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According to the rules of the Collatz Conjecture if $n$ is odd then execute $3n + 1$ and when $n$ is even execute $n/2$. Repeat until (supposedly) reaching $n = 1$.

examples:

$1 → 4 → 2 → 1$

$13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1$

In this experiment I have taken a set of the first $2500$ consecutive odd numbers and named an element in that set $x$. I have then checked what are the common endings, and these are the first few results:

When $x ≥ 1$ will always end in  $4 → 2 → 1$ (2500/2500 times)

When $x ≥ 3$ will always end in $16 → 8 → 4 → 2 → 1$ (2499/2500 times)

When $x ≥ 5$ will end in $40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 �� 1$ (2351/2500 times)

OR

When $x ≥ 5$ will end in $64 → 32 → 16 → 8 → 4 → 2 → 1$  (146/2500 times)

When $x ≥ 7$ will end in $160 → 80 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1$ (1200/2500 times)

OR

When $z ≥ 7$ will end in  $52 → 26 → 13 → 40 → 20 → 10 → 5 → 16 → 8 →4 → 2 → 1$ (1150/2500 times)

OR

When $x ≥ 7$ will end in $256 → 128 → 64 → 32 → 16 → 8 → 4 → 2 → 1$  (145/2500 times)

...

*Note the $3$ ending possibilities of $x ≥ 7$ are extensions of the $2$ the ending possibilities of $x ≥ 5$, and the $2$ ending possibilities of $x ≥ 5$ are extensions of the $1$ ending possibility of $x ≥ 3$, and the $1$ ending possibility of $x ≥ 3$ is an extension of the $1$ ending possibility of $x ≥ 1$.

The heuristic shows that for $x ≥ y$, where $y$ is any natural odd number, $x$ will have a limited amount of ending possibilities, that are extensions of the ending possibilities when $x ≥ y-2$.

My set is a finite set of only the first $2500$ odd numbers, I am definitely not claiming proof or even a direction for a proof (way above my level). I am trying to learn what needs to be proved, besides the obvious that we need to prove that the end result should always be $1$.

The heuristic shows $x ≥ y$, $x$ can only end in a limited amount of ending possibilities depending on $y$ and these endings possibilities are extensions of the the ending possibilities when $x ≥ y-2$, would it be sufficient to prove the conjecture is true?

In other words, If I start with $1$ and mapping backwards all the possibilities permitted by the Collatz rules, will I eventually hit all the numbers?

Every Time I will hit a new number. It means that this number has no loops, since we got to this number from $1$,(that number would not have been able to appear earlier without ending in $1$.)

So if we present the conjecture in such way, we don't have to prove possible loops?

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    $\begingroup$ Not quite sure of the question. A proof that any starting point eventually iterates into some fixed interval (or other set) where all numbers in the interval are (somehow) already known to get to $1$ would be a proof of the conjecture. One can identify many sets, including large or infinite sets, whose members are known to get to $1$. The hard part is finding one where one can also prove everything else will wind up there. Roughly speaking, "does everything eventually get into [some big set I where I know everything eventually gets to $1$]" seems no easier than "does everything get to $1$." $\endgroup$
    – leslie townes
    Commented Feb 24, 2021 at 6:27
  • $\begingroup$ @leslie townes But what about a proof that shows that all numbers have a finite way to get to $1$ ? $\endgroup$
    – Isaac Brenig
    Commented Feb 24, 2021 at 6:41
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    $\begingroup$ The proof requires - there are two challenges: 1. Show that there is always a "limit circle" - a closed list of numbers - no divergence. 2. Show that there is only one such cycle containing the vale 1. $\endgroup$
    – Moti
    Commented Feb 24, 2021 at 6:54
  • $\begingroup$ @Moti but if i can prove that a number has only a finite amount of possible endings, then i can prove that the Collatz Conjecture is true because these endings always lead to the vale 1. $\endgroup$
    – Isaac Brenig
    Commented Feb 24, 2021 at 7:08
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    $\begingroup$ If you can prove convergence than you still need to prove that there is only ONE cycle. Try to define under what conditions a cycle exist and than that there is only one solution. $\endgroup$
    – Moti
    Commented Feb 25, 2021 at 3:21

1 Answer 1

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While the ending possibilities may be finite, only if the beginnings don't diverge or loop will the ending possibilities matter.

We can look and see, all even numbers fall to their odd parts. Odd numbers of form $4k+1$ go through $4k+1\to 12k+4\to 6k+2\to 3k+1$ which shows they can't be the lowest member, of a lowest escaping or looping sequence. However numbers of form $4k+3$ go through: $$4k+3\to 12k+10\to 6k+5\to 18k+16\to 9k+8$$ Now we have a problem , if $k$ is even we can divide by 2 to get $9j+4$ (for which we have a similar conundrum) otherwise, we go to $27k+25$ which will be down to $27m+26$ (for which the conundrum shows up again)

so in most of these cases we need to mod by a higher power of $2$ to get anywhere. parity falls to the last term any time the modulus is even, hence the self similarity of the Collatz graph.

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