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

1,3

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,....

Links

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

Example

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.

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}]

Crossrefs

Cf. A006577.

Sequence in context: A368437 A258852 A021706 * A131953 A167508 A167507

Adjacent sequences: A224180 A224181 A224182 * A224184 A224185 A224186

Keyword

nonn

Author

Michel Lagneau, Apr 01 2013

Extensions

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

Status

approved