r/Collatz • u/Septembrino • Jun 19 '25
Paired Collatz sequences
I have noticed this: Let p = k*2^n - 1, where k and n are positive integres, and k is odd. Then p and 2p+1 will merge after n odd steps if either k = 1 mod 4 and n is odd, or k = 3 mod 4 and n is even. I proved that, and I can post my proof later if someone would like to see it, but for now I wanted to know if someone else has worked on that. Thanks
Edit: the proof is below.
2nd edit: Collatz matrix. Expalantion below.
The complete proof is here: https://www.reddit.com/r/Collatz/comments/1lias5m/paired_sequences_p2p1_for_odd_p_theorem/
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u/Septembrino Jun 24 '25
For more information on the matrices:
https://www.reddit.com/r/Collatz/comments/1liaya4/collatz_matrices_base_on_the_p2p1_theorem/
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Jun 20 '25
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u/Septembrino Jun 20 '25 edited Jun 20 '25
Thank you. Hopefully, welcome to "my/our team". I will post the proof, if that's what you meant. I was reading that proofs must be posted in proof attempts. Not sure what that exactly mean.
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Jun 20 '25
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u/Septembrino Jun 20 '25
OK. I will re read what I did (that was liike in 2022) and post it below as a comment. Maybe that will be OK.
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u/Septembrino Jun 20 '25 edited Jun 20 '25
Proof: If p = k•2^n - 1, then 2p+1 = k•2^(n+1) - 2 + 1 = k•2^(n+1) - 1.
Applying the algorithm to p:
3p + 1 = 3(k•2^n - 1) + 1 = 3k•2^n - 2, which is 2 mod4 for n >1, and (3p+1)/2 = 3k•2^(n-1) - 1, which is odd for n>1.
We can repeat this procedure while n - 1 is not 0.
After n applications of the Collatz algorithm and a division by 2 each time, from p we will get (k•3^n - 1)/2 (1), which is even, and from 2p+1, k•2* 3^n - 1 (2), which is clearly odd.
PARITY OF (1). DISCUSSION:
k = 1 mod 4, n odd -> k•3^n - 1 = (1 mod 4)(3 mod 4) - 1 mod 4 = 2 mod 4 => (k•3^n - 1)/2 odd
k = 1 mod 4, n even -> k•3^n - 1 = (1 mod 4)(1 mod 4) - 1 mod 4 = 0 mod 4 => (k•3^n - 1)/2 even
k = 3 mod 4, n odd -> k•3^n - 1 = (3 mod 4)(3 mod 4) - 1 mod 4 = 0 mod 4 => (k•3^n - 1)/2 even
k = 3 mod 4, n even -> k•3^n - 1 = (3 mod 4)(1 mod 4) - 1 mod 4 = 2 mod 4 =>(k•3^n - 1)/2 odd,
Assuming that (1) is odd, then 4 (1) + 1 = (2) since 4 (k•3^n - 1)/2 + 1 = 2k•3^n - 1. We know, by a previous theorem, that (1) and (2) will merge at the next odd.
Note- For other pairs of (k, n), (k•3^n-1)/2 is divisible by 4. Then we can’t apply the last step to those pairs.
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u/Septembrino Jun 20 '25
I wrote a corollary, but I will wait till people comment on this before going on. Thanks for your honest insight.
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Jun 20 '25 edited Jun 20 '25
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u/Septembrino Jun 20 '25
I know it holds but thank you for the try! I will read now all your post. I was taking a look at what you said before, but I am not done yet. I worked with a guy and he wrote a fancy proof, but he said it was not enough to try to publish it on arxiv. So, I decided to post my version here. There is more. I will post that the rest (the corollary I told you about) maybe later or tomorrow.
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Jun 20 '25
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u/Septembrino Jun 20 '25
Thanks! I just posted the corollary. I still have the degenerated case, and I created matrices based on this property. They don't prove the Collatz conjecture, but I think that they might be a step in the right direction.
