[seqfan] Re: This sequence proves the Collatz Conjecture

[Alex Meiburg]

Dear Ali,

I've recently become quite engrossed with Lean:
https://leanprover-community.github.io/
Lean is a proof assistant that produces formally verified proofs. If you
submit a proof of the Collatz conjecture
which Lean correctly verifies (a check anyone could run in a matter of ~one
minute, by downloading and
"running" your proof), then I think essentially any mathematician would
trust it.

If you believe that your proof is correct, I strongly encourage you to try
formalizing it in this way.
Although it may not be the fastest way to write up a proof (it essentially
requires learning a new programming
language), it may well be the fastest way to *convince* anyone that your
proof is correct. This is not a fault
of you or your proof; view it as a product of the fact that there are many,
many, many wrong proofs of
the Collatz conjecture, which essentially means (by way of sad necessity)
that many mathematicians are
unwilling to read any proof that isn't written by someone who is already a
very well established
mathematician. It's reasonable to call this ridiculous and unfair. I hope
that you also understand that
it is necessary, given the volume of claims.

But if you successfully formalize it in Lean, you will definitely get the
attention of many people quickly!
Even Terence Tao(!) has recently begun using Lean for his work -- indeed,
he is quite social on the Lean
chat communities and everyone learns from each other.

And whether your proof turns out to be correct or not, I personally believe
that learning Lean gives one
a great deal of practice in formal mathematical reasoning. I believe it has
turned me into a better
mathematician myself, forcing myself to think through steps more carefully
and what I really mean.

-- Alexander Meiburg


On Sun, Apr 14, 2024 at 6:00 AM Ali Sada via SeqFan <seqfan at list.seqfan.eu>
wrote:

> Hi everyone,
> It turns out the array (H(i,j)) is already in the OEIS (A257852.) That
> made the analysis shorter. And I have made additional changes and the text
> now is 1 page.
> Also, I have added the sequence to the OEIS. It's A371949. However, I was
> told it could be deleted.
>
> Best,
> Ali
>
>
>
> Let's call 3x+1 the I operation, and let's call applying m +(3m+1) * (1,
> 4,16..,4^(m-1)) on each even number, Odd * 2^m, the A operation.The IA
> operation is a combination of the two, in this order. If we can provethat
> by applying the IA operation several times on any odd number weunavoidably
> reach A002450, then the Collatz is proven.
>
> A096773 is combination of A002450 and A072197. And since termsof A072197
> reach A00245 in a single IA operation (in other words reach a powerof 2 in
> two Collatz iterations), that means that if applying the IA operation
> multipletimes on any odd number leads unavoidably to A096773, then the
> Collatz isproven. Let’s call A096773 A1, and A079319 A2.
>
> To prove that let’s start with A257852, which is an arraythat contains all
> odd numbers. Let’s call it H(i,j).
>
> Applying the IA operation once on any H(i,j):
>
> IA(H(i,j)) = 3 * 2^j * H(i,j)+ A2(j) = 3 * H(i,j) *2^j + 2^j+ (4 ^ j-1)/3
>
> Let d1(x)​denote the number of A1​terms that H(i,j) jumps following one IA
> operation. Let d2(x)​ represent the differencebetween the last number
> resulting from the jump and the nearest preceding A1​ term that is less
> than or equalto IA(H(i,j)).
>
> Assume H(i,j) falls within the interval 2^k1 and 2^(k1+1). Withinthis
> range, the number yielding the highest result after one IA operation isthe
> largest number from sequence A096773 that fits within this interval.
> Thisnumber is referred to as the “peak” or p(x,x+1).
>
> p(k1,k1+1) = (2^(k1+2)-1) /3 if k1 is even, and (5*2^k1-1)/3 if k1 is odd.
>
> When the IA operation is applied once to p(k1,k+1), theoutcome is
> (2^(6+4*(Int(k1/2)))-1)/3​,regardless of k1 being odd or even. After one IA
> iteration on the peak we get:
>
> d1(p(k1,k1+1)) = 4 if k1 = 0
>
> d1(p(k1,k1+1)) = 3 if k1 = 1
>
> d1(p(k1,k1+1)) = k1+5 if k1 is an odd number >1
>
> d1(p(k1,k1+1)) = k1+3 if k1 is an even number > 0
>
> And for any peak, d2(p(k1,k1+1)) equals zero.
>
> When we apply the IA on H(i,j), which is any other number inthe range,
> that’s not a member of A1 already, d1(H(i,j)) is less than
> d1(p(k1,k1+1)),however, d2(H(I,j) is > zero. Which means that d1(H(i’,j’)
> will graduallyincrease with multiple IA iteration until it reaches k1+3,
> meaning H(i’,j’)itself became a peak.
>
>
>
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>