/sci/ - Science & Math

>no cycles are allowed
What did he mean by this?

>>10734226
1, 2, 4, 1, 2, 4, 1.... is a cycle as it loops back to the first term. If any number has this property, then the conjecture should be disproven

Another interesting point:

it seems that all numbers 2^n - 1 will triple n amount of times before halving twice in a row.

1 -> 4 -> 2 -> 1
3 -> 10 -> 5 -> 16 -> 8 -> 4
7 -> 22 -> 11 -> 34 -> 17 -> 52 -> 26 -> 13

etc

>>10734214
In other words, prove that the sequence's prime decomposition always becomes a power of 2.
I did some thinking about this a while back.
I am going to paste my notes.
The Collatz path is generated as follows. Take a number n. Then:

1) if n is even, divide it by 2.
2) if n is odd, multiply it by 3 and add 1.

Repeat this and, the conjecture says, you will end up at 1.

Step 1 removes all powers of 2 from the prime factorization, so we only need to worry about odd numbers.
Step 2 adds a power of 3 to the prime factorization, but takes the number after that for the next iteration round

3*odd + 1 = odd*odd + 1 = odd + 1 = even
If every odd number after the division is less than the original odd number, the proof is complete
(3x+1)/2 < x if dy/dx((3x+1)/2) < dy/dx(x)
3/2 > 1, so this is NEVER true for any positive integer!
This means that eventually, this will need to have enough powers of 2 in the prime factorization of a given 3x+1 to cancel out all of the stuff accumulated.
This means there exists odd numbers x such that 3x+1 has an arbitrary number of powers of powers of 2 in its PF, and that the Collatz sequences will always reach such numbers.
The intersection at 1 may mean something in this graph (Figured it out, It’s just due to the 1->4->1 cycle being divided by 2^2)
Would there be some similar intersection for other cycles?

Note: iff a sequence reaches 1, then 3x+1 must eventually become a power of 2
Which odd #s turn into powers of 2? (i.e. 3x+1 = 2^y) We could work the cycle backwards for proof!
3x+1 = 2^y: x = (2^y- 1)/3
x and y both need to be integers
This appears to be true whenever y is even
Let’s call these numbers “First-order odds”
The numbers which will turn into First-order odds are of the form (2^n/3)(2^y-1)

To solve, is there a way to find the prime factorization of the next number in a sequence?
Each number has a unique PF, so every PF corresponds to a number. Check for patterns of each permutation of primes with a sieve

>>10734315
I bring this up as it complicates the strategy of proving that N must eventually reach a number less than itself (that is known to reach 1), since these numbers triple as much as they half, and so for a large n, it becomes difficult to pinpoint why the number grows so large and comes back around itself.

>>10734387
How do you plan to prove that all numbers will eventually reach x, such that 3x + 1 = 2^y?

I call a quick transform: Q(n)=(3n+1)/

If Q^x(n) is Q applied to n, x times, then Q^x(n) = n(3^x)/(2^x) + sum(j in 0->(x-1), (3/2)^j)

Therefore, for 2^x-1, we eventually get ½Q^x(2^x-1)

I think we have to prove that for every odd n, given a natural number z less than n, there is an x such that Q^x(n) = z2^r where r is a natural number

>>10734509
my head hurts

>>10734214
By induction it's true for n so it's true for 2n, and so on QED

>>10734648
what do you plan to do with the prize money?

>>10734679
bunch of hookers and cocaine

>>10734694
not very interesting desu

>>10734700
cause you don't get the reference

>>10734391
>since these numbers triple as much as they half
No, since every tripling is always followed by a halving.

>>10734805
yeah that’s what I meant

triple, half, triple, half....triple, half, half....

>>10734509
Simplified, if you do x triple-halfs on n, it is equal to: (n+1)(3^x)/(2^x)-1

So if n is 2^x-1, you'd get:
(2^x-1+1)(3^x)/(2^x)-1 = 3^x-1
Which proves what 10734315 said

>>10735655
plug in values. I don’t think that’s correct

>>10734214
This problem comes heavily down to a prime factorization problem. You remove all factors of 2 from any number and are left with an arbitrary odd number, then you must determine if the where that with an extra factor of three lies, and then make some progess.

Let's break it down into an important, and digestible problem: What does adding one do to a number's prime factor composition? Is there any algorithmic way faster and/or more general than plugging and chugging to determine this?

>>10735802
You're right

>>10736109
about time

>>10736110
Hold up
https://www.csun.edu/~vcmth02i/Collatz.pdf#page=4&zoom=auto,-49,751

>>10734214
Hint: draw a transit map with the powers of 2 as a backbone

I know a prof that is working on the collatz conjecture and he's using discrete dynamical systems and entropy, that's all I'm allowed to say... I just wanted to help you a lil bit brainlets

>>10736405
>viewing Collatz as a discrete dynamical system
Wow, how original. Bet nobody's thought of that before

If a number can grow so much, then what exactly is preventing the sequence touch upon infinite growing numbers without ever going down?