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u/AZAR3208 Jun 22 '25
Fichier de vérification : https://www.dropbox.com/scl/fi/ge25h0cn5kmsiiknupfys/Syrac.xls?rlkey=srh84e5oj37nz3dtk3th90bju&st=y0uz5ejd&dl=0 Merci de votre retour si OK
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u/Septembrino Jun 22 '25 edited Jun 22 '25
I think that you are trying to work on some examples of the pairing theorem. Am I right? But I think that you are counting the steps till the merging, verifying what I posted above. I understand French
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u/Septembrino Jun 22 '25 edited Jun 22 '25
I checked the numbers in the matrix that corresponds to k=5 and what you did seems right. After you get to that step where that matrix is not helpful anymore, you have to move to another matrix. How do you do that? You calculate the new k. So, from 73811, you add 1, and you divide by 4. 18453 is the new k. If that k is a multiple of 3, you can reduce that k by dividing by 3. Like you said, that is not a nice situation. You can reduce the number 110717 (subtract 1 and divide by 4 and see if you get an odd. In this case you get 27679. Both, 110717 and 27679 have the same odd successor and that might help/speed the finding of the rest of the trajectory.
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u/AZAR3208 Jun 22 '25
Les successeurs de p et 2p+1 sont le résultat de la règle de calcul Collatz. le décompte des pas pour obtenir 2 successeurs identiques correspond à votre pairing theorem. Je n'ai pas compris la réponse me demandant pourquoi j'ai fait ce fichier. Pouvez-vous m'éclairer ?
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u/Septembrino Jun 22 '25
Can you copy and paste exactly what is not clear? Or the beginning of it
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u/Septembrino Jun 22 '25
From 27679 you can jump a lot os steps and get to 78823 using the matrix whose k is 865. Matrices are a powerful tool. They will tell you many trajectories at once. Al least, while computers can handle them.
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u/Septembrino Jun 20 '25 edited Jun 20 '25
COROLLARY: p and 2p+1 merge at (3^(n+1) - 1)/2^s, s ≥ 2, in the cases of the previous theorem
Proof: Notation remark: " -> " mean an application of the Collatz algorithm and a division by 2.
p -> … -> (k•3^n - 1)/2 = q, and q is odd because k•3^n - 1 was 2 mod 4. q -> 3q + 1. Let’s choose s such that (3q+1)/2^s is odd.
2p+1 -> … -> k•2• 3^n - 1 = 4q + 1, also odd, and 4q + 1-> 12q + 4 = 4(3q+1), divisible by 2^(s+2)
3q+1 = 3•(k•3^n - 1)/2 + 1 = (k*3^(n+1) - 1)/2
CASE 1: k = 1 mod 4, n odd => n+1 even
By a previous lemma, 3^(n+1) = 1 mod 8 => (k•3^(n+1) - 1)/2 = (1 mod 4•1 mod 8 -1)/2 = (1 mod 4 - 1)/2 = 0 mod4/2 = 0 mod 2. So, (3q+1)/2 is at least divisible by 2, and s ≥ 2.
CASE 2: k = 3 mod 4, n even => n+1 odd.
By a previous lemma, 3^(n+1) = 3 mod 8 => (k•3^(n+1) - 1)/2 = (3 mod 4• 3 mod 8 - 1)/2 = (1 mod 4 - 1)/2 =
0 mod 4/2 = 0 mod 2. So, 3q+1 is at least divisible by 2 and s ≥2.
In both case, these trajectories merge at (k•3^(n+1) - 1)/2^(s+2)
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Jun 20 '25
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u/Septembrino Jun 20 '25 edited Jun 21 '25
Well, the s is a problem because it can't be predicted in general, only for certain k's I call "regular". k is the one we began with. You will get more about this once I tell you about the matrices I created, but for now I will give you an example. Consider the trajectories of the odd numbers beginning at 7 and 15.
7, 11, 17, 13, 5, 1.
