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).

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

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.

Links

Antti Karttunen, Table of n, a(n) for n = 1..65537

Index entries for sequences related to 3x+1 (or Collatz) problem

Prog

(PARI)
A006519(n) = (1<<valuation(n, 2));
f(n) = ((3*((n-1)/2))+2)/A006519((3*((n-1)/2))+2); \\ Defined only for odd n. Cf. A075677.
A159885(n) = { my(w=hammingweight(n), n = (n+n+1)); for(k=1, oo, n = f(n); if(hammingweight(n) <= w, return(k))); }; \\ Antti Karttunen, Sep 22 2018

Crossrefs

Cf. A006519, A007814, A000120, A159945, A160198, A160266.

Sequence in context: A364513 A144358 A049404 * A178803 A292901 A083773

Adjacent sequences: A159882 A159883 A159884 * A159886 A159887 A159888

Keyword

nonn,look

Author

Vladimir Shevelev, 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, May 15 2009

Status

approved