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%I #6 Apr 03 2013 11:52:39
%S 1,2,5,10,5,9,15,5,18,37,14,8,25,10,37,39,25,24,56,21,29,78,21,26,94,
%T 10,101,30,38,39,79,18,37,121,8,61,100,17,55,68,11,107,185,44,75,154,
%U 52,17,85,35,11,213,13,171,209,27,61,58,9,58,93,54,86,59,149
%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 value is 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 A210471 with n negative.
%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.
%F a(n) = A224299(n) + 1.
%e For n = 3, a(3) = 10 because the corresponding trajectory of -1/7 requires 10 iterations (the first term -1/7 is counted) 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+1,{n,-2,-100,-1}]] (* program from T. D. Noe, adapted for this sequence - see A210471 *).
%Y Cf. A210468, A210471, A224299.
%K nonn
%O 1,2
%A _Michel Lagneau_, Apr 03 2013
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