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

1,3

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.

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.

Example

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.

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 *).

Crossrefs

Cf. A210468, A210471.

Sequence in context: A068950 A021673 A372912 * A141653 A336051 A071793

Adjacent sequences: A224296 A224297 A224298 * A224300 A224301 A224302

Keyword

nonn

Author

Michel Lagneau, Apr 03 2013

Status

approved