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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)

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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; listen)

	OFFSET	`1,1`
	COMMENT	`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 this_sequence A083773 A129116 A096179` `Adjacent sequences: A159882 A159883 A159884 this_sequence 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 November 25 20:09 EST 2009. Contains 167514 sequences.

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