A224254 Full cycle lengths in the Collatz (3x+1) problem when the negative integers are used.

2, 2, 2, 2, 5, 2, 5, 2, 5, 5, 2, 2, 5, 5, 2, 2, 18, 5, 5, 5, 18, 2, 18, 2, 18, 5, 5, 5, 2, 2, 18, 2, 18, 18, 5, 5, 18, 5, 2, 5, 18, 18, 2, 2, 18, 18, 5, 2, 18, 18, 5, 5, 2, 5, 18, 5, 2, 2, 2, 2, 18, 18, 5, 2, 2, 18, 18, 18, 2, 5, 2, 5, 18, 18, 5, 5, 2, 2, 2, 5, 5

Offset: 1

Author

Michel Lagneau, Apr 02 2013

Keywords

• nonn

Comments

There are other cycles of lengths 2, 5 and 18 if negative integers are used. In Z, it is conjectured that the five values of cycle are 1, 2, 3, 5 and 18 (see A121510).

Examples

			a(1) = 2 because the cycle -1 -> -2 -> -1... contains 2 distinct terms;
a(5) = 5 because the cycle -5 -> -14 -> -7->-20 -> -5 ... contains 5 distinct terms;
a(17) = 18 because the cycle -17 -> -50 -> -25->-74 -> -37 -> -110 -> -55->-164 -> -82 -> -41 -> -122->-61 -> -182 -> -91 -> -272->-136 -> -68 -> -34 -> -17... contains 18 distinct terms.
		

Links

• Wikipedia, Collatz conjecture

Crossrefs

Cf. A121510, A224166, A224183.

Programs

Python

import sympy
def A224254(n):
  return next(sympy.cycle_length(lambda x:3*x+1 if x%2 else x//2,-n))[0] # Pontus von Brömssen, Jan 24 2021

Extensions

Data corrected by Pontus von Brömssen, Jan 24 2021