Is this a valid interpretation???

[alfansome]

I have wondered about the 1->4->2->1 description as a "loop" and have
come to the conclusion that in fact this is not a good description.
Since the problem states that repeated applications of the algorithm
eventually lead to 1, my interpretation is that it halts there.

It may be that removing the "loop" description is of some assistance
in this problem. Maybe not. But I think it is a distraction.

Other thoughts?

[mensa...@aol.com]

On Feb 12, 1:10 pm, alfansome <alfans...@yahoo.com> wrote:
> I have wondered about the 1->4->2->1 description as a "loop" and have
> come to the conclusion that in fact this is not a good description.
> Since the problem states that repeated applications of the algorithm
> eventually lead to 1, my interpretation is that it halts there.

I believe the original conjecture says "halt at 1".
The issue of "loop" arose when someone asked "what happens
if I don't halt at 1?".

>
> It may be that removing the "loop" description is of some assistance
> in this problem.

It's more important than that.

> Maybe not. But I think it is a distraction.

Perhaps, if all you ever consider is 3n+1. But to
understand 3n+1, you really need to look at it in
the context of 3n+C (where C=1).

When we look at it that way, we see that "1" is
inconsequential, that the conjecture should state
"for all n in 3n+C, the sequence leads to C, which
then loops C->4C->2C->C".

>
> Other thoughts?

This then brings up the concept of "Trivial Loop"
vs. "Non-Trivial Loop" (of which you have never
even seen one in the 3n+1 positive domain).

But Non-Trivial Loops abound! In all systems where
C isn't a power of 3 and in the negative domain.

In that light, it is better to state the conjecture
as "the only loop in 3n+1 positive domain is the
Trivial Loop".

That then leads to the question "WHY are there no
Non-Trivial Loops in the positive domain? Is it
because they CAN'T happen or, due to the Law of Small
Numbers, we simply haven't seen one? If you can answer
"yes" to the first part, you have effectively proved
the conjecture. You CAN prove it for SOME cases, but
you CANNOT prove it for ALL cases, thus the conjecture
remains unsolved.

Note that you don't HAVE to prove it for ALL cases,
but you must then prove that the single Trivial loop
in the positive domain is the ONLY loop.

[alfansome]

really? it seems like there is a geometric argument that can be made
here in terms of an n-dimensional space, the only possible loop is the
trivial loop as described. I would further guess this might be in base
4.

Al

[mensa...@aol.com]

Well, you can't prove the factors of the numerator can't
cancel the factors of the denominator in the Crossover Point
because a counterexample to that exists. It's in the negative
domain, so it's not a counterexample to Collatz per se.

But factoring is not a domain issue, so you can't rule out
the possibility that one exists in the positive domain. You
have to expilicitly prove that none exist in the positive
domain, not simply that they don't exist period.

> it seems like there is a geometric argument that can be made
> here in terms of an n-dimensional space, the only possible loop is the
> trivial loop as described. I would further guess this might be in base
> 4.

I wouldn't know about that. Just be careful that that argument
doesn't prove they CAN'T exist, which would be false. I've seen
arguments that claim Non-Trivial loops can't exist. Such arguments
are obviously false since Non-Trivial loops are known to exist.

>
> Al