[seqfan] Re: This sequence proves the Collatz Conjecture

Wed Apr 10 19:33:09 CEST 2024

  Robert Munafo dug more deeply than I, turning up a relevant original
"What's new" page at

https://web.archive.org/web/20200813035618/https://terrytao.wordpress.com/advice-on-writing-papers/on-local-and-global-errors-in-mathematical-papers-and-how-to-detect-them/
complete with all TT's recommended site list, including my missing
reference
  Scott Aaronson "Ten Signs a Claimed Mathematical Breakthrough is Wrong"
at
    https://www.scottaaronson.com/blog/?p=304

  Thankyou RM!

WFL

On Wed, Apr 10, 2024 at 3:55 PM Fred Lunnon <fred.lunnon at gmail.com> wrote:

>
>   Much advice to potential authors is available online; perhaps too much,
> and not all of it useful or technically relevant to mathematics.  But the
> people who ought to be reading and digesting it are the authors, who are
> all much too busy cooking up their manuscripts and falling headlong into
> well-known elephant traps.  'Twas ever thus!
>
>   Terry Tao's original (sadly missed) Wordpress layout used to include a
> lengthy list of recommended reading, including a critique entitled along
> the lines "How to tell if a submitted manuscript is probably wrong".
> However, author and reference now lie beyond my powers of retrieval.
>
>   A hoary anecdote that just seems worthy of frequent repetition.
> The number-theorist Edmund Landau was rumoured to maintain
> preprinted forms paraphrasing "Dear ___ ,  Thankyou for your proof
> of Fermat's last theorem.  The first mistake is on page ___ at line ___",
> to be passed together with the MS concerned to some hapless postgrad
> for processing.  That always appeared to me an excellent way to kill two
> birds with one stone; assuming a supply of suitable stones, naturally.
>
>   My only personal involvement with a Collatz claim was mercifully
> brief.  At the end of a few pages of perfectly respectable manipulation,
> the author had derived an equivalent conjecture (concerning a slightly
> different sequence), which on some inexplicable grounds was then
> declared obviously true.  It cost only a few minutes to draft the computer
> program which quickly found a counterexample ...
>
>   I would be certainly be favourably impressed by an author asserting
> only a weaker result (Tao on Collatz?) or a more general one (Wiles on
> FLT?) of whatever famous problem has ensnared their attention.  Such
> features distinguish serious researchers who have at least taken the
> trouble
> to familiarise themselves initially with the relationships of their topic
> to the
> mathematical world at large!
>
> WFL
>
>
>
>
> On Wed, Apr 10, 2024 at 7:23 AM jean-paul allouche <
> jean-paul.allouche at imj-prg.fr> wrote:
>
>> 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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>> --
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>