A224300 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).

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, 10, 101, 30, 38, 39, 79, 18, 37, 121, 8, 61, 100, 17, 55, 68, 11, 107, 185, 44, 75, 154, 52, 17, 85, 35, 11, 213, 13, 171, 209, 27, 61, 58, 9, 58, 93, 54, 86, 59, 149

Offset

1,2

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 A210471 with n negative.

Links

Table of n, a(n) for n=1..65.

J. C. Lagarias, The set of rational cycles for the 3x+1 problem, Acta Arith. 56 (1990), 33-53.

Formula

a(n) = A224299(n) + 1.

Example

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: -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+1, {n, -2, -100, -1}]] (* program from T. D. Noe, adapted for this sequence - see A210471 *)

Crossrefs

Cf. A210468, A210471, A224299.

Sequence in context: A352199 A194356 A227317 * A175467 A361806 A188525

Adjacent sequences: A224297 A224298 A224299 * A224301 A224302 A224303

Keyword

nonn

Author

Michel Lagneau, Apr 03 2013

Status

approved