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

Author

Vladimir Shevelev, Apr 25 2009, Apr 27 2009

Keywords

• nonn
• look

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

Crossrefs

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

Programs

PARI

A006519(n) = (1<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

Extensions

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

a(25) corrected, sequence extended by R. J. Mathar, May 15 2009