>>10734387
>only need to worry about odd numbers

no, say you find an odd number n that never reaches 1
2n is even and also never reaches one, and there are infinitely many even numbers (keep doubling) that also dont

did you guys do it

>>10738817
I am not saying that there are no even solutions, I am saying that all even solutions can be trivially generated using the prime factorization of an odd number solution because first step results in a loss of all powers of two in the prime factorization of such a solution.
Finding all odd solutions therefore directly solves all even number solutions as well, since every even solution is just an odd solution multiplied by 2^n.

Oh my gosh I think I just solved it

>>10739071
nvm I made a bad step. I was excited for a moment

I’ve found that if a solution N exists, then it is possible for all other solutions N * 2^n to only be reached by the number twice its size, if N is of the form 12k + 3. For when this is multiplied by any 2^n, and 1 is subtracted, it is never divisible by 3.

this place will never amount to anything
especially when it comes to fucking solving problems like this
only a deluded retard would unironically believe that this shithole will EVER achieve anything of value

>>10739448
Lurk moar

>>10739456
ooops little brainlet incel got triggered that his online support community that he spends his entire day on is absolutely useless and unproductive
protip: /sci/ has been having an "ok today's the day we finally solve the RH or AI or *insert random popsci problem*" pretty much every fucking week since it came out in 2010
the people here are all brainlets and the ones who aren't are so lazy and dysfunctional that their societal contribution is less than that of the average brainlet
you will NEVER amount to anything, ever. The guy who accidentally solved the permutations problem is just a very lucky exception and he's very likely not browsing this shithole anymore, as don't the 80% of people who have left as well which you can see from the constantly-dropping post activity ever since 2012

>>10739467
Lurk moar lol, there's a mathematical paper with a contributor of "anonymous 4chan user"

>>10739467
>wastes 3 minutes on pointless post trying to troll
This isn't /b/

Just prove that it's possible for every number

>>10739487
>there's a mathematical paper with a contributor of "anonymous 4chan user"
Which was a run-of-the-mill contributed-talk-tier obscure combinatorics result. The most interesting feature was that it said 4chan on it, a fact the other (jewish) authors tried to milk for all it was worth

>>10734315
2^n - 1
-> 3(2^n - 1)+1 = 3*2^n -2
-> 3*2^(n-1) - 1
-> 3(3*2^(n-1) - 1)+1 = 3^2 * 2^(n-1) - 2
-> 3^2 * 2^(n-2) - 1
-> 3^3 * 2^(n-2) - 2
-> 3^3 * 2^(n-3) - 1
->...
-> 3^k * 2^(n-k+1) - 2
-> 3^k * 2^(n-k) - 1
->...
-> 3^n * 2 - 2
-> 3^n - 1, which is even so it will halve again

>>10740445
Now here’s an interesting problem: can an integer 2^n - 1 reach 2^m - 1 such that m > n, and so on? If this were possible then the sequence could go on forever it seems.

The proof is trivial.

>>10740955
a true intellectual

>>10739467
>waiting ever so patiently for someone to take his bait
>replies 9 minutes later so as not to look *too* desperate
>posts the next iteration in his prized "laughing frog" collection
>all according to plan
>the master baiter in action
lol

>>10739560
what the fuck does this even mean

>>10739490
>>10741660
the fact that both of you legitimately think that I'm trolling and that I couldn't possibly be telling the truth says enough about the state of this place - you are so invested into the delusion that you're onto something great, in the exact same way that /pol/ legitimately believes that they're the enlightened redpill saviors of humanity, that you cannot accept that someone else wouldn't believe it either so your defense mechanisms flare up the moment he does and you proceed to dissect his posts and try to come up with any proof that proves to you that the mean poster didn't mean it and was only joking. You're basically one step away from becoming /pol/ and accusing me of being a jewish shill

Listen I'd love to have this place do something relevant so I can justify the 12+ years spent here, but it ain't happening. It doesn't have the organization, it doesn't have the dedication and it doesn't have the brain power necessary for something like this. At best, the majority of posters here are depressed adhd incel teenagers and the handful of people who aren't either that or that but just the adult version are too uninterested in wasting their time with this place. If you want it to finally evolve out of the standard disorganized forum structure into something higher that can solve problems like this, you must find a way to motivate the handful of smart people here into wasting the majority of their time in researching that problem but also exclusively posting their conclusions here and nowhere else, and good fucking luck with that because every time such a person starts contributing and the Tooker-tier people start posting with an equal voice right next to his posts, he'd conclude that this board is a schizo shithole and quit right away in the same way that many people into politics quit /pol/ after they realized that it's a group of meme-spouting teenagers and schizoid boomers

>>10742052
>high effort troll
Would you mind to go somwhere else?

>>10742052
okay, i believe you. i have no argument against this.