Collatz Conjecture proof?
Mensanator
is not about the Wikipedia article, but about a supposed proof.
datta...@gmail.com writes:
> ON A STEADY NOTE
>
> Sorry,I made few mistakes.
>
> in the 1st line,I need to add that every positive odd numbers
> can be represented as
>
> (2^p)*x-1
>
> where x is an odd number and p is any positive integer.
>
> If we perform Collatz process in it, it will become
>
> (3^1)*(2^p)*x-3+1 = (3^1)(2^p)*x-2
>
> which is an even number. then by performing the process we have
>
> (3^1)*(2^p-1)*x-1,
>
> which is clearly an odd number..eventually after 2p steps
> it will become
>
> (3^p)*x-1.
>
> Now if I increase the value of the positive integer p,
> then after a few many steps, one may find that the value of p
> can still be increased atleast by 1. so there is no upper boundary
> for the term p as well as for the term (3^p)*x-1.
>
> then,the term can never be closer to 1. [as (3^p)*x-1>(2^p)*x-1]
>
> Still there remains a fair chance of increasing the value of p again.
> So by increasing the value of p continuously, the process will never
> results 1.
>
> [Foot note]:
> We are using (3^p)*x-1 as a term which still can be processed by
> Collatz process and somehow it can become 1. But thats not our goal
> we are trying 2 show that, for increasing value of p, this term
> can never be reach 1]
Mensanator
So? The process of continuously increasing p isn't the Collatz
process.
>
> > [Foot note]:
> > We are using (3^p)*x-1 as a term which still can be processed by
> > Collatz process and somehow it can become 1. But thats not our goal
Which means your goal doesn't prove the conjecture.
> > we are trying 2 show that, for increasing value of p, this term
> > can never be reach 1]
The conjecture doesn't say that THAT term must reach 1. The conjecture
states that if you continue past that term you eventually reach 1.
datta...@gmail.com
>
> - Show quoted text -
datta...@gmail.com
>
> - Show quoted text -
yes..if I continuously procces it..
but as we know that there exist infinite number of positive integers.
so we can increaase the value of p infinite times.and still it will
be positive integer ..(2^p)x-1 also be positive integer too.But if we
process collatz on it...its value will increase 2*infinite
times=infinite times...and become closer to 3^p*x-1.which clearly not
1 and much more than (2^p)*x-1.
In this case (2^p)*x-1 not becoming 1 in finite time..It is increasing
infinitely..which is opposing collatz assumption.
pls do not delete post if I answer lately...as it is expensive for me
to come cyber cafe and become online everyday..
pls wait for my answer..
anyway thank u for giving me ur precious time.
Mensanator
Yes, an infinite number OF them, but none ARE infinity.
> so we can increaase the value of p infinite times.and still it will
> be positive integer ..(2^p)x-1 also be positive integer too.
But ALWAYS finite.
> But if we
> process collatz on it...its value will increase 2*infinite
> times=infinite times...
No, that would require a number with an infinite number of bits.
There is NO such number. All numbers have a finite number of bits.
> and become closer to 3^p*x-1.which
... will always be finite also.
> clearly not 1 and much more than (2^p)*x-1.
Irrelevant. Although that number isn't 1, it WILL be reached
in a finite number of steps and then will proceed from that
number to 1 in a further finite number of steps.
Neither the run-up to that number nor the glide back down to 1
can have an infinite number of steps.
> In this case (2^p)*x-1 not becoming 1 in finite time..
False.
> It is increasing infinitely..
False.
> which is opposing collatz assumption.
Conclusion based on false premise.
>
> pls do not delete post if I answer lately...
Don't worry. Unlike Wikipedia, posts can't be deleted from Usenet.
That's one reason I moved it here lest some other Wikipedia editor
were to simply delete it.
> as it is expensive for me
> to come cyber cafe and become online everyday..
No problem. Even though I can, I may not have the time everyday.
Without new posts, your last message will sink deeper into the
archive. But it can ALWAYS be found easily. And the next post will
cause it to rise to the top again.
> pls wait for my answer..
Sure, take your time. When you're ready, I'll see your new post
at the top of the heap again.
>
> anyway thank u for giving me ur precious time.
Your ideas aren't without merit. Once you get past the idea of
an infinite integer, there are some things one can learn from
this. Unfortuneately, a proof isn't one of them. Yet I find
the whole thing endlessly facinating.
