[seqfan] Re: This sequence proves the Collatz Conjecture

[jean-paul allouche]

Dear Ali, dear all

Thank you for your views on Collatz. I must say that I have decided a 
few years ago
to stop looking at /any/ proof of this conjecture: I have received so 
many "proofs" on
which I have spent so much time, to finally discover that what I 
suspected was true
(namely the proof was wrong). One of the reasons it was time-consuming 
is that
all these proofs began with a super-long list of definitions, which were 
exhausting me
very quickly. One of the other reasons is that as soon I had found some 
mistake somewhere,
the author(s) sent an /ad hoc/ modification that either was clearly 
wrong or needed
a few more weeks to be proved wrong (etc.: iterative process).

The only exception in my staying away from all the proofs or "proofs" 
has been the
fantastic paper of Tao. When I see that he used several dozens of pages 
and an
incredibly technical arsenal to finally obtain a result both marvellous 
and weak when
compared to the conjecture, I am more and more convinced that there 
cannot be
a "simple" proof.

So Ali please excuse me that I will not read your proof even if it might 
look a priori
appealing.

best wishes
jean-paul



Le 10/04/2024 à 00:36, Ali Sada a écrit :
> Forgot to add the OEIS email!
>
>
>
> Dear Neil, Olivier, Allan, Maximilian, Robert, Jean-Paul, Jon, Michel, 
> Antti, and all,
>
>
>
> Hope everything is going well with you. As I consider you my 
> professors and the OEIS my scientific institution, it would be an 
> honor and a pleasure to submit this analysis for your review. It will 
> only take a couple of minutes to read, and I hope you find it 
> enjoyable. Since this is my first attempt to document this analysis, I 
> kindly ask for your forgiveness if I haven't explained certain aspects 
> adequately.
>
> Please note that since I have a severe problem with notations, 
> sometimes I will explain with numbers. The rules could be generalized 
> for any range.
>
> I will use the terms below in the coming text:
>
> A1 is A002450 (a(n) = (4^n-1)/3
>
> A2 is A072197 (a(n) = 3+ 10* (4^(n-1)-1)/3)
>
> A3 is A096773, a combination of A1 and A2, starting with 3.
>
> 3, 1, 13, 5, 53, 21, 213, 85,…
>
> A4 is A079319 (a(n) = 2^n + (4^n-1)/3)
>
> m is an odd positive integer.
>
> n, i, j, l, k, t, q are positive integers.
>
> E is an even positive integer
>
> O is an odd positive integer
>
> The I operation, or I(m), means 3*m+1
>
> The R operation, or R(m), means 4m+1
>
> To start the proof, I will use the word “equivalent” to refer to the 
> relationship between m, 4m+1, 4(4m+1)+1, etc.
>
> For example, in the Collatz world, 7, 29, 117, 469, …are all 
> equivalent to each other.
>
> 7*3+1 = 22, 22/2 = 11
>
> 29*3+1 = 88, 88/8 =11
>
> 117*3+1= 352, 352/32 = 11
>
> (Say m=2k+1. I(m) = 6k+4. 4m+1 = 8k+5. I(8k+5) = 3*(8k+5)+1 = 24k+16 
> and when we divide by 4 we also get 6k+4)
>
> Then, we define the following operation, let’s call it Operation “A”:
>
> If E is an even number, we write it in this form O * 2^k
>
> To obtain A(E), we repeat the operation R on O k times.
>
> For example to find 8 =1 * 2^3, so, to find A(8), we repeat R on 1 
> three times
>
> 1*4+1 = 5, 5*4+1 = 21, 21* 4+1 = 85. So, A(8) = 85.
>
> Another example, 60 = 15 *2*2, to find A(60), we repeat R on 15 twice
>
> 15*4+1 = 61, 61*4+1 = 245. So, A(60) = 245.
>
> Now, we define the IA operation, which is a combination of the I 
> operation and the A operation, in this order.
>
> IA(m,t) means we repeat the IA operation t times.
>
> For example, IA(7,3) is done line this
>
> 7*3+1 = 22, A(22) = 45, 45*3+1 = 136, A(136) = 1109, 1109*3+1 = 873813
