I was looking into solving the Collatz Conjecture and I came up with a couple of simple tools and programs to help out.

# Collatz Conjecture table of solutions/series produced by a PHP script.

n (n base 2) | solutions to 3n+1 and n/2 as a series (in base 2) |

1 (1) | 1 ( 1 ) |

2 (10) | 2, 1 ( 10, 1 ) |

3 (11) | 3, 10, 5, 16, 8, 4, 2, 1 ( 11, 1010, 101, 10000, 1000, 100, 10, 1 ) |

4 (100) | 4, 2, 1 ( 100, 10, 1 ) |

5 (101) | 5, 16, 8, 4, 2, 1 ( 101, 10000, 1000, 100, 10, 1 ) |

6 (110) | 6, 3, 10, 5, 16, 8, 4, 2, 1 ( 110, 11, 1010, 101, 10000, 1000, 100, 10, 1 ) |

7 (111) | 7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1 ( 111, 10110, 1011, 100010, 10001, 110100, 11010, 1101, 101000, 10100, 1010, 101, 10000, 1000, 100, 10, 1 ) |

8 (1000) | 8, 4, 2, 1 ( 1000, 100, 10, 1 ) |

9 (1001) | 9, 28, 14, 7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1 ( 1001, 11100, 1110, 111, 10110, 1011, 100010, 10001, 110100, 11010, 1101, 101000, 10100, 1010, 101, 10000, 1000, 100, 10, 1 ) |

10 (1010) | 10, 5, 16, 8, 4, 2, 1 ( 1010, 101, 10000, 1000, 100, 10, 1 ) |

11 (1011) | 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1 ( 1011, 100010, 10001, 110100, 11010, 1101, 101000, 10100, 1010, 101, 10000, 1000, 100, 10, 1 ) |

12 (1100) | 12, 6, 3, 10, 5, 16, 8, 4, 2, 1 ( 1100, 110, 11, 1010, 101, 10000, 1000, 100, 10, 1 ) |

13 (1101) | 13, 40, 20, 10, 5, 16, 8, 4, 2, 1 ( 1101, 101000, 10100, 1010, 101, 10000, 1000, 100, 10, 1 ) |

14 (1110) | 14, 7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1 ( 1110, 111, 10110, 1011, 100010, 10001, 110100, 11010, 1101, 101000, 10100, 1010, 101, 10000, 1000, 100, 10, 1 ) |

15 (1111) | 15, 46, 23, 70, 35, 106, 53, 160, 80, 40, 20, 10, 5, 16, 8, 4, 2, 1 ( 1111, 101110, 10111, 1000110, 100011, 1101010, 110101, 10100000, 1010000, 101000, 10100, 1010, 101, 10000, 1000, 100, 10, 1 ) |

16 (10000) | 16, 8, 4, 2, 1 ( 10000, 1000, 100, 10, 1 ) |

17 (10001) | 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1 ( 10001, 110100, 11010, 1101, 101000, 10100, 1010, 101, 10000, 1000, 100, 10, 1 ) |

18 (10010) | 18, 9, 28, 14, 7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1 ( 10010, 1001, 11100, 1110, 111, 10110, 1011, 100010, 10001, 110100, 11010, 1101, 101000, 10100, 1010, 101, 10000, 1000, 100, 10, 1 ) |

19 (10011) | 19, 58, 29, 88, 44, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1 ( 10011, 111010, 11101, 1011000, 101100, 10110, 1011, 100010, 10001, 110100, 11010, 1101, 101000, 10100, 1010, 101, 10000, 1000, 100, 10, 1 ) |

20 (10100) | 20, 10, 5, 16, 8, 4, 2, 1 ( 10100, 1010, 101, 10000, 1000, 100, 10, 1 ) |

21 (10101) | 21, 64, 32, 16, 8, 4, 2, 1 ( 10101, 1000000, 100000, 10000, 1000, 100, 10, 1 ) |

22 (10110) | 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1 ( 10110, 1011, 100010, 10001, 110100, 11010, 1101, 101000, 10100, 1010, 101, 10000, 1000, 100, 10, 1 ) |

23 (10111) | 23, 70, 35, 106, 53, 160, 80, 40, 20, 10, 5, 16, 8, 4, 2, 1 ( 10111, 1000110, 100011, 1101010, 110101, 10100000, 1010000, 101000, 10100, 1010, 101, 10000, 1000, 100, 10, 1 ) |

2 axis graph of Collatz conjecture.

6 axis graph of Collatz conjecture.

You can see that the number of steps to get to the final value of 1 depends mostly on how many 1 digits there are in binary value.

The more 0 digits there are the more times it can be divided evenly by 2 which leads to the final value of 1.

Look at the series produced by n=3, n=7 or n=15

1. The reason these values take the most steps to solve is because anything with a 0 can be divided by 2.

For example: 10, 100, 1000, 10000, 100000 will become 1 by the collatz conjecture quickly using the function/rule: n/2 for n%2=0

## Step by step solution to n=7. In base 2 and base 10.

base 2 | base 10 |

111 | 7 |

1110+111+1 | 3×7+1 |

10110 | 22 |

1011 | 22/2 |

10110+1011+1 | 3×11+1 |

100010 | 34/2 |

10001 | 17 |

100010+10001+1 | 3×17+1 |

110100 | 52 |

11010 | 52/2 |

1101 | 26/2 |

11010+1101+1 | 3×13+1 |

101000 | 40 |

10100 | 40/2 |

1010 | 20/2 |

101 | 10/2 |

1010+101+1 | 5×3+1 |

10000 | 16/2 |

1000 | 8/2 |

100 | 4/2 |

10 | 2/2 |

1 | 1 |

After finding the binary approach to the collatz conjecture and making my graphs,

I found out that others have gone there before:

http://www.journalrepository.org/media/journals/BJMCS_6/2014/Aug/Nag4212014BJMCS12538_1.pdf’

Unfortunately the argument of the paper is that no formal proof to the Collatz conjecture exists due to Hasse’s algorithm.

Here’s a really good video explaining more on collatz conjecture.