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Nice find, also don't mind other people too much.

Good video.i think one of the best video for beginners becouse you explain very slowly,wich for beginner is a good king.i also saw your video on the conjecure and i like you pattern.i am also working at the collatz conjecture knowing that i probably not solve it,but is fun finding these pattern.i hope you will upload new videos😊

Thank you for sharing this game! We need more!

Good

quadrillions of numbers have been analyzed so no it's not one step forward because your script checked 10000 numbers. also if the number is odd you can do (3x+1)/2 in one step because every odd number becomes even.

So basically what you were trying to show is that this function is strictly decreasing, and therefore theres no number which will diverge to infinity. Even if you somehow managed to prove this (which no one has yet), this is still not a complete proof, because there is still the possibility that the number forms a closed loop, and you also have to prove that too

Your phrasing is slightly wrong. The conjecture is that it will go to 1. Not that it might go to infinity, because, there is a possibility it might get into a loop.

REEEEEEEEEEE

I’ve been working on this problem algebraicly so much sometimes I forget what the actual conjecture is

I feel like the Collatz Conjecture would be fun for people, even those who don't like math, if it could be gamified in a fun way. Might be something I look into in the future.

Mt Graham 3267 high it's prominence is 1926. Mathew 19:26 with God all things are possible. Mt Graham is the holy Mt where the Vatican observatory is. There is a list of that particular number 3267 is.

It's a biblical algorithm. It is equal to 3267. It's a Easter date. All roads lead to Jesus. Pi is 3261967. Fine Structure Constant 1967. Pascals triangle 3267. John 1.1 3267 one parsec is 3.26 at 6700000mph is 3261967. Jesus was 33 1967 plus 33 is 2000 years. Subtract 26% from 186000 mps is 137640 fsc. It is an Easter date. All roads lead to Jesus.

I'm convinced that trying to investigate this just gives you schizophrenia

remember: there is no better discovery in math than rediscovery, for rediscovery must mean you took steps in the right direction

the terminating sequence could be other than 4->2->1

The "piano" regularity is easy to explain: one needs two pair numbers to go lower than one odd. Once in a while, the rest of this difference interferes.

That's exactly the approach I took when I tried to look for a way to make progress on this. If you only focus on odd numbers, you can write them in the form of 4n+k, with k = 1 or 3. 4k+1 numbers are trivial to handle, but 4k+3 cannot be handled in the general sense, you always will see numbers which do not conform to a any rule.

Enjoyed the video. But 'the amount of steps' is as painful to the ear as 'the number of milk'. Then again, the piano part more than made up for that.

I think it's really simple. 3n+1 can only apply once then the number is even. /2 can be applied as many times as needed to get to uneven. The probability of 3n+1 against /2 can easily be calculated and it turns out that /2 just outperforms 3n+1 so that you always end up with 1. I guess that 5n+1 would always lead to infinity but I have not tried it yet.

For all of you saying this has been done before... care to cite your sources?

Your graph looks a lot like a drum pattern )

It's a logistics problem, so it will find a solution within logic and we got logic that's how we construct abstract thinking...

Unfortunately, I think there is an error with the final implication that the pattern being finite proves the collatz conjecture. Even if the pattern held, it wouldn't prove the collatz conjecture. Assuming the 2,3 pattern holds, does not necessarily imply that every number reduces in size. It just means that if a number reduces in size, then it must do so according to the 2,3 pattern. There still might exist a number that blows up to infinity and doesn't apply to the pattern even if all other numbers reduce in size. Basically, you still need to prove that the pattern applies to all natural numbers and not just a specific subset of natural numbers. Still though, the 2,3 pattern is very interesting

The 3n+1 problem has been solved, the Collatz Conjecture is true and it has been proven. th-cam.com/video/ewZnfVO3wX8/w-d-xo.html

Remember, proving it doesn't go to infinity is not enough. You have to prove there is no loop other Than 4 2 1 4 2 1 also.

What if proving 3kn+3^(k-1)+...3^0 always reaches a power of 2... sure that assumes division doesn't occur in even cases but that might not be so important. The conjecture always feels like inevitable power of 2 is the solution.

You know, something interesting about the collatz problem is that if you reduce the problem to even -> n/2 and odd -> (n3+1)/2, you can entirely predict a numbers path for a number of steps m by looking at it mod 2^m

Thanks for this. I like what you have done here. Am I correct in stating that, an actual Proof would need to demonstrate the inner logic as to why ""any number reverts back to 4,2,1"" ??

Your claim at 4:00 can be mathematically proven, but going from that to a useful bound on the number of steps for the flight to drop sadly doesn't work (or at least not in any obvious way). About this last part, it sadly wouldn't be enough to prove the conjecture: this would give you a formula for the number of "3x+1" operations knowing the lengths of the flight in altitude, but not way to show for a given number that its flight in altitude has a length. Though maybe I misunderstood your explanations, so don't hesitate to correct me, or to ask if you want clues about how to prove your 4:00 hypothesis.

I don't see it stated this way, but the conjecture is equivalent to saying every series eventually produces a power of 2. THe powers of2 that you observed in the differences is intriguing.

