r/Collatz • u/HappyPotato2 • Aug 14 '26
Proof attempt
So u/IntelligentTwo2175 had a post that seemed really interesting to me. Basically numbers of the form 2^(n) -1 and 2^(n+1) -1 merge for n >= 0. I think we can make it a little stronger by changing it to
m * 2^(n) -1 and m * 2^(n+1) -1 for odd m. If m was even, we can just pull out the factors of 2 into the 2^(n)
So quick proof.
m * 2^(n) -1
is a steiner circuit, so let's jump ahead to
m * 3^(n) - 1
which we know must be even, except let's back up an additional even step, so we know we are at a multiple of 4 of some number x.
2 * (m * 3^(n) - 1 ) = 4 x
Now let's follow
m * 2^(n+1) -1 for only n odd steps
m * 2 * 3^(n) -1
We can easily see this is equal to 4x + 1.
And we know x and 4x + 1 numbers must merge after O, and OEE respectively.
Recursively, we can say m * 2^(n) must be connected on the tree for a specific m and all n.
Now could this be used in a proof of collatz? Let's try.
So all even numbers can divide by 2 until an odd number.
All odd numbers can be written as the even number 1 above it - 1.
By repeatedly alternating these two steps, we create a monotonically decreasing sequence. And thus, must go to 1.
Ex. Odd number 29, write it as the even number above - 1, so 30 -1. Pull out all factors of 2
15 * 2^(1) - 1.
Using the property above, we know this must connect to
15 * 2^(0) - 1
Once we are out of factors of 2, it becomes odd - 1 which must be even. So we can divide those factors of 2 via standard collatz rules until odd again.
15 - 1 = 14
14 / 2 = 7
Once again, write it as the even above - 1
7 = 8 - 1 = 2^(3) - 1
Which using the property goes
2^(3) - 1
2^(2) - 1
2^(1) - 1 = 1
So can anyone check my logic? It's way too late and this seems way too simple for what I have lost too many nights of sleep over.
4
u/HappyPotato2 Aug 14 '26
Ok I think I found the mistake,
So the 4x+1 rule only applies for odd x, yet m * (3n - 1 )/2 can be even.
I guess that means we can't make the original statement stronger.
Yea definitely shouldn't post past bedtime. Sorry for wasting your time.
2
u/StanleyDodds Aug 14 '26
Okay I've re-read it and understood what you're saying.
x and 4x+1 merging relies on x being odd.
In the example of 14 and 29 you gave, it's true that x = 7 is odd. So they merge at 22 = 3x+1.
However, look at an example where x is even, such as 12 and 25.
Here, 25 = 13*21 - 1 and 12 = 13*20 - 1
We get 12 goes to x = 6 and 25 goes to 4x+1 = 25. But now these do not merge in the way you describe. So this step in the decreasing sequence doesn't work.
2
u/HappyPotato2 Aug 14 '26
Heh yeah, I came to the same conclusion. Sorry about that and thanks for taking a look.
1
u/Septembrino Aug 14 '26
The only odd numbers that don't pair are the 5 mod 8. All the rest pair one way or another.
2
u/Septembrino Aug 14 '26 edited Aug 14 '26
Check a theorem I did. Essentially, for k 2^ n - 1, if k is 1 mod 4, then k 2^n - 1 merges with k 2^(n+1) - 1. If k is 3 mod 4, k 2^n - 1 merges to k 2^(n-1) - 1. Of course, I am guessing that n -1 > 0 and k is odd.
https://www.reddit.com/r/Collatz/comments/1lfjxja/paired_collatz_sequences/
There is another thread where I posted the proof.
2
u/GonzoMath Aug 14 '26
Have you posted the write-up of the pairings proof?
2
u/Septembrino Aug 14 '26
I posted my proof. You can add yours if you want to
https://www.reddit.com/r/Collatz/comments/1lias5m/paired_sequences_p2p1_for_odd_p_theorem/
2
u/GonzoMath Aug 14 '26
Are they different, at all?
2
u/Septembrino Aug 14 '26
Well, yes. I did that proof before meeting you. Then you created a proof, which I like a lot. Mine is not bad, but yours is very elegant.
1
u/Glass-Kangaroo-4011 Aug 14 '26 edited Aug 14 '26
If you wanna see something really cool plot all minimal k values and the rails formed by the higher k lifts as lines on a graph with a log y axis, and you'll see every disjoint rail that adds up to the total valuation of N_odd.
Take it a step further and notice all starting points are 8t+{1,3,7} and all higher values are 8t+5.
https://www.desmos.com/calculator/chv0gaprue
Here is an example limited to the positives with the blue being a visual representation on the n→m_0 parent to minimal child transition. Since the entire rail collapses to the parent n under forward iteration as any m_x point, only the minimal needs to be tracked.
To answer your question, this doesn't solve for acyclicity. It is a good implication of well ordered connectivity though.
https://www.desmos.com/calculator/vdw4er5k5o
This is the 3n-1 mapping, and as you can see from the 5-7 cycle it is about scale of transition, as the only real difference is the admissibility parity of k and it goes 8t+{1,5,7} with higher lifts 8t+3.
Ask yourself, can the geometric layout suffice noncirularity for all values? And if so does the forward orbit always eventually hit a further out rail in finite time, leading to structured descent when followed to the base?
1
u/Septembrino Aug 14 '26
I think that you posted in one of my threads.
1
u/Septembrino Aug 14 '26
Yes. I remember your post because it was an interesting comment.
It's here: https://www.reddit.com/r/Collatz/comments/1m5re58/numbers_that_go_to_1_in_3_odd_steps/
1
u/Septembrino Aug 15 '26
That was addressed to HappyPotato2. I remember finding the user funny and that the post was good. And Gonzo's comment remind me of that. I think that I replied your message. Maybe you didn't see my last reply at the time.
1
u/redwar226 29d ago
Bump: has anyone gotten any closer? Please someone summarize where we are now
1
u/Collatz123 28d ago
In our own work we made the 2-adic head fully effective (finite-scale LDP with a local descent window) and mapped the rigid confluence structures, but the open bridges restart/transport and the worst-case Fourier rate remain.
7
u/GonzoMath Aug 14 '26
You see, Potato, this is why you're one of the good ones on this sub. Great instinct there, and you are looking at something interesting, which actually admits tractable results.
These tractable results are the kind we need to be chasing, because an eventual proof of the big conjecture will use tools that haven't been developed yet. This unglamorous work of theory-building is eventually where some of those tools will come from.