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2012-05-26, 20:48   #12
Raman
Noodles

"Mr. Tuch"
Dec 2007
Chennai, India

3·419 Posts

Quote:
 Originally Posted by Raman Thus, it translates into "Has the Collatz Conjecture been proved? Is it being true? It states that any starting number when repeatedly multiplied by three, and then added one, terminates in one. It appears to me that it is being very likely for this(it) to be true." For example, for some starting number N, repeatedly iterating the 3x+1 function one can try out to prove off something thereby establishing an upper bound result that the intermediate number value will not (cannot) go beyond N.2k at all.
Oops, sorry, a major correction please.

It states that if the number is odd, then multiply by 3, and then add 1,
if it is even, then divide by 2.
If this process is being continued again and again, then taking any starting number will eventually end in 1 always.

I think that it is not being very tough in order to prove this fact,

There are more being chances for this to be true, in turn
that's what I will feel it repeatedly, in fact

Quote:
 Take any starting number. If it is odd, then multiply by three and then add one. If it is even, then divide by two. If this process is continued, every number will eventually terminate in one. I feel that it is very likely for this to be true. I think that it is very easy to prove this, as well.
முக்கியமான திருத்தம்

ஏதாவது எண்ணை எடுத்துக்கொண்டு தொடங்குங்கள். அதை ஒற்றைப்படையாக இருந்தால், மூன்றால் பெருக்கிவிட்டு, ஒன்றை சேருங்கள். இரட்டைப்படையாக இருந்தால், இரண்டால் வகுங்கள். இந்த செயலை தொடர்ந்தால், எல்லாம் எண்களும் எப்பொழுதும் இறுதியில் ஒன்றில் வந்து முடிவடையும் என்று இந்த ஊகம் குறிக்கிறது. நான் இதை உண்மையாக இருக்க அதிக வாய்ப்புகள் இருக்கிறது என்று நினைக்கிறேன். அது போல், இதை நிரூபிக்கவும் மிகவும் எளிதாக இருக்கும் என்று எனக்கு தோன்றுகிறது.

சரி?

Last fiddled with by Raman on 2012-05-26 at 21:12

2012-05-26, 21:16   #13
Dubslow

"Bunslow the Bold"
Jun 2011
40<A<43 -89<O<-88

3×29×83 Posts

Quote:
 Originally Posted by Raman Oops, sorry, a major correction please. It states that if the number is odd, then multiply by 3, and then add 1, if it is even, then divide by 2. If this process is being continued again and again, then taking any starting number will eventually end in 1 always. I think that it is not being very tough in order to prove this fact, There are more being chances for this to be true, in turn that's what I will feel it repeatedly, in fact முக்கியமான திருத்தம் ஏதாவது எண்ணை எடுத்துக்கொண்டு தொடங்குங்கள். அதை ஒற்றைப்படையாக இருந்தால், மூன்றால் பெருக்கிவிட்டு, ஒன்றை சேருங்கள். இரட்டைப்படையாக இருந்தால், இரண்டால் வகுங்கள். இந்த செயலை தொடர்ந்தால், எல்லாம் எண்களும் எப்பொழுதும் இறுதியில் ஒன்றில் வந்து முடிவடையும் என்று இந்த ஊகம் குறிக்கிறது. நான் இதை உண்மையாக இருக்க அதிக வாய்ப்புகள் இருக்கிறது என்று நினைக்கிறேன். அது போல், இதை நிரூபிக்கவும் மிகவும் எளிதாக இருக்கும் என்று எனக்கு தோன்றுகிறது. சரி?
If it was easy, someone would have done it.

2012-05-26, 21:23   #14
garambois

"Garambois Jean-Luc"
Oct 2011
France

100010000012 Posts

Quote:
 Originally Posted by R. Gerbicz There are serious problems with the proof of lemma 5 (2nd article) even for cases where m is prime. As I can see you are trying to determine the complementer event: if q==-1 mod m and q|n and q^2 doesn't divide n then m|sigma(n). But there are cases you left out in the counting: let m=5 and 2^3|n and 2^4 doesn't divide n you can get that sigma(2^3)=15 divides sigma(n), so 5 also divides sigma(n) but 2==-1 mod 5 is not true. For composites m there are much more problems with the proof.

You are wrong !
It's a problem because the demonstration is in french. And the translation on the forum is wrong.

If q divides n and q don't divide n^2 then :
If n=qA and q and A are coprime

Then :
sigma(n)=(1+q) sigma(A)

And because q==-1[m] then m will divide (q+1)
So m will divide sigma(n)

You have to take q==-1[m], it is written in the french demonstration.

Did you read the french demonstration or the english demonstration on the forum before posting this message ???

Jean-Luc

 2012-05-26, 21:46 #15 Dubslow Basketry That Evening!     "Bunslow the Bold" Jun 2011 40=2 _along with_ an m \in A_k"? Another note: In the first paragraph, it defines A_k for all l >= 1, but I don't see an l in the definition of the set. (Also, I think there's a typo 'pout' -> 'pour' in there.) Also, it's slow going because half the time is spent reproducing the tex.
2012-05-26, 22:21   #16
R. Gerbicz

"Robert Gerbicz"
Oct 2005
Hungary

5×17×19 Posts

Quote:
 Originally Posted by garambois It's a problem because the demonstration is in french. And the translation on the forum is wrong.
OK, your proof is correct, though not checked every line and my French is poor. But I understand your idea, its nice.

