mersenneforum.org 3n+1. a.k.a. The Collatz Conjecture.
 User Name Remember Me? Password
 Register FAQ Search Today's Posts Mark Forums Read

2022-11-04, 18:45   #12
chalsall
If I May

"Chris Halsall"
Sep 2002
Barbados

2×5,531 Posts

Quote:
 Originally Posted by science_man_88 Fair. I'm not good at either.
There are four (4#) elements in that entity-relationship graph. Two Venn spaces, admittedly...

2022-11-04, 18:50   #13
mart_r

Dec 2008
you know...around...

24×53 Posts

Quote:
 Originally Posted by LaurV The way they "proved" it is just like that: assume there is a smallest number which never converge to 1, let's cal it s, or "seed" and let's see what this number can be. [...] And so on, doubling the power of two every time, you get to 2^68, and you will just have "few" remaining cases which you can test "deeper". Like for example, under 128=2^7, you only need to test 13 sequences, i.e. those starting with 27, 31, 39, 47, 63, 71, 79, 91, 95, 103, 111, 123, 127.
Never seen this approach described so nicely before - then again, I never looked that closely into Collatz. Interesting though, and I might add the pertinent sequence A076227 here for further reading.

2022-11-04, 19:18   #14
science_man_88

"Forget I exist"
Jul 2009
Dartmouth NS

2×3×23×61 Posts

Quote:
 Originally Posted by mart_r Never seen this approach described so nicely before - then again, I never looked that closely into Collatz. Interesting though, and I might add the pertinent sequence A076227 here for further reading.
I thought about n, n+2 approach but it doesn't yield much n+2 goes to 3n+7, and n goes to 3n+1, then they divide by 2 giving a gap of 3, before meaning applying the n+2 to n comparison again we get n+4 and n yield a a gap of 6 which can be yielding a gap of three on division by 2 again.

2022-11-04, 19:42   #15
chalsall
If I May

"Chris Halsall"
Sep 2002
Barbados

2·5,531 Posts

Quote:
 Originally Posted by science_man_88 I thought about n, n+2 approach but it doesn't yield much n+2 goes to 3n+7, and n goes to 3n+1, then they divide by 2 giving a gap of 3, before meaning applying the n+2 to n comparison again we get n+4 and n yield a a gap of 6 which can be yielding a gap of three on division by 2 again.
I really don't understand why we all waste our time. Beyond wondering what babies see...

@SM88: Based on your observations, what might you logically work with? Or, more accurately, what can you use to *predict* a statistically significant signal?

This is a serious question. I doubt you'll be able to answer. Which is part of the point...

2022-11-04, 19:48   #16
science_man_88

"Forget I exist"
Jul 2009
Dartmouth NS

2·3·23·61 Posts

Quote:
 Originally Posted by chalsall I really don't understand why we all waste our time. Beyond wondering what babies see... @SM88: Based on your observations, what might you logically work with? Or, more accurately, what can you use to *predict* a statistically significant signal? This is a serious question. I doubt you'll be able to answer. Which is part of the point...

We get examples: 3 and 5 lead to 3(3)+1=10, 3(5)+1=16, then both fail to 5, and 8 which have difference 3. using that pattern again we can predict 7 will go to 11 as that's 3 away from 8. We then use it again to show 9 goes to 14. 14 and 8 are 6 apart. They are both on division by 2 and yield a distance apart of 3.

2022-11-04, 19:56   #17
chalsall
If I May

"Chris Halsall"
Sep 2002
Barbados

2×5,531 Posts

Quote:
 Originally Posted by science_man_88 They are both on division by 2 and yield a distance apart of 3.
Which was observed and documented thousands of years ago.

We tend to work with borderline seminal around here. Please try to keep up. Or, even better, stop wasting our time.

2022-11-04, 20:00   #18
science_man_88

"Forget I exist"
Jul 2009
Dartmouth NS

841810 Posts

Quote:
 Originally Posted by chalsall Which was observed and documented thousands of years ago. We tend to work with borderline seminal around here. Please try to keep up. Or, even better, stop wasting our time.
If the fact you can get two sequences In effect arbitrarily close apart from odd numbers then what is the sticking point.

2022-11-04, 20:07   #19
chalsall
If I May

"Chris Halsall"
Sep 2002
Barbados

2·5,531 Posts

Quote:
 Originally Posted by science_man_88;617169t ...then what is the sticking point.
Absolutely none. A stone played almost in sympathy of putting the opponent out of their misery... (Go metaphor, just in case you get the reference).

2022-11-04, 20:47   #20
science_man_88

"Forget I exist"
Jul 2009
Dartmouth NS

2×3×23×61 Posts

Quote:
 Originally Posted by chalsall Absolutely none. A stone played almost in sympathy of putting the opponent out of their misery... (Go metaphor, just in case you get the reference).
My point is it's bound by certain quantity difference. I said it got nowhere effectively. The difference has a bound. I helped with semiprime searching, Pari gp use improvements , even attempts to change the very test you to check for primes. I hardly think these are seminal ideas though.

