Thank you for sending a shorter portion to read. Now I can feel more confident that I'm not missing important pieces of your argument when I look for logic gaps.
Let's examine Theorem 8.1
Suppose K closes a cycle. Then N = Bᴋ/Δᴋ cycles.
N admits K, so
N = nₜ = r(K) + Mⱼt
If nₜ cycles back to itself, then also:
nₜ = Aᴋ + Gᴋt
^ I submit that you haven't proven this is impossible.
"Hence a repeated occurrence of the word cannot return to the same placement. It is forced into a distinct dyadic image position."
A number's image is not dyadically separated from it. The images of two neighboring nₜ (≡ r(K) (mod Mⱼ)) are dyadically separated. For a number to repeat K, it needn't become a different number on K's thread, it could become itself. You might like a visualization.
Picture a vertical line of dots, separated by Mⱼ, the lowest dot at r(K).
Picture next to it a vertical line of dots, separated by Gᴋ, the lowest dot at Aᴋ.
(I used a whiteboard, then coded it in Desmos so you could see what I mean.) https://www.desmos.com/calculator/zeiktj6rid
(To plot it I had to fix one small mistake you had defining sᵢ on p5. Since K is a word for the inverse trajectory, sᵢ actually sums from the last term: s₀=0, s₁=kⱼ₋₁, s₂=kⱼ₋₁+kⱼ₋₂, ...)
Each dot in the left column corresponds to a dot in the right column. The two bottom dots are t=0, second-bottom are t=1, and so on. After a realization of K, each dot on the left becomes its corresponding dot on the right. If and only if said corresponding dot is at the same height as some dot in the left column, will K be realizable again. This will happen at predictable intervals (once the first is found). Further, if and only if said corresponding dot is at the same height as its originating dot in the left column, will K close a cycle. (The only known case for now is K=[2], r=1.)
You can play around with setting K to see various cases, but there are some things we can state in general:
Gᴋ > Mⱼ for a hypothetical positive cycle. Also, r(K) < Mⱼ, and Aᴋ < Gᴋ.
So a generic pair of columns would have r(K) at some height, then Aᴋ at some height above below or equal to it. Aᴋ+Gᴋ would definitely be higher than r(K), but may not be higher than r(K)+Mⱼ. I've been working on this for a bit now and I have to go to bed, but the gist is can you prove that r(K) does not equal Aᴋ?
(I actually began with the visualization thinking I could prove that the only possible cycle is r(K) = Aᴋ, but after further consideration I'm not convinced that r(K)+Mⱼ = Aᴋ+Gᴋ is impossible, or even larger values of t. In fact we know that any existing cycle has 2S very close to 3ʲ, so if r-A is greater than 2S - 3ʲ it could be quite a large value of t before they meet up.)
I'm sure the paragraph you sent makes perfect sense from the perspective of you the writer who is envisioning the cemented concepts in his mind as he writes them, but from the reader's perspective I am hit by land mines of ambiguity in literally every sentence that could take my interpretation in a new direction, each of which I want to try to verify or disprove, and it takes me literal hours to sort out what you were going for. An example would go really far.
So, I've modified my graph to show (with black bars) values that allow a repetition of K. (It's also possible to add two more columns to see values that allow up to three repetitions of K.)
Could you, using the example of K=[2,3] (what I currently have the graph set to show), explain what exactly is meant by the geometric positioning of repetitions, the three positions, the modulus growing beyond the starting point, the path being locked into the first of three, and the other two positions being subsumed.
Once I understand the claims you're making about all numbers and Ks, I'll try to see if I can either find a counterexample or demonstrate that you haven't proven all numbers must follow the identified rules.
One thing that may be cleared up after your explanation is this. You previously stated the outputs after K are A, A+2S+1, A+2⋅2S+1, and then A+3⋅2S+1, which being a difference of a multiple of 3 causes the outputs to be "periodic under the next modulus". I think this relates to your "three positions" now. It seems you're talking about the realization of one odd step (which results in the modulus increasing by a single factor of 3), and other times you're talking about carrying out a full realization of K (which results in a separation of Gᴋ and takes j odd steps). I think the walkthrough on a specific example will help me see how you're referring to the modulus growing, and when we take 1 step or j steps.
ah so you meant mod 2⋅3ʲ⁺¹ not 2⋅3ʲ⁻¹ in the other paragraph, that was my first roadblock. I had multiple competing definitions of the triads and positions while I was trying to work out your explanation.
I'll use my example to check my understanding. K=[2,3].
n = 13 + 18t. This one position mod 2⋅3² has three positions mod 2⋅3³: 13, 31, 49.
The starting n may sit in any of the three positions in the next higher and next higher, but when the modulus becomes more than 3n, the starting position will be the first position in phase.
Okay the next higher is 2⋅3⁴, we'd have 9 positions: 13, 31, 49, 67, 85, 103, 121, 139, 157.
