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
KEYWORD
nonn
AUTHOR
Michel Lagneau, Apr 03 2013
STATUS
approved