>
> So, IA(7,3) = 87813.
>
>
>
> To construct the main sequence in this analysis, let’s call it S(n), 
> we take the positive integers sequence and replace each even number 
> with its A form. Odd numbers remain the same.
>
> We get:
>
> 1, 5, 3, 21, 5, 13, 7, 85, 9, 21, 11, 53, 13, 29, 15, 341,…
>
>
>
> Now, if we can prove that applying the IA operation on any odd number 
> would lead eventually to A1, then the Collatz conjecture is proven. 
> And since A2 is one step away from A1 (Collatz-wise), then the 
> conjecture is proven if m reaches A3 through multiple IA operations.
>
>
>
> To do that, let’s put all odd numbers in an array, let’s call H(i,j).
>
> The first column is numbers congruent to A3(j) (mod 2^(j+1))
>
> The first column will be numbers congruent to 3 (mod 4)
>
> 3, 7, 11, 15, 19, 23, ……
>
> The second column will be numbers congruent to 1 (mod 8)
>
> 1, 9, 17, 25, 33, …
>
> The third column will be numbers congruent to 13 (mod 16)
>
> 13, 29, 45, 61, …
>
> And so on.
> In Summary, we want to prove that m, after multiple IA operations, 
> will reach the top of a column in the H array.
>
>
>
> Now if we apply the IA operation one time on H(i,j)
>
> IA(H(i,1)) = 3 * 2^j + A4 = 3 * 2^j + 2^j + (4 ^ j-1)/3 …………(1)
>
> Let’s call it Equation 1.
>
> For example, if we apply the IA operation in the first column we get 6 
> * H(i,1) +3, and if we we apply the IA operation on the second column 
> we get 12*H(i,2) + 9, and so on.
>
>
>
> To prove that all off numbers will reach A3 if we apply the IA 
> operation enough times, let’s take the odd numbers in the range 
> between 2^k and 2^(k+1)
>
> When we apply the IA operation once, each number “jumps” to the left, 
> based on Eq. 1.
>
> There are two important numbers here.
>
> D is the difference between IA(m,1) and the largest A3 term that is 
> less than IA(m,1).
>
> For example, when we apply the IA on 33 once, we get 405. The largest 
> A3 term smaller than 405 is 341. So, D(IA(33,1)) = 405-341 = 64.
>
> The second number is d, which is the number of A3 terms that m jumps 
> over after one IA operation.
>
> For example, after one IA operation, 33 jumps over 53, 85, 213, and 
> 341. So, d(IA(33,1)) = 4
>
> The highest jump will be for the members of A3 in the group, let’s 
> call them “Qabsus” (centers in Akkadian).
>
> Between 2^k and 2^(k+1), the qabsu is (4^((k-1)/2)-1)/3 if k is odd, 
> and 3+ 10* (4^(k/2)-1/3) if k is even, where k1 = k/2 if k is even, 
> and k1=(k-1)/2 if k is odd.
>
>
>
> For example, let’s take the odd numbers in the range between 32 and 
> 64. The qabsu is 53.
>
> For this range, the highest qabsu belongs to 53. D(IA(53,1))= 
> 5461-3413= 2048. In general, D(q) = 2^(2k+1). And in each IA 
> iteration, k increases to 2k+2 if k is odd, and 2k+4 if k is even.
>
> As for d(q), it is k+2 for any k.
>
> For all other numbers in the range, k, D and d increase, but less than 
> k(q), D(q) and d(q) respectively.
>
> However, in each iteration, d increases compared to its range. For 
> example,
>
> IA(27,1) = 165 and IA(27,2)= 8021 .  k increased from 4 to 7, to 13. 
> This rate of change will increase, slowly but surely, until it reaches 
> the rate of A3 terms.
>
>
>
> The same thing with D. It was 6 (2*3) , then it became 80 (5*16), then 
> it became 2560 (5*512).  This will continue with each iteration, and 
> eventually D must become a “pure” power of 2 and joins A3.
>
>
>
> d also has no other option but to increase until it reaches a range 
> between 2^kf and 2^(kf+1) where its growth reaches the same level as 
> A3 terms.
>
>
>
> Best,
>
>
>
> Ali
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