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 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). 2
 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 (list; graph; refs; listen; history; text; internal format)
 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 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: A252867 A194356 A227317 * A175467 A188525 A324503 Adjacent sequences:  A224297 A224298 A224299 * A224301 A224302 A224303 KEYWORD nonn AUTHOR Michel Lagneau, Apr 03 2013 STATUS approved

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Last modified February 26 12:33 EST 2020. Contains 332279 sequences. (Running on oeis4.)