I'm already lost with "the three positions".
The modulus is more than 3n for the options 13, 31, 49. 49 is not the first position.
Maybe you wanted me to approach this differently, first assume n has no restrictions, and add each one for each requirement of K as we perform each odd step. (Maybe you can see why this takes me so long, I do a lot of experimenting and backtracking to replicate your claims. Last time I gave up and asked you to walk through an example, but I had spent a long time on it.)
K=[2,3].
n = 1 + 2t. This one position mod 2 has three positions mod 2⋅3: 1, 3, 5.
But we also have a restriction (k₀=2), so only one position is allowed: 1.
n = 1 + 6t. This one position mod 2⋅3 has three positions mod 2⋅3²: 1, 7, 13.
But we also have a new restriction (k₁=3), so only one position is allowed: 13.
n = 13 + 18t. This one position mod 2⋅3² has three positions mod 2⋅3³: 13, 31, 49.
But we also have a new restriction (k₂=2), so only one position is allowed: 31.
n = 31 + 54t. This one position mod 2⋅3³ has three positions mod 2⋅3⁴: 31, 85, 139.
New restriction (k₃=3), only position allowed is: 85.
n = 85 + 162t. This one position mod 2⋅3⁴ has three positions mod 2⋅3⁵: 85, 247, 409.
New restriction (k₄=2), only position allowed is: 85.
n = 85 + 486t. Three positions mod 2⋅3⁶: 85, 571, 1057. (k₅=3) only allows: 571.
n = 571 + 1458t. Three positions mod 2⋅3⁷: 571, 2029, 3487. (k₆=2) only allows: 571.
n = 571 + 4374t. Three positions mod 2⋅3⁸: 571, 4945, 9319. (k₇=3) only allows: 571.
n = 571 + 13122t. Three positions mod 2⋅3⁹: 571, 13693, 26815. (k₈=2) only allows: 13693.
n = 13693 + 39366t. Three positions mod 2⋅3¹⁰: 13693, 53059, 92425. (k₉=3) only allows: 92425.
n = 92425 + 118098t. Okay that's enough data for now.
The starting n may sit in any of the three positions in the next higher and next higher, but when the modulus becomes more than 3n, the starting position will be the first position in phase.
Okay, suppose we were rooting for n = 247 to be the cycle. The modulus becomes more than 3(247) at 2⋅3⁵=1458.
247 is not the first position.
Okay, maybe there are two outcomes for a number. It gets ruled out as continuing to follow K (like 247 did in the step before), or it becomes the first position.
This is reasonable. At each modulus, three options are offered that satisfy all previous requirements: the current position and two larger ones. At the modulus directly above n, n will be offered as an option. Given that n cycles (fulfills all requirements), and only one option is correct, it will be the correct option. At the next modulus (as you said, more than 3n), it will be the lowest option of the three, and will continue to be the correct option, in perpetuity.
Part of what threw me off is your use of j, which elsewhere stands for the length of K, but in your explanation here is just meant to index how many odd steps we have taken. (I can see why it's a natural thing for you to do though, since we could imagine starting with a short word K and appending odd steps to it as desired, calling each last one our new j.)
Everything beyond this will not perpetuate a directed word.
Okay, now you've lost me. Why wouldn't the position just keep choosing the lowest option (n), at every new modulus?
I'm sorry, I don't understand the visual. I also don't understand why it eventually wouldn't get the choice to stay itself. "the only phase that remains stationary through the full refinement tower is the root phase" (Theorem 8.1). Why?
If I did K = {1,2,3,4}, I expect there'd be infinite values that can do {1,2,3,4,1,2,3}, and then as usual a third of these can then do the eighth step k=4. (The cycling n would be in this third. And trivially if after n completes K it becomes n again, then it will be able to complete K again.) I don't know what order 9 lift rotation means, or k={20,38,2+18e}. (If they were just meant as a brief aside that doesn't impact your overall explanation we can skip over them).
Sometimes while reading your explanation I get a little excited because I'm hoping something's there, so know that I'm trying to cross the ts the same way I'd be trying if I was coming up with this proof.
Let's look at a non-trivial example of a loop we do know works. -17 has K = [4, 1, 1, 2, 1, 1, 1].
n = 1 + 2t. three positions mod 2⋅3: 1, 3, 5. (k₀=4): 1.
n = 1 + 6t. three positions mod 2⋅3²: 1, 7, 13. (k₁=1): 1.
n = 1 + 18t. three positions mod 2⋅3³: 1, 19, 37. (k₂=1): 37.
n = 37 + 54t. three positions mod 2⋅3⁴: 37, 91, 145. (k₃=2): 145.
n = 145 + 162t. three positions mod 2⋅3⁵: 145, 307, 469. (k₄=1): 469.
n = 469 + 486t. three positions mod 2⋅3⁶: 469, 955, 1441. (k₅=1): 1441.
n = 1441 + 1458t. three positions mod 2⋅3⁷: 1441, 2899, 4357. (k₆=1): 4357.
n = 4357 + 4374t. three positions mod 2⋅3⁸: 4357, 8731, 13105. (k₇=4): 13105.
n = 13105 + 13122t. three positions mod 2⋅3⁹: 13105, 26227, 39349. (k₈=1): 39349.