Last fiddled with by science_man_88 on 2022-11-04 at 21:24

2022-11-05, 00:28   #21
mathwiz

Mar 2019

2×163 Posts

Quote:
 Originally Posted by science_man_88 If the fact you can get two sequences In effect arbitrarily close apart from odd numbers then what is the sticking point.
Try to think of it this way:

Reasoning about different $$kn+c$$ sequences is all well and good, and you can use a finite number of them to show the conjecture holds up to a certain $$n$$ limit.

But if you want to use this approach to show the conjecture holds for all $$n$$ then you'll need an infinite number of $$k$$ and/or $$c$$ to do it.

2022-11-05, 02:08   #22
R. Gerbicz

"Robert Gerbicz"
Oct 2005
Hungary

32·179 Posts

Quote:
 Originally Posted by LaurV Maybe your post is sarcastic and we don't catch it, but that is how they "proved" it to 2^68 (currently, it may be much larger, I assume 2^75 or so).
Wikipedia says it is 2^68, so that could be true. Someone from this forum has done it.

Quote:
 Originally Posted by LaurV You don't imagine they tested so many numbers, do you? 2^68 is 295'147'905'179'352'825'856. Even testing a million sequences per second (which is impossible, some sequences need days, and few need weeks to be tested) and using 10 thousand computers, you would still need about 900 years to go to 2^68.
It could be boring, but actually they have done this. Fix d (or not fix) and see if n=k*2^d+t then could we reach m<n, if yes then throw away the residue class completely, otherwise test all numbers from this class. A trivial speed up here is that you can do multiple iterations at once, say if n%256 is known, use a precomputed table for this.
And not forget that the maximal conjectured number of iterations is O(log(n)^2) if the starter is n, so these are not taking weeks (???). The average number of steps is even smaller, maybe only O(log(n)) but I could be wrong.

 Thread Tools

 Similar Threads Thread Thread Starter Forum Replies Last Post RMLabrador Miscellaneous Math 12 2022-12-26 18:58 Steve One Miscellaneous Math 21 2018-03-08 08:18 MattcAnderson MattcAnderson 16 2018-02-28 19:58 MattcAnderson MattcAnderson 4 2017-03-12 07:39 nibble4bits Math 1 2007-08-04 07:09

All times are UTC. The time now is 23:45.

Fri Jan 27 23:45:12 UTC 2023 up 162 days, 21:13, 0 users, load averages: 1.52, 1.11, 1.09

Powered by vBulletin® Version 3.8.11
Copyright ©2000 - 2023, Jelsoft Enterprises Ltd.

This forum has received and complied with 0 (zero) government requests for information.

Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation.
A copy of the license is included in the FAQ.

≠ ± ∓ ÷ × · − √ ‰ ⊗ ⊕ ⊖ ⊘ ⊙ ≤ ≥ ≦ ≧ ≨ ≩ ≺ ≻ ≼ ≽ ⊏ ⊐ ⊑ ⊒ ² ³ °
∠ ∟ ° ≅ ~ ‖ ⟂ ⫛
≡ ≜ ≈ ∝ ∞ ≪ ≫ ⌊⌋ ⌈⌉ ∘ ∏ ∐ ∑ ∧ ∨ ∩ ∪ ⨀ ⊕ ⊗ 𝖕 𝖖 𝖗 ⊲ ⊳
∅ ∖ ∁ ↦ ↣ ∩ ∪ ⊆ ⊂ ⊄ ⊊ ⊇ ⊃ ⊅ ⊋ ⊖ ∈ ∉ ∋ ∌ ℕ ℤ ℚ ℝ ℂ ℵ ℶ ℷ ℸ 𝓟
¬ ∨ ∧ ⊕ → ← ⇒ ⇐ ⇔ ∀ ∃ ∄ ∴ ∵ ⊤ ⊥ ⊢ ⊨ ⫤ ⊣ … ⋯ ⋮ ⋰ ⋱
∫ ∬ ∭ ∮ ∯ ∰ ∇ ∆ δ ∂ ℱ ℒ ℓ
𝛢𝛼 𝛣𝛽 𝛤𝛾 𝛥𝛿 𝛦𝜀𝜖 𝛧𝜁 𝛨𝜂 𝛩𝜃𝜗 𝛪𝜄 𝛫𝜅 𝛬𝜆 𝛭𝜇 𝛮𝜈 𝛯𝜉 𝛰𝜊 𝛱𝜋 𝛲𝜌 𝛴𝜎𝜍 𝛵𝜏 𝛶𝜐 𝛷𝜙𝜑 𝛸𝜒 𝛹𝜓 𝛺𝜔