Editing Talk:710: Collatz Conjecture

Technically, the 1 should lead to the four, causeing a loop. Because (1*3)+1=4, 4/2=2, 2/2=1 xD {{unsigned|‎87.242.215.66}}
:I'm pretty sure (follow the Wiki link, perhaps, although I haven't yet) there's an implicit "...eventually lead to one, then you can stop" for this process.
:And (again without checking the Wiki link), I suppose the conjecture could fail in one of two cases.  Firstly if multiplying by three and adding one would take us to another odd number.  Which cannot happen, because (odd*3) will be odd, so (odd*3)+1 is even.  Which leads us to the possibility that the even number leads back to a prior odd number, to circle around again.  There's no trivial case of (n*3)+1 => m, (m/2) => n, although there is the case of (n*3)+1 => m, (m/2) => o, (o/2) => n, for n=1.  How, though, could we evaluate f<sub>1|2</sub>(x<sub>i</sub>) => x<sub>i+1</sub> for all countably finite limits to i, given the rules of which f() function to use, to ensure that x<sub>i</sub> never equals x<sub>0</sub>.  Now, ''that's'' a question and a half. Which I suspect has already been asked. And a half. [[Special:Contributions/31.111.50.225|31.111.50.225]] 21:36, 7 May 2013 (UTC)
:: Wonderful! "Without checking the wiki link"... The guy from [[675: Revolutionary]] in real life![[User:Mumiemonstret|Mumiemonstret]] ([[User talk:Mumiemonstret|talk]]) 15:14, 17 October 2014 (UTC)
::: It's pretty obvious the person you replied to has a head for mathematics. If you really think this is at all similar to the Goatee Guy from [[675: Revolutionary]], then you are sorely mistaken. [[Special:Contributions/172.68.59.84|172.68.59.84]] 13:40, 22 November 2018 (UTC)
:: It seems way too general to be much more than "asked," and I am sure that it has been addressed in its simpler forms. In any case, there is enough amateur, recreational, and serious mathematical literature on it to find out that there are indeed two failure cases: a starting Collatz number results in an infinitely increasing sequence, or a loop exists apart from the 4-2-1 loop. (Curiously enough, some loops exist when negative numbers are allowed.) Stuff like this and Goldbach made me realize just how hard simple things can get. --[[User:Quicksilver|Quicksilver]] ([[User talk:Quicksilver|talk]]) 03:04, 20 August 2013 (UTC)


Just for fun, here's 710's Collatz Trajectory. 
     0: 710
     1: 355
     2: 1066
     3: 533
     4: 1600
     5: 800
     6: 400
     7: 200
     8: 100
     9: 50
    10: 25
    11: 76
    12: 38
    13: 19
    14: 58
    15: 29
    16: 88
    17: 44
    18: 22
    19: 11
    20: 34
    21: 17
    22: 52
    23: 26
    24: 13
    25: 40
    26: 20
    27: 10
    28: 5
    29: 16
    30: 8
    31: 4
    32: 2
    33: 1
[[User:International Space Station|International Space Station]] ([[User talk:International Space Station|talk]]) 09:36, 25 January 2016 (UTC)

I wrote a program to calculate the tree (Python 3.9), but you have to install treelib. Do this by typing 'pip install treelib' in the terminal. The program:

 from treelib import *
 tree=Tree()
 levels=int(input('Levels of tree? '))
 start=int(input('Starting number? '))
 tree.create_node(start, start)
 def f(x):
     global tree
     if tree.depth(x)==tree.depth():
         return True
     else:
         return False
 def collatz(x):
     if x%2==0 and x%3==1 and x!=4:
         return [(x-1)//3,x*2]
     else:
         return [x*2]
 for i in range(levels):
     N = list(tree.filter_nodes(f))
     for i in N:
         c=collatz(i.tag)
         for j in c:
             tree.create_node(j,j,i)
 tree.show(line_type='ascii-ex')

--[[Special:Contributions/198.41.238.106|198.41.238.106]] 01:31, 28 July 2021 (UTC)

:If we're sharing code, I made a Python script that calculates a string for any given number and then graphs it:

 import matplotlib.pyplot as plt
 start_number = int(input("Enter a number:\n"))
 loop_count = 0
 full_list = [start_number]
 while start_number != 1:
   loop_count += 1
   if start_number % 2 == 0:
     start_number = int(start_number / 2)
     print(start_number)
     full_list.append(start_number)
   elif start_number % 2 == 1:
     start_number = int(start_number * 3 + 1)
     print(start_number)
     full_list.append(start_number)
   else:
     print("Error")
 print("Iterations: " + str(loop_count))
 plt.plot(full_list)
 plt.title('Graph (Linear)')
 plt.xlabel('Steps')
 plt.ylabel('Hight')
 plt.show()