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%I #6 Apr 03 2013 11:51:43
%S 0,1,4,9,4,8,14,4,17,36,13,7,24,9,36,38,24,23,55,20,28,77,20,25,93,9,
%T 100,29,37,38,78,17,36,120,7,60,99,16,54,67,9,106,184,43,74,153,51,16,
%U 84,34,10,212,12,170,208,26,60,57,8,57,92,53,85,58,148,52
%N Collatz problem with rational negative numbers: number of steps to reach the end of the cycle starting with 1/(2n+1) where n is negative (the initial term is not counted).
%C This variation of the "3x+1" problem with a class of rational negative numbers is as follows: start with any number 1/(2n+1), n= -1, -2, -3, .... If the numerator is even, divide it by 2, otherwise multiply it by 3 and add 1. Do we always reach the end of a cycle with a rational number? It is conjectured that the answer is yes. This sequence is an extension of A210468 with n negative.
%H Michel Lagneau, <a href="/A224299/b224299.txt">Table of n, a(n) for n = 1..10000</a>
%H J. C. Lagarias, <a href="http://pldml.icm.edu.pl:80/mathbwn/element/bwmeta1.element.bwnjournal-article-aav56i1p33bwm?q=bwmeta1.element.bwnjournal-number-aa-1990-56-1&qt=CHILDREN-STATELESS">The set of rational cycles for the 3x+1 problem,</a> Acta Arith. 56 (1990), 33-53.
%e For n = 3, a(3) = 9 because the corresponding trajectory of -1/7 requires 9 iterations to reach the last term of the cycle:
%e -1/7 -> 4/7 -> 2/7 -> 1/7 -> 10/7 -> 5/7 -> 22/7 -> 11/7 -> 40/7 -> 20/7 and 20/7 is the last term because 20/7 -> 10/7 is already in the trajectory.
%t Collatz[n_]:=NestWhileList[If[EvenQ[Numerator[-#]],#/2,3 #+1]&,n,UnsameQ,All];Join[{0},Table[s=Collatz[1/(2 n+1)];len=Length[s]-2;If[s[[-1]]==2,len=len-1];len,{n,-2,-100,-1}]] (* program from T.D. Noe, adapted for this sequence - see A210468 *).
%Y Cf. A210468, A210471.
%K nonn
%O 1,3
%A _Michel Lagneau_, Apr 03 2013
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