A224299 - OEIS

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, 10, 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

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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

Falk Hüffner, Table of n, a(n) for n = 1..10000 (correcting table by Michel Lagneau)

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

EXAMPLE

For n = 4, a(4) = 9 because the corresponding trajectory of 1/(2*(-4)+1) = -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. [Corrected by Sean A. Irvine, Mar 19 2026]

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 *)

PROG

(Python)

from fractions import Fraction

def A224299(n):

if n == 0: return 0

x = Fraction(1, 2*-n - 1)

seen = set()

while x not in seen:

seen.add(x)

x = x/2 if x.numerator % 2 == 0 else 3*x + 1

return len(seen) - 1 - (x == 2) # Falk Hüffner, Mar 19 2026

CROSSREFS