

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).


4



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
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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, `, d1): 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: A222641 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 BrÃ¶mssen, Jan 24 2021


STATUS

approved



