%I A119733
%S A119733 0,1,2,5,4,7,10,19,8,11,14,23,20,29,38,65,16,19,22,31,28,37,46,73,40,49,
%T A119733 58,85,76,103,130,211,32,35,38,47,44,53,62,89,56,65,74,101,92,119,146,
%U A119733 227,80,89,98,125,116,143,170,251,152,179,206,287,260,341,422,665,64,67
%N A119733 Offsets of the terms of the nodes of the reverse Collatz function.
%C A119733 Create a binary tree starting with x. To follow 0 from the root, apply 
               f(x)=2x. To follow 1, apply g(x)=(2x-1)/3. For example, starting 
               with x, the string 010 {also known as f(g(f(x))) }, you would get 
               (8x-2)/3. These expressions represent the reverse Collatz function 
               and will provide numbers who's Collatz path may include x. These 
               expressions will all be of the form (2^a*x-b)/3^c. This sequence 
               concerns b. What makes b interesting is that if you draw the tree, 
               each level of the tree will have the same sequence of values for 
               b. The root of the tree x, can be written as (2^0*x-0)/3^0, which 
               has the first value for b. Each subsequent level contains twice as 
               many values of b.
%F A119733 a(0)=0, a(2i)=2*a(i)+3^[the number 1's in the binary representation of 
               i], a(2i-1)=2*a(i)
%e A119733 a(2)=1=2*0+3^0 since 0 written in binary contains no 1's.
%o A119733 # perl, sorry :-( # call with n to get 2^n values $depth=shift; sub funct 
               { my ($i, $b, $c) = @_; if ($i < $depth) { funct($i+1, $b*2, $c); 
               funct($i+1, 2*$b+$c, $c*3); } else { print "$b, "; } } funct(0, 0, 
               1); print " ";
%Y A119733 Sequence in context: A154811 A036237 A015948 this_sequence A111570 A057954 
               A155896
%Y A119733 Adjacent sequences: A119730 A119731 A119732 this_sequence A119734 A119735 
               A119736
%K A119733 nonn,uned
%O A119733 0,3
%A A119733 William Entriken (att(AT)phor.net), Jun 14 2006