A159885 - OEIS

%I #14 Sep 23 2018 20:58:05

%S 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,

%T 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,

%U 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

%N 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).

%C 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.

%H Antti Karttunen, <a href="/A159885/b159885.txt">Table of n, a(n) for n = 1..65537</a>

%H <a href="/index/3#3x1">Index entries for sequences related to 3x+1 (or Collatz) problem</a>

%o (PARI)

%o A006519(n) = (1<<valuation(n, 2));

%o f(n) = ((3*((n-1)/2))+2)/A006519((3*((n-1)/2))+2); \\ Defined only for odd n. Cf. A075677.

%o 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

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

%K nonn,look

%O 1,1

%A _Vladimir Shevelev_, Apr 25 2009, Apr 27 2009

%E Edited by _N. J. A. Sloane_, May 03 2009

%E a(25) corrected, sequence extended by _R. J. Mathar_, May 15 2009