Is it possible to narrow down a domain of possible counter-examples to the Collatz Conjecture?

Question

First of all, I am not trying to prove the Collatz Conjecture.

I want to know if it is possible to rule out certain values of a counter-example.

Suppose $k \in \Bbb Z^+$ is the lowest counter-example to the Collatz Conjecture. Notice that because it is the lowest such number any number $N<k$ automatically goes to $1$.

If $k$ is even:$$\frac{k}{2}<k$$

Therefore $k$ is odd. $k$ is also in the form $4l-1$ because:

$$3(4l + 1) + 1$$ $$12l+4$$ $$3l+1<4l+1$$

Therefore: $$3k+1$$

This is always even as: $$3(2n + 1) +1$$ $$6n+4$$ $$2(3n +2)$$

Therefore, $$\frac{3k+1}{2}$$ is odd, as: $$\frac{3k+1}{4}<k$$ And so: $$\frac{9k+5}{4}$$ Here, however, there is no definite path to take as $\frac{9k+5}{4}$ could be even or odd. If we assume that it is even, (and there is a reason I do this as I shall later show,) then: $$\frac{9k+5}{8}$$ must be odd. Therefore: $$\frac{27k+23}{16}$$ is odd. Therefore: $$\frac{81k+85}{32}$$ Now here again, we reach a point of indecision. As with last time, I shall assume that the above is even. Therefore: $$\frac{81k+85}{64}$$ is odd. $$\frac{243k+319}{128}$$ This can be re-arranged into a Diophantine equation in terms of $k,m \in \Bbb Z^+$. $$243k +319 = 128m$$ Solving this gives values of $k$ in the form $128l -5 ,l \in \Bbb Z$.

However, if $k$ is not in this form, then either it went into the region below $k$, where we know it will go to $1$, or one of our "assumptions" was incorrect.

Solving these allows us to eventually get to the forms:

$$k = 16l-1$$ Or $$k = 32l+7$$ Or $$k = 64l +27$$ Or $$k = 128l-5$$

So my question is:

Is this a valid method for decreasing the domain of $k$ and, also, is it possible to show that $k$ in these forms will go to $1$?

Thanks.

Answer 1

Remember that the natural numbers are well-ordered: if a subset of $\mathbb{N}$ is non-empty, then it has a least element. Consequently, if there is a natural number that, regardless of the number of Collatz iterates, fails to be mapped to something less than itself, then there must be a least such example.

By confining the list of counterexamples to classes of integers of the form $ax + b$, where $a$ and $b$ grow unboundedly, one could indeed prove Collatz, as the minimum natural numbers of these classes grows to $\infty$ as $b \to \infty$. But, some things have to be observed.

1. It is not sufficient to list out a few such refinements and say "and so on". You need to prove (probably inductively) that this method will continue to work.
2. More importantly, it's vital that you show that the minimum example of this class of numbers grows without bound. You don't want your method to start eventually leaving out some very large numbers.
3. Note the branching here. You haven't just shown the Collatz counterexamples must be of the form $ax + b$, but you've narrowed it down to one of four possibilities. You need to ensure that the minimum number satisfying one of these possibilities grows. So far, you've shown that the minimum counterexample must be at least $27$, which is a fairly low number given the amount of steps and the fact that you've branched to $4$ possibilities already!