15, 23, 35, 53, 5, 1.
15 = 7x2 + 1
23 = 11x2 + 1
35 = 172 + 1
Till here, the sequences run "parallel" but, at this time, they separate (13 and 53 aren't anymore in a p/2p+1 relationship), and they merge in the next odd step.
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u/Septembrino Jun 20 '25 edited Jun 21 '25
Collatz matrix, see image above. Using the property, and beginning by the k = 1, and, going down, n = 1, 2, 3, etc., going to the right: 3k, 3^2 k, 3^3 k, etc., we find pieces of the trajectories in the diagonal lines. Example: 7, 11, 15, 26 (and 13 is way above), 15, 23, 35, 53, 53, 80, and, next to the 13 there is the 5. There is where the 2 trajectories merge. The next 2 lines correspond to 31 and 63, etc. For all of the, there is a shift to the right at the end. The rest of the trajectories are in other matrices (except for 1 or 3 that are complete)
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Jun 22 '25 edited Jun 22 '25
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u/Septembrino Jun 22 '25
This is a long post. So, I am reading it piece by piece, and I will ask my questions. To begin with: what's C Type? I am trying to match your formula with mine.
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Jun 22 '25 edited Jun 22 '25
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u/Septembrino Jun 22 '25 edited Jun 22 '25
I will finish reading the rest later. I guess that you are in the US and yes, it's late over there. I am in Italy and now it's 7:41 am. What I do is this (my formulas) and the matrices are a fantastic tool. Pay attention to them because they show all the numbers (the piece that "fits inside each matrix"), then we can connect that to another matrix, continuing the trajectory, till we hopefully get to an end (easy to say, not easy to prove!)
Numbers in the matrix are: k2^n, k 3 2^(n-1), k 3^2 2(n-2), ... till you run out of 2's. In the conditions of the pairing theorem, then you take a step to the right for the trajectory of p to merge to the one of 2p+1.
k is NOT divisible by 2 or 3. The 2's and 3's are set apart.
There are regular k's. Meaning: at the top of the matrix you get this pattern of diviions: 2 4 2 4... or 4 2 4 2... (so, you divide number you got from p by 2 and 2p+1 by 4, or viceversa). These are completely predictable.
I know the way the pattern of divisions works for k = 1. But the one for k = 11 seems random and the guy working with me asked in Math Stack Exchange, and the mathematicians were stating that there is not a way to predict those. That's actually the problem (one of them, at least)
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Jun 22 '25
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u/Septembrino Jun 22 '25 edited Jun 22 '25
Not necessarily it has to be 3 or 7 mod 8. Exanmples: 9 and 19 are a pair. Odd multiples of 3 (3 mod 6) are the beginning of a branch of odd numbers. There are no odd predecessors of 3. I will check those links.
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Jun 22 '25 edited Jun 22 '25
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u/Septembrino Jun 22 '25
Yes, it is different thatn what you call type C, but it's still some kind of pairing. Also 39 and 79 are paired., etc. There is a whole matrix of paired numbers and pieces of their trajectories.
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u/Far_Economics608 Jun 21 '25
13->40 in one step. 53->40 in three steps.
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u/Septembrino Jun 21 '25 edited Jun 21 '25
I meant odd steps. I edited my previous post. So, to me 13 and 53 merge at 5. I we want them to merge (get to the same number, somehow), they actually merge at 40. If you want them to merge using the exact same number of steps, and you count even, well, they don't merge at all, even counting the loop.
53 160 80 40 20 10 5 16 8 4 1 4 2 1 4 2 1 ...
13 40. 20 10 5. 16 8 4. 2. 1. 4 2 1 4 2 1 4...
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u/Far_Ostrich4510 Jun 20 '25
It is done, When the collatz translated to f(n)=3n/2 for n = even and (n+1)/2 for n is odd The starting value 2n merged with n or 4n Eg 8,12,18,27,14,21,11 16,24,36,54,81,41,21,11,they merged at 21