 2, 4, 1 (list; graph; refs; listen; history; internal format)
OFFSET

1,1

COMMENTS

Conjecture: a(n) exists for every n>=1. It is easy to see that this conjecture is equivalent to the well-known Collatz 3x+1 conjecture.

CROSSREFS

Cf. A006519, A007814, A000120, A160198

Sequence in context: A144824 A144358 A049404 * A178803 A083773 A129116

Adjacent sequences:  A159882 A159883 A159884 * A159886 A159887 A159888

KEYWORD

nonn

AUTHOR

Vladimir Shevelev (shevelev(AT)bgu.ac.il), Apr 25 2009, Apr 27 2009

EXTENSIONS

Edited by N. J. A. Sloane, May 03 2009

a(25) corrected, sequence extended by R. J. Mathar (mathar(AT)strw.leidenuniv.nl), May 15 2009

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Last modified February 17 16:49 EST 2012. Contains 206058 sequences.