Srijit
>
> - Show quoted text -
If I increase value of a integer infinite times = p tends to infinity.
its means positive integer will never reach infinity.
but infinite times of increasing = a never ending process.
if I stop the process and then count it.it will may accept collatz
assumption.
but if I continuously process, then there is infinite number of
positive integer exist which will not accept collatz assumption.
Mensanator
> its means positive integer will never reach infinity.
That's right, "tends" to infinity, no integer ever IS infinity.
> but infinite times of increasing = a never ending process.
Right again. But that ISN'T the Collatz Process. Collatz does
NOT say that THIS process must be finite or that THIS integer
ever becomes 1.
THIS process does NOT prove Collatz false.
>
> if I stop the process and then count it.it will may accept collatz
> assumption.
The conjecture is that whenever you stop the process, the integeger
you stop on will reach 1 in a finite number of iterations of (3n+1):(n/
2).
>
> but if I continuously process, then there is infinite number of
> positive integer exist which will not accept collatz assumption.
Really? Name one that doesn't. Surely you can identify one out the the
infinite multitude.
Let's look at it this way. Let's call (2^p)*x-1 the starting point
of Process A: A(p). And the process of determining the successor
of A(p) to be (2^p+1)*x-1.
The Process A sequence is thus:
A(p) => A(p,1) => A(p,2) . . . A(p,m) . . .
The dots imply this sequence is infinite.
Now we also have Process B where, starting from B(n), the successors
are
B(n) => B(n,1) => B(n,2) => . . . B(n,k)
where Process B itself is the usual (3n+1):(n/2) stuff. There are no
trailng dots because the process is defined to halt when B(n,k)=1.
Combining Process A & Process B two dimensionally, we have:
A(p) => B(A(p),1) => B(A(p),2) => . . . B(A(p),k)
A(p,1) => B(A(p,1),1) => B(A(p,1),2) => . . . B(A(p,1),k)
A(p,2) => B(A(p,2),1) => B(A(p,2),2) => . . . B(A(p,2),k)
.
.
.
A(p,m) => B(A(p,m),1) => B(A(p,m),2) => . . . B(A(p,m),k)
.
.
.
Collatz says ONLY that ON ANY GIVEN horizontal line, k is finite and
B(A(p,m),k)=1.
It does NOT say the A(p,m) sequence is finite, NOR does it say that
A(p,m) must equal 1.
Do you understand?
Would an example help?
Suppose you set x=1 in (2^p)*x-1. starting with p=1, Process A gives
the Mersenne Numbers: 1, 3, 7, 15, 31, 63, . . .
In Binary (base 2), these would be
1 = 1
3 = 11
7 = 111
15 = 1111
31 = 11111
63 = 111111
and the successor function in base 2 is simply append a 1 to the
predecessor.
So we can find A(p,177149) by simply writing 11111...<177139 digits>...
11111.
We can now calculate what k is in Process B because there are formulae
to
determine k based on the PATTERN of base 2 digits! For the "all 1s"
pattern,
ODD = 4.819 * m
which means given an m-bit Mersenne Number, ODD is the count of odd
numbers
in the Collatz sequence. And there should be ODD*2 even numbers in
the
sequence. So k should be 3*(4.819*m).
This is, of course, based on statistics. Which means "your mileage
will vary".
See <http://members.aol.com/mensanator666/Page.htm>
Of course, once you get to an A(p,m) with of a statitically
significant size,
the estimate is pretty damn good.
>>> import collatz_functions
>>> m = 2**177149 - 1
>>> c = collatz_functions.collatz(m,0)
>>> print c # [EVEN,ODD]
[1531812, 854697]
>>> k = sum(c)
>>> print k
2386509
>>> kk = 3*(4.819*177149)
>>> print kk
2561043.093
>>> print k/kk
0.93185038804
Only 7% error between the predicted vs actual values of k.
Nifty, eh?
Note that the only way k could be infinite would be if m were
infinity. But there are no integers with an infinite number of
bits. Look at the base 2 successor process. If the predecessor
has a finite number of bits (m), then the successor must also
have a finite number of bits (m+1). No successor will ever have
an infinite number of bits.
Srijit
Yes I understand.
thank you..
It was nice talking to you.
phyti
Mensanator
> This is my view on solving this problem.http://uts.awardspace.info
Looks like all you've done is show that the only loop sequence
that contains 1 is the 4-2-1-4 loop. That does NOT prove,
however, that that is the ONLY loop cycle (and if there were
another, it certainly wouldn't contain 1, so you're correct
about (1,1) being unique).
