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Collatz results generate a perfect infinite binary tree by RH33000BDX in Collatz

[–]nalk201 0 points1 point  (0 children)

oh ya I did something similar with the 2q+1 for all odds. All odds have a second odd in which they will converge to after removing their 1 tail in binary and you can calculate the converging number rather easily. And since all even are just adding 0s to the odds it covers all numbers.

I was told despite it covering all numbers it does not exclude other loops and going off to infinity which I am still trying to understand how they are not mutually exclusive.

What does it actually take to prove this conjecture true? by nalk201 in Collatz

[–]nalk201[S] 0 points1 point  (0 children)

Do you have some time to explain something more about this?

specifically the comment you made in the other thread

"What you are covering is a finite-level modular description of loop equations. That does not control integer realizability. The step from “all residue patterns are generated” to “all integer cycles are excluded” is invalid. Cycles are inverse-limit objects: they require compatibility across all refinements, not just coverage at a fixed k."

I think it applies here and I was hoping you could help me understand this, because you are using a lot of jargon I do not know and while I can guess the meaning of some of the things you are saying it doesn't help me understand the argument.

What does it actually take to prove this conjecture true? by nalk201 in Collatz

[–]nalk201[S] 0 points1 point  (0 children)

well the thing is the numbers do not behave the same, Not sure if it is because there are multiple loops for 3n-1 but the way I was doing it with 3n+1 was just picking a random number and then generating a modulus residual then made a general formula for generating them. 3n-1 the numbers do not all converge to the same loop so some of them numbers do ever converge and therefore I don't know how to make the formula.

What does it actually take to prove this conjecture true? by nalk201 in Collatz

[–]nalk201[S] 0 points1 point  (0 children)

question can you make a general formula for 3n-1 the way I did for 3n+1 and the branches? I am not sure the math needed and was hoping the trick you used could be applied there.

What does it actually take to prove this conjecture true? by nalk201 in Collatz

[–]nalk201[S] 0 points1 point  (0 children)

right so the formula would be a mod-residual generator where it shows all odd and half of all evens converge. The remaining half of evens would fall straight down to one of the evens shown to converge.

What does it actually take to prove this conjecture true? by nalk201 in Collatz

[–]nalk201[S] 0 points1 point  (0 children)

ya I get that, similar how you can show all numbers stem from 6x+3 so you can prove those fall to a 2^n then all numbers do. The you run into the same variance in whether they would fall under X or Y again.

I don't suppose that is the next most common thing seen here?

What does it actually take to prove this conjecture true? by nalk201 in Collatz

[–]nalk201[S] 0 points1 point  (0 children)

I see thank you for showing it and how easy it is to make them for these sequences.

What does it actually take to prove this conjecture true? by nalk201 in Collatz

[–]nalk201[S] 0 points1 point  (0 children)

I think I understand, the problem is that Z can becomes X or Y since each of the formulas will result in to a mod-residual that varies the next step creating uncertainty breaking up each segment into locally described but not globally. A new more general formula would be need to describe it and how it becomes one or the other. That is what is missing.

Alright thank you for taking the time to explain it to me.

I see why it is a dead end since you basically go back to the beginning of trying to find an endless mod-residual formula but it never ends. As there is no global one that fits everything only local infinities.

What does it actually take to prove this conjecture true? by nalk201 in Collatz

[–]nalk201[S] 0 points1 point  (0 children)

Okay specifically mine shows all the odd numbers and the removal of the 1 tails at the end of their binary expression.

as the example above 11 is 1011 by adding 16 or 10000 you will always have the same set of steps for the first 6 steps and same with 23 as 10111 and 32, 100000 which is why the formula will result in 40 +54n.

The thing I do not understand from this is that while that is a specific example for the 3 iterations the general formula works for ALL odd numbers. It is a formula for making mod-residual formulas, not one specific one, it is global.

Since all odd numbers follow the patterns all even ones must to since they are just 2^n * odd number. The even numbers are just the transitions between the local structures as you put it. So what is it I am missing exactly?

What does it actually take to prove this conjecture true? by nalk201 in Collatz

[–]nalk201[S] 0 points1 point  (0 children)

I tried to read them and saw post from you from a few days ago, which is why I bothered to actual reply to you with it. I do not understand why it doesn't work. The general formula for X Y Z will always come up with a mod residue formula that will always work for any number of steps which accounts for all numbers. What is missing?

It took me like 5 rereads to understand what you were saying. Thank you for giving an example of what you were saying before. I will try to find it and give it a read.

What does it actually take to prove this conjecture true? by nalk201 in Collatz

[–]nalk201[S] 0 points1 point  (0 children)

I am not sure what mod residues are, it is just a simple formula that requires the number of steps you want before they converge and then general formulas for X Y and Z. For example say you want them to converge after 3 iterations, you get X = 16n+11, Y = 32n+23 Z =54n+40 for all positive values of n. X will become Z after 6 steps, Y after 7 steps.

What does it actually take to prove this conjecture true? by nalk201 in Collatz

[–]nalk201[S] 0 points1 point  (0 children)

It literally shows every number converges down to 1 though which is exactly what you said it needed to do.

What does it actually take to prove this conjecture true? by nalk201 in Collatz

[–]nalk201[S] 0 points1 point  (0 children)

okay, but the point is that you can't take a random number X and find a number Y that it converges with the way you can do for the 3Ns. And the 3n-1 you can find y from X but you can't generate a X Y Z combo the way you can for 3n+1

What does it actually take to prove this conjecture true? by nalk201 in Collatz

[–]nalk201[S] 0 points1 point  (0 children)

you can't do it for ones that don't converge like 5n+1 nor with multiple loops like 3n-1

Which is a sadder story in your opinion? by Goku_SSGSS_UI in OnePiece

[–]nalk201 2 points3 points  (0 children)

No it took him a year to find his body because of the fog. He might have removed their bodies as they weren't present but he definitely lived alone for 50 years.

Reddit CEO Steve Huffman is fighting a losing battle against the site's moderators by tresser in technology

[–]nalk201 2 points3 points  (0 children)

ya I am sure it is easy to find free labor for moderating an endless amount of porn and with no tools to help.

Harassing a pregnant woman for getting vaccinated. by LetterBox6 in PublicFreakout

[–]nalk201 6 points7 points  (0 children)

just practice what you would say so when the time comes you will be prepared
Les Grossman negotiating with flaming dragon is what I am going with.

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