A224299 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).

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, 100, 29, 37, 38, 78, 17, 36, 120, 7, 60, 99, 16, 54, 67, 9, 106, 184, 43, 74, 153, 51, 16, 84, 34, 10, 212, 12, 170, 208, 26, 60, 57, 8, 57, 92, 53, 85, 58, 148, 52

Offset: 1

Author

Michel Lagneau, Apr 03 2013

Keywords

• nonn

Comments

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.

Examples

			For n = 3, a(3) = 9 because the corresponding trajectory of -1/7 requires 9 iterations to reach the last term of the cycle: -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.
		

Links

• Michel Lagneau, Table of n, a(n) for n = 1..10000
• J. C. Lagarias, The set of rational cycles for the 3x+1 problem, Acta Arith. 56 (1990), 33-53.

Crossrefs

Cf. A210468, A210471.

Programs

Mathematica

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 *)