Your statement:
Assume an arbitrary sequence S with a repeated term r (1, p, q, r,
s, t, r)
is a false premise, because if the sequence contains (r,s,t,r), it
cannot also contain (1,p,q), so we can dismiss all the derivations
that follow.
And this conclusion:
There are no loops in the sequences, except (1, 1).
Is simply false. Even though you THINK you're working in
the positive domain, it actuality, you aren't. You're
working in the integers (positive and negative) and in the
integers, there are ALWAYS four loop cycles for 3n+C:
+C, -C, -5C and -17C.
Any system that doesn't explain why there are 4 and why there
is only one in the positive domain is wanting and will never
be a proof.
btv2...@gmail.com
Let me lay out my simple train of thoughts and views first.
1. When even do n/2 is a reduction operation which results in a prime or an even number. With 2 being the final which would give 1 in the end.
2. When odd do 3n+1 is a multiplying operation which always gives an even number or a prime or a multiple of a prime
Since both operation are a lot of time in a direct cycle for these cases they could be simplified to (3n+1)/2 ~ 2. So these will always end up with 1.
All the other cycles will have to be due to their specific starting number or results which give rise to their specific type of loops. This means all that have to be done is group them according to type of loop cycles and determine the commonalities of these numbers. From above it's clear they must be primes and or certain multiple constellation of primes.
dyachenk...@gmail.com
A. The following work DOI 10.5281/zenodo.3630682 proves that the Collatz transformation leads the “length of a number” indicated in the system q = 4∩3 to a unit length.
The main idea is that reduction of the "length of the number" occurs when converting oddness of the form 4k+1 and preservation of the "length of the number" when converting oddness of the form 4k+3.
Since the transformation 4k+3 cannot be stored indefinitely, periodically the "length of the number" decreases.
As a result of consequent iterations the number transforms into the form 2^p/3^q .
B. Simultaneously, during the Collatz transformation of the number (position A), it appears in the system qi = 2∩3 .
C. The record of the number in system qi = 2∩3 (B) consists of the sum of elementary numbers in form {(2^a(ⅈ) -2^b(ⅈ) )/3^ }
D. Example: number 27 = 3/3⋅q^9+2/3⋅q^8+1/3 q^7+0⋅q^6+2/3 q^5+0⋅q^4+1/3 q^3+1/3 q^2+0⋅q^1+1/3 q^0 and its transformation (see table)
i n(i) r(i) -r(i)>1 4k(i)+1 4k(i)+3 k(i) P(i)
0 27 -1 4*6+3 14+13 even 1
1 41 -2 2 4*10 + 1
2 31 -1 4*7+3 16+15 odd 4
3 47 -1 4*11+3 odd
4 71 -1 4*17+3 odd
5 107 -1 4*26+3 even
6 161 -2 2 4*40+1
7 121 -2 2 4*30+1
8 91 -1 4*22+3 46+45 even 1
9 137 -2 2 4*34+1
10 103 -1 4*25+3 52+51 odd 2
11 155 -1 4*38+3 even
12 233 -2 2 4*58+1
13 175 -1 4*43+3 88+87 odd 3
14 263 -1 4*65+3 odd
15 395 -1 4*98+3 even
16 593 -2 2 4*148+1
17 445 -3 3 4*111+1
18 167 -1 4*41+3 84+83 odd 2
19 251 -1 4*62+3 even
20 377 -2 2 4*94+1
21 283 -1 4*70+3 213+212 even 1
22 425 -2 2 4*106+1
23 319 -1 4*79+3 160+159 odd 5
24 479 -1 4*119+3 odd
25 719 -1 4*179+3 odd
26 1079 -1 4*269+3 odd
27 1619 -1 4*404+3 even
28 2429 -3 3 4*607+1
29 911 -1 4*227+3 456+455 odd 3
30 1367 -1 4*341+3 odd
31 2051 -1 4*512+3 even
32 3077 -4 4 4*769+1
33 577 -2 2 4*144+1
34 433 -2 2 4*108+1
35 325 -4 4 4*81+1
36 61 -3 3 4*15+1
37 23 -1 4*5+3 12+11 odd 2
38 35 -1 4*8+3 even
39 53 -5 5 4*13+1
40 5 -4 4 4*1+1
41 1 0
balance -70 46 24
E Find complete prove here (https://zenodo.org/record/3630682#.XjagkGhKjIU).