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 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). 5
 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 (list; graph; refs; listen; history; text; internal format)
 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

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Last modified September 14 03:32 EDT 2024. Contains 375911 sequences. (Running on oeis4.)