 2012-05-26, 23:19 #17 Dubslow Basketry That Evening!     "Bunslow the Bold" Jun 2011 401$ and all $l\in \mathbb{N}$, there exists an Aliquot sequence such that for $l$ consecutive iterations, $\frac {s(m)} {m} > k$. Proof: We use Mertens' formula $\sum_{p\quad prime,\quad p\le x}\frac {1} {p}\quad \sim\quad log\quad log\quad x$. In particular, $\sum_{p\quad prime} \frac{1}{p}$ diverges and the product $\prod_{p\quad prime}(1+\frac{1}{p})$ goes to infinity. Therefore there exists an $n \in \mathbb{N}$ such that $\prod_{i=1}^{n}(1+\frac{1}{p_i})>k+1$. We choose an element $m_0$ of $A_{n+l}$. By Lensta's theorem, the first $l+1$ iterations of the Aliquot sequence starting at $m_0$ are respectively in $A_{n+l},\quad A_{n+l-1},...,A_n$. Let $k \in [0,l]$; then $m\quad:=s^k(m_0)$ is in $A_{n+k}$ and thus can be written as $p^{t_1}_1...p^{t_{n+k}}_{n+k}B$ for an integer $B$ coprime to $p_1,...,p_{n+k}$. We have $\frac {s(m)}{m}=\frac{\sigma(m)}{m}-1$ and $\frac{\sigma(m)}{m}=\frac{\sigma(B)}{B}\prod_{i=1}^{n+k}(1+\frac{1}{p_i}+...+\frac{1}{p^{t_i}_i})\quad\ge\quad \frac{\sigma(B)}{B}\prod_{i=1}^{n+k}(1+\frac{1}{p_i})\quad \ge\quad \frac{\sigma(B)}{B}\prod_{i=1}^n (1+\frac{1}{p_i})$ which is larger than $\frac{\sigma(B)}{B}(k+1)$ by our choice of $n$. Finally, $\sigma(B)$ is the sum of the divisors of $B$, here including $B$, so $\sigma(B)>B$. Thus $\frac{\sigma(m)}{m}>k+1$, whence we conclude $\frac{s(m)}{m}>k$. Last fiddled with by Dubslow on 2012-05-26 at 23:33
2012-05-26, 23:45   #18
R. Gerbicz

"Robert Gerbicz"
Oct 2005
Hungary

5×17×19 Posts

Quote:
 Originally Posted by Dubslow http://www.aliquotes.com/Aliquote.pdf I can't offer any commentary at all on the mathematics of it; I'm just a messenger (I still don't really understand the first sentence of the proof of Theorem 1.)
Thanks for the translation! So val_{p_i}(m)=t_i means that the exact p_i power divisor of m is p_i^t_i. You give no definition just explaining it with the driver.

The proof is good, but if you know Lenstra's theorem then this proof is really triviality.

 2012-05-27, 05:36 #19 Dubslow Basketry That Evening!     "Bunslow the Bold" Jun 2011 40 "et on définit" 3 Theorem 1 "The (asymptotic) density of the integers n divisible by m such that m doesn't divide $\sigma^\prime(n)$ ( $=\sigma(n)-n$) is null (0). Further, we give the following [asymptotic density]|[asymptotic complexity] for any sufficiently large real number x: $A_m(x)\quad := card\left{ n\le x\quad such\quad that\quad m|n\quad and\quad m\dagger\sigma^\prime(n)\right}=\mathcal{O}\left(\frac{x}{\left(\ln\ln x\right)^{\frac{1}{\phi(m)}}}\right)$. TN: Sorry about the dagger, I couldn't find a "does not divide" symbol that the forum would render. TN2: I guess I'll slowly translate the thing over the course of a few days or something, then copy and paste the tex into one post. Last fiddled with by Dubslow on 2012-05-27 at 05:55 Reason: adding part 2
 2012-05-27, 05:53 #20 firejuggler     "Vincent" Apr 2010 Over the rainbow 22×7×103 Posts As I said earlier, I shouldn't try to translate math-involving thing.
2012-05-27, 05:55   #21
Dubslow

"Bunslow the Bold"
Jun 2011
40<A<43 -89<O<-88

3×29×83 Posts

Quote:
 Originally Posted by firejuggler As I said earlier, I shouldn't try to translate math-involving thing.
I don't understand the math either.
Quote:
 Originally Posted by R. Gerbicz So val_{p_i}(m)=t_i means that the exact p_i power divisor of m is p_i^t_i. You give no definition just explaining it with the driver.
I assume you're talking to the OP?
Quote:
 Originally Posted by R. Gerbicz The proof is good, but if you know Lenstra's theorem then this proof is really triviality.
I suppose that's why he called it a corollary

Last fiddled with by Dubslow on 2012-05-27 at 05:56

2012-05-27, 10:40   #22
garambois

"Garambois Jean-Luc"
Oct 2011
France

100010000012 Posts

Quote:
 Originally Posted by garambois You are wrong ! It's a problem because the demonstration is in french. And the translation on the forum is wrong. If q divides n and q^2 don't divide n then : If n=qA and q and A are coprime Then : sigma(n)=(1+q) sigma(A) And because q==-1[m] then m will divide (q+1) So m will divide sigma(n) You have to take q==-1[m], it is written in the french demonstration. Did you read the french demonstration or the english demonstration on the forum before posting this message ??? Jean-Luc
I'm sorry but it was an error in the text !

Wrong (old post) :
If q divides n and q don't divide n^2 then :

Right :
If q divides n and q^2 don't divide n then :

Jean-Luc

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