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A210471 Collatz (3x+1) problem with rational numbers: number of steps to reach the end of the cycle starting with 1/(2n+1).

%I A210471 #12 Jan 22 2013 19:32:37
%S A210471 1,3,4,7,6,8,5,18,9,32,11,77,25,9,6,36,29,18,17,12,28,14,23,45,73,55,
%T A210471 91,16,17,39,7,36,40,114,87,100,93,34,54,64,14,55,171,80,57,72,42,108,
%U A210471 24,12,97,68,31,159,88,10,41,50,23,117,63,61,8,55,72,45,68
%N A210471 Collatz (3x+1) problem with rational numbers: number of steps to reach the end of the cycle starting with 1/(2n+1).
%C A210471 In this sequence, the initial value is counted and a(n) = A210468(n) + 1.
%C A210471 This variation of the "3x+1" problem with a class of rational numbers is as follows: start with any number 1/(2n+1). 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.
%H A210471 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&amp;qt=CHILDREN-STATELESS">The set of rational cycles for the 3x+1 problem,</a> Acta Arith. 56 (1990), 33-53.
%F A210471 a(n) = A210468(n) + 1.
%t A210471 Collatz[n_]:=NestWhileList[If[EvenQ[Numerator[#]], #/2, 3 #+1]&, n, UnsameQ, All]; Join[{1}, Table[s=Collatz[1/(2*n+1)]; len=Length[s]-1; If[s[[-1]]==2, len=len-1]; len, {n,100}]]
%Y A210471 Cf. A210468.
%K A210471 nonn
%O A210471 0,2
%A A210471 _Michel Lagneau_, Jan 22 2013