Of course, if a series doesn't end at 1, does not imply that it increases without bound.

1:00 no, that is not true ("In other words that should go off to infinite"). It can have another loop somewhere else

cool result man. i love finding this kind of pattern. 🎉

It makes sense that 2^n would show up in multiple places when looking at this function. Because an entailment of the Collatz conjecture is that, if you the function for long enough, it will always hit a power of 2 in a finite amount of steps.

I love this insight! It's a unique way of seeing how the numbers are working together to build this complicated, but simple pattern.

The Collatz Conjecture is not really a true equation or a true algorithm because there is an "if" within it... it is a "function" though, but that is NOT nessisarily an equation. Equations may have an "if" outside the equation, like: x+1=1 if and only if x=0. In that case the "if" is not within the equation but rather an external "requirement" of the equation. What this Collatz Conjecture function is saying (in pseudo code): x={if odd(x)=true then: x=x*3+1 else: x=x/2} So, it is closer to a system of equations rather than a pure equation or pure algorithm. Look at the Wikipedia page he shows, see the "if"s? The equation has two "elements" to it... that's the giveaway. So trying to resolve this for infinity is pointless, because that would say that ALL systems of equations resolve for ALL inputs, no matter how MANY other elements are within it... that is: an infinite number of equations, with an infinite number of elements, with an infinite number of results. We can make conclusions that there are certain functions that provide predictable results for an infinite series inputs, but once the factors are "multiples" of infinity (like what I stated above which is infinity^3) then you can no longer make any proof or assumption about any single infinity - like a function that has infinite elements within it. So it's pointless. Like trying to prove: 1=1 | if over an infinite number of evaluations. ...well, duh! We assume that 1=1 period... forever in the past, now, and forever in the future. So if you want to waste your time trying to resolve the Collatz Conjecture, may I suggest you first prove: not(1=1)=true | if "ever" over an infinite number of evaluations. Hmmm, the Madelbrot set is jumping out at me on this. I wonder if "order out of chaos" could come out of the Collatz Conjecture if plotted by color vs. resolution rate.

It doesn't stop at one. It reaches an equilibrium, hypothetically.

Oh for fuck sake this is solved, powers of two cause a cascading reduction to a 421 loop.

I like the 2-adic extensions of the Collatz function. Basically since even numbers always become smaller, consider a function defined from positive odd integers to positive odd integers: f(x)=(3x+1)/2^a where a is the largest power of 2 that divide 3x+1. This can be restated as 3x+1 divided by the even part of 3x+1. Division by the even part of a number is known as the 2-adic norm ||•||_2. So, f(x) is easily extended as f(x)=(3x+1)||3x+1||_2 which can be applied to all 2-adic rationals.

This explains why The Guardian of this realm seems to favour The Collatz' Conjecture, aswell as the timeless design of pianos. They are, from The Guardian's perspective, one and the same. I salute your discovery, brother!

The reason it looks like a piano is because in music the 3/2 ratio is also important. 19 is the difference (or sum of differences) over 12 steps. 2^19=524288 is close to 3^12=531441 but not exactly so it eventually diverges.

As a next step, you can consider 'what does it mean that the pattern repeats every 32', and make the parallel that this means 'the last five bits, if the number is written in binary'. Then you might think how that related to the underlying 3k+1, k/2 mechanism, and think not of steps, but how many divisions by 2 you encounter. The patterns then become even more apparent...

I did my own experiments using values up to 4.000.000. Cutting at every 12th- step, I got the following "piano"-patterns: [2, 3, 2, 3, 2, 3, 3, 2, 3, 2, 3, 3] [2, 3, 2, 3, 2, 3, 3, 2, 3, 2, 3, 3] [2, 3, 2, 3, 2, 3, 3, 2, 3, 2, 3, 3] [2, 3, 2, 3, 2, 3, 3, 2, 3, 2, 3, 3] [2, 3, 2, 3, 3, 2, 3, 2, 3, 2, 3, 3] [2, 3, 2, 3, 3, 2, 3, 2, 3, 2, 3, 3] [2, 3, 2, 3, 3, 2, 3, 2, 3, 2, 3, 3] [2, 3, 2, 3, 3, 2, 3, 2, 3, 2, 3, 3] [2, 3, 2, 3, 3, 2, 3, 2, 3, 3, 2, 5] [3, 2, 3, 3, 2, 3, 2, 3, 3, 12, 8, 21] (The vaules in the last two patterns are propably incorrect, since they still can change if we calculate more values) Note you get only 4 "piano"-patterns starting on the note f#. Then it changes to 4 "piano" -patterns starting on the note c#. Then it propably changes to a g# scale. So both times it went up 7 halfnotes (a perfect fith). Don't think this pattern continues forever thogh, as my mathematical instincts that the periodicity would be linked to log_2(3) being rational, which it is not. It'd be funny if the collatz-sereies generated a full cricle of fiths before the pattern breaks, though :D

I thought you were gonna play it on a piano software. It’s never to late to plug it in