A224183 Number of halving and tripling steps to reach the last number of the cycle in the Collatz (3x+1) problem for the negative integers (the initial term is not counted).

1, 1, 4, 2, 4, 5, 4, 3, 11, 4, 6, 6, 9, 4, 9, 4, 17, 12, 7, 4, 23, 7, 18, 7, 17, 10, 7, 5, 10, 10, 21, 5, 28, 17, 13, 13, 17, 8, 13, 5, 17, 24, 8, 8, 22, 19, 16, 8, 26, 17, 11, 11, 16, 8, 17, 6, 29, 11, 11, 11, 17, 22, 19, 6, 37, 29, 20, 17, 19, 14, 19, 14, 24

Offset: 1

Author

Michel Lagneau, Apr 01 2013

Keywords

• nonn

Comments

The Collatz problem with negative numbers is as follows: start with any number n = -1, -2, -3, ... If n is even, divide it by 2, otherwise multiply it by 3 and add 1. Do we always reach a last number of each cycle? It is conjectured that the answer is yes.

A majority of last numbers of each cycle is -1, but it is possible to obtain also -5, -10, -14, -20, -34,....

Examples

			a(10) = 4 because the trajectory of -10 is -10 -> -5 -> -14 -> -7 -> -20 -> -10 and -10 is the last term of the cycle, hence 4 iterations.
		

Links

• Michel Lagneau, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A006577.

Programs

Maple

printf(`%d, `,1): for n from -2 by -1 to -100 do:x:=n:lst:={n}:for it from 1 to 100 do:if irem(x,2)=0 then x := x/2: lst:=lst union{x} :else x := 3*x+1:  lst:=lst union{x}fi:od:d:=nops(lst): printf(`%d, `,d-1): od:

Mathematica

Collatz[n_] := NestWhileList[If[EvenQ[#], #/2, 3 # + 1] &, n, UnsameQ, All]; Table[s = Collatz[-n]; len = Length[s] - 2; If[s[[-1]] == 2, len = len - 1]; len, {n, 1,100}]

Extensions

a(1) changed to 1 by Pontus von Brömssen, Jan 24 2021