Source
r/Collatz•
D2cookie

Well, this might work?

https://www.youtube.com/watch?v=oXR7v_zUM_g

Don't mind my inability to explain my thoughts, if I'm wrong I'd be wasting more time, or this could be trivial and everyone knows it or this is pretty much it.

tl;dw put the tree in base 4, and just look at it, the solution will be obvious as daylight.

If you can solve the number 2 and 10(4), then you solved 2, 10, 20, 100, 200, 1000, ...

If you solved 11 (5 base 10), and you double it to get 22 then you solved 110, 1100, 11000, and 220, 2200, 22000, 220000, ...

But, since you can get from the base of 1 path to the next one with just slapping a 1 to the end that means you also solved:

  • 111

  • 1111

  • 11111

  • 111111

And, that should include all the 2x of those bases, and any amount of 0's you could attach to the end of those numbers.

If you solve 3, that means you immediately solved 31, 311, 3111, ...

Anyways you can remove almost all the multiplication / division down to just addition / subtraction and slapping on / removing digits.

It would mean that both loops and numbers escaping to infinity are impossible.

I was gonna do more optimizations if needed, but I could either be wrong, am doing something trivial, or am beating up a dead horse.

Posted Aug 31, 2023, 11:00 AM UTC

115
D2cookieOP•

Another thing you can notice with the prime factors of the bottoms of each tower, let's say you were on 23 * 2^n , you drop down to 23, which is 11 in decimal.

You now know that you can never get back onto [11, 2^x ], decimal, but you also can't can never get back onto:

  • [231, 2^y ] <=> [45, 2^y] <=> [3^2 , 5, 2^y ]
  • [2311, 2^y ] <=> [181, 2^z]
  • [2311, 2^y ] <=> [725, 2^w ] <=> [5^2 , 29, 2^w ]

...

But, that doesn't mean you can't get down to [5, 2^k ], But specifically the prime or composite odd number combination at the bottom of each tower can never be reached again.

1
deabag•

Im coding with Bard and am using Collatz on prime numbers. I posted a code that creates its own sieve to GitHub/johnnydtennessee , and am working on the "whole enchilada." I've already made progress. Bard is very smart, best to stock to one topic IMO but maybe not, he is a good communicator and calculator.

1
D2cookieOP•

Thank you, I don't have too much time this week, but given my lack of patience i just uploaded a video with almost no notes or anything to the side as I thought this might either work or I'm wasting my time.

I'm not a mathematician, but I'll try my best to make it a bit more clear, and I'll see if I can find more tricks.

1
deabag•

I have another one that is better, similar to yours even it will be efficient at higher numbers for doing the slow calculations first, and using the factors later. Does yours do that, sieve the hard way by creating an internal sieve instead of an external one to apply evenly, then benefit later for large numbers? I haven't tested it in large numbers yet.

1
D2cookieOP•

No, I haven't thought up a way yet of properly doing it forwards yet.

1
elowells•

Everything that follows is in base ten. You are saying that for odd integer x:

  1. If x iterates to 1, then 2^(p)x also iterates to 1 where p is any positive integer. Sure.

  2. If x iterates to 1, then 4x+1 also iterates to 1. True since 3(4x+1) + 1 = 4(3x+1), that is the next odd integer following x is the same as the next odd integer following 4x +1. It follows that if x iterates to 1 then 4(4x+1)+1 etc also iterates to 1.

But you've left out a lot of numbers such as all numbers that are of the form 4x+3. For any x, the spacing between successive odd integers of the form (4x+1), 4(4x+1)+1 etc grows exponentially as 4^(n.) If you have shown that all odd integers x<xmax iterate to 1, then the fractional coverage grows as xmax*4^(-n), that is the fraction of all odd integers covered goes to 0.

1
D2cookieOP•

O yea, you're right, I did actually miss those ones. I'll see if I can fix it.

1
elowells•

Unless there is some meaning of "edge case" that I am unfamiliar with, those aren't edge cases. Those are most numbers. If you show that all odd integers x<=N iterate to 1, the probability that an odd integer x'>N is of the form 4x+1 or 4(4x+1)+1, etc for some x<=N goes to zero as x' increases. Even if you could show that if x iterates to 1 implies that 4x+3 iterates to 1 (it doesn't), that still wouldn't help since 4x+3 still has exponentially sparse coverage since the spacing between successive numbers grows exponentially, so you would still only cover an infinitesimal fraction of numbers.

1
D2cookieOP•

I'm sorry, can you give me concrete examples, I can't see which numbers you're talking about, I'm looking but I can't see

1
elowells•

I need a picture (please don't point me to your video). Are you making an inverse tree starting at 1?

1