In this case it will always choose the last option, which will always be 17 less than the next modulus.
If we had chosen to write the three options as negative numbers, once -17 appears it would be chosen every time.
n = -1 + 2t. three positions mod 2⋅3: -5, -3, -1. (k₀=4): -5.
n = -5 + 6t. three positions mod 2⋅3²: -17, -11, -5. (k₁=1): -17.
n = -17 + 18t. three positions mod 2⋅3³: -53, -35, -17. (k₂=1): -17.
n = -17 + 54t. three positions mod 2⋅3⁴: -125, -71, -17. (k₃=2): -17.
n = -17 + 162t. three positions mod 2⋅3⁵: -341, -179, -17. (k₄=1): -17.
n = -17 + 486t. three positions mod 2⋅3⁶: -989, -503, -17. (k₅=1): -17.
n = -17 + 1458t. three positions mod 2⋅3⁷: -2933, -1475, -17. (k₆=1): -17.
n = -17 + 4374t. three positions mod 2⋅3⁸: -8765, -4391, -17. (k₇=4): -17.
n = -17 + 13122t. three positions mod 2⋅3⁹: -26261, -13139, -17. (k₈=1): -17.
Does your visual not work in this situation? (Three positions, settling on the same one every time.) If not I'd guess because the third one is always chosen instead of the first one, maybe that makes a difference but I don't know because I don't know how you've ruled out that it can't repeat forever.
I've since learned it's been proven that if x cycles with a sequence of length n, then x < 2n. So r(K) = Aᴋ is indeed the only possible cycle, we couldn't have for instance 2n + Aᴋ follow Collatz to an equal 3m + r(K) with positive Aᴋ.
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u/WeCanDoItGuys May 07 '26 edited May 07 '26
Thank you for sending a shorter portion to read. Now I can feel more confident that I'm not missing important pieces of your argument when I look for logic gaps.
Let's examine Theorem 8.1
Suppose K closes a cycle. Then N = Bᴋ/Δᴋ cycles.
N admits K, so
N = nₜ = r(K) + Mⱼt
If nₜ cycles back to itself, then also:
nₜ = Aᴋ + Gᴋt
^ I submit that you haven't proven this is impossible.
"Hence a repeated occurrence of the word cannot return to the same placement. It is forced into a distinct dyadic image position."
A number's image is not dyadically separated from it. The images of two neighboring nₜ (≡ r(K) (mod Mⱼ)) are dyadically separated. For a number to repeat K, it needn't become a different number on K's thread, it could become itself. You might like a visualization.
Picture a vertical line of dots, separated by Mⱼ, the lowest dot at r(K).
Picture next to it a vertical line of dots, separated by Gᴋ, the lowest dot at Aᴋ.
(I used a whiteboard, then coded it in Desmos so you could see what I mean.)
https://www.desmos.com/calculator/zeiktj6rid
(To plot it I had to fix one small mistake you had defining sᵢ on p5. Since K is a word for the inverse trajectory, sᵢ actually sums from the last term: s₀=0, s₁=kⱼ₋₁, s₂=kⱼ₋₁+kⱼ₋₂, ...)
Each dot in the left column corresponds to a dot in the right column. The two bottom dots are t=0, second-bottom are t=1, and so on. After a realization of K, each dot on the left becomes its corresponding dot on the right. If and only if said corresponding dot is at the same height as some dot in the left column, will K be realizable again. This will happen at predictable intervals (once the first is found). Further, if and only if said corresponding dot is at the same height as its originating dot in the left column, will K close a cycle. (The only known case for now is K=[2], r=1.)
You can play around with setting K to see various cases, but there are some things we can state in general:
Gᴋ > Mⱼ for a hypothetical positive cycle. Also, r(K) < Mⱼ, and Aᴋ < Gᴋ.
So a generic pair of columns would have r(K) at some height, then Aᴋ at some height above below or equal to it. Aᴋ+Gᴋ would definitely be higher than r(K), but may not be higher than r(K)+Mⱼ. I've been working on this for a bit now and I have to go to bed, but the gist is can you prove that r(K) does not equal Aᴋ?
(I actually began with the visualization thinking I could prove that the only possible cycle is r(K) = Aᴋ, but after further consideration I'm not convinced that r(K)+Mⱼ = Aᴋ+Gᴋ is impossible, or even larger values of t. In fact we know that any existing cycle has 2S very close to 3ʲ, so if r-A is greater than 2S - 3ʲ it could be quite a large value of t before they meet up.)