%I A050603
%S A050603 1,1,2,2,1,1,3,3,1,1,2,2,1,1,4,4,1,1,2,2,1,1,3,3,1,1,2,2,1,1,5,5,1,1,2,
               2,1,
%T A050603 1,3,3,1,1,2,2,1,1,4,4,1,1,2,2,1,1,3,3,1,1,2,2,1,1,6,6,1,1,2,2,1,1,3,3,
               1,1,
%U A050603 2,2,1,1,4,4,1,1,2,2,1,1,3,3,1,1,2,2,1,1,5,5,1,1,2,2,1,1,3,3,1,1,2,2,1,
               1,4
%N A050603 Column 2 of A050600: a(n) = add1c(n,2).
%C A050603 Absolute values of A094267.
%C A050603 Consider the Collatz (or 3x+1) problem and the iterative sequence c(k) 
               where c(0)=n is a positive integer and c(k+1)=c(k)/2 if c(k) is even, 
               c(k+1)=(3*c(k)+1)/2 if c(k) is odd. Then a(n) is the minimum number 
               of iterations in order to have c(a(n)) odd if n is even or c(a(n)) 
               even if n is odd. - Benoit Cloitre (benoit7848c(AT)orange.fr), Nov 
               16 2001
%F A050603 Equals A053398(2, n).
%F A050603 G.f.: (1+x)/x^2 * Sum(k>=1, x^(2^k)/(1-x^(2^k))). - Ralf Stephan (ralf(AT)ark.in-berlin.de), 
               Apr 12 2002
%F A050603 a(n) = A136480(n+1). - Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), 
               Dec 31 2007
%Y A050603 Bisection gives column 1 of A050600: A001511.
%Y A050603 a(n)=A007814(n+1)+A007814(n+2).
%Y A050603 Sequence in context: A003638 A094267 A136480 this_sequence A037162 A027358 
               A155092
%Y A050603 Adjacent sequences: A050600 A050601 A050602 this_sequence A050604 A050605 
               A050606
%K A050603 nonn
%O A050603 0,3
%A A050603 Antti Karttunen Jun 22 1999