[seqfan] Re: This sequence proves the Collatz Conjecture

[Ali Sada]

Hi Fred,

Thank you very much for your thoughtful response; I really appreciate it.

I admit I am not well-versed in the Collatz literature. I watched 2 or 3 YouTube videos and read the introduction of one paper. However, even with a small sample like this, the most important question about the Collatz was not asked. It is the center of the conjecture and should have been mentioned, at least. I figured out the question, and the sequence built itself.

You mentioned that counterexamples existed previously. So, I will donate $1,000 to the OEIS in your name if you could find one example where d or D doesn't exceed or falls short of their A3 counterparts. (And yes, I am luring you to read the analysis!)

Best,

Ali




    On Wednesday, April 10, 2024 at 10:56:20 AM EDT, 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
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
>
>
> --
> Seqfan Mailing list - http://list.seqfan.eu/
>

--
Seqfan Mailing list - http://list.seqfan.eu/