Collatz Conjecture in Color - Numberphile 

I'd say more like 7 billion, because everyone has their own favorite thing that is the most famous to them :)

IceMetalPunk some people may share things though so it must be MUCH less than 7billion

not even close to 7 billion, you dont think at least 1 overlapping of favorite concepts has occured?

But maybe they are all collapsing to one...

60²

Can you share the logic?

Never have i felt so understood. I too, have tried my hand at it...

You gave it a go! You gave it a Parker! That's the important thing.

I too, like probably many numberphiles, often dream of stumbling my way to solving an a open problem in mathematics. It would be incredible! I think it would most definitely require divine intervention or inspiration to solve something that has escaped the reach of professional mathematicians who dedicate their life, to take that immense creative leap that finally let's you see the problem in a way that is solvable. Has this happened before? Has an amateur solved an open problem the escaped the professionals? The best candidate I can think of is Ramanujan, who was self taught and isolated from the math would, iirc, and sent G.H. Hardy a number of incredible results?

@@kaziaburousan166 Kazi, I'm sorry, but I have forgotten any details of my false proof. I'm sure it was not salvageable.

fatsquirrel75 - yea, it's beautiful

Mathsgear pleasssse!

I'd like a print of that, put it on me wall!

Or a print. I'd have it on my wall in a flash.

i would love to jave a print of it indeed

I'm not sure the overlaps would allow you to use only 4 colors without coloring neighboring branches the same way.

He was probably refering to the four colour map theorem, check out that vid

The four color map theorem doesn't apply if there are exclaves so TiagoTiago is right. This map might very well not work for the four color theorem

You could colour it in 4 colours, but it might be impossible to colour one string in one colour, because sometimes a string overlaps another string and one string has to be drawn in 2 parts.

That's what jogiff meant by exclaves. An exclave is basically two regions which are disconnected but must have the same colour. This additional restriction means that a 4-colouring might be impossible.

To add to that, I think, that any loop would have to be outside the main tree (the tree with root 1). Reason is that applying the rules forwards (i.e. if n is even, next is n/2, if n is odd...) never diverges to two branches. To have a loop or cycle in the graph presented here you would need a node diverge in the forward direction, but they can only diverge in the backward direction. 1-2-4 is the only possible loop within the presented graph. Correct me if I'm wrong. That being said there could be numbers a, b, c, ... that form a closed loop but are separate from the main tree. It would be that not all n reach 1, but also don't go off to infinity. Could there be branches coming out of the loop?

@@josevillegas5243 it's been proven that if a loop exists it'd have to be at least size 17087915 outside the main branch (of course)

@@josevillegas5243 There are 2 ways the conjecture could be false, either there exists a loop other than 1, 4, 2 or there is a sequence that eventually goes to infinity. In either case there then exist a load of other numbers that drop down into those sequences. At a minimum there are all numbers that are a power of 2 times members of these sequences. Since it is deterministic once you get in a loop you stay in that loop.

Mister Bateman are you by any chance a descendant of Fermat?

Mister Bateman I have an original joke that the margin is too narrow to contain

You are closer than you realise. The proof is related to fermats theorum. IMO.

Shut up Fermat! You have caused so many mathematical headaches already LOL

*Bateman's Last Theorem* :)

Thanks for the reference! it sounds interesting :)

Yes! The issue is that we don't really know which kind of complexity class the collatz conjecture is in.... - On the inner digits is repeated multiplication by 3 in base 6, which scrambles everything so it's a chaos-dominated random cellular automata.... - On the outer digits, it's the loss of the last digit every time the last number is even, eating up digits, so it's a decay-to-fixed-state cellular automata.... - There's no proof that it's impossible to produce complex, localized structures like the famously Turing-complete rule 110 either....

That sounds like it is measuring "failure to exponentiate". Basically a "general resistance" in a general environment.

Yes, but why would you need to do that, if you're not constantly switching between them?

Monica, is that you ? :-)

I was thinking about this, too since those markers are fricking expensive.

... because he has gotten rich from youtube videos.

to divide the time taken by 3

They absolutely are. The universe is nothing but emergence, and it's beautiful in that :)

Layer upon layer of emergence creates some incredibly beautiful things indeed.

You seem to have it backwards. There is nothing simple or simplistic about these maths. They are what you might call infinite amounts of information condensed into a single phrase or word of semiotic linguistics. You only think that they are simple because you view them as their starting point. An acorn is very simple, but it contains all the vast amounts of information to form a tree, given the proper context. When you view the acorn and the tree as the same thing, expanded across time, you start to realize that it's not simple at all, but rather it is the single most complex system in all of existence. Other things in life are the same way. Human traditions, for example, seem simple and silly at times. But when you realize that Tradition is actually Democracy extended through time, handed down from generation to generation, you start realizing that traditions are vastly complex things which involved thousands or maybe even millions of people. A pencil is another example of something that is remarkably complex, but is a very simple object. It would be very difficult for any human being to produce a pencil by himself, without anybody helping him. It takes woodcutters and coal miners and all sorts of manufacturing plants just to get the raw materials and turn them into a pencil. Be very cautious not to mistake simple DESCRIPTIONS of something for the thing itself being simple. Being able to encode information about an object in very compact coded ways doesn't mean the object itself is simple.

Triumvirate888 - indeed. Feynmann said if you knew the names of all the birds you still wouldn't actually *know* anything about those birds. You just simply memorized the names. You have no wisdom. Identifiers, names, these things are not real wisdom, just rote knowledge.

Joshua Green I think it was from a previous video where it said they'd computed that all numbers up to 2^60 go down to 1

It is proven, that every number up to 2^60 goes down to 1, including eg (2^60)-1.

Joshua Green I think they meant "All numbers up to 2^60 go down to 1"

A Kaizofan All numbers that can be written in the form 2^n, where n is a positive integer, will go down to 1. (2^n)/2 = (2^n)/(2^1) = 2^(n-1)

It seems like these things will meander around but eventually they hit a factor of 2 and go to 1 pretty quickly

yes but it would not look 3d.

it's not a map, each line is continuous and goes under other lines. if it was shaded with only 4 colors it wouldn't look right.

I feel like it would be difficult to prove that all numbers will do this. I feel like somewhere there could be an odd number which, when doing this, it ends up with 1 0 to cancel, then the new number with only 1 0 to cancel, and so on, allowing it to be infinitely big. I feel like proving this would be equally difficult as proving the conjecture (especially considering that I kinda doubt no-one has thought of this until now)

19 = 10011 (3 ones) 19 * 3 + 1 = 58 = 111010 (4 ones) When 19 is multiplied by 3 and you add 1, you actually increase the number of ones. Additionally, 58 ends in only one zero. When you eliminate the zero, you get 29, which is greater than 19. There might be a number that does this forever, increasing to infinity. There could also be a cycle, where eventually you end up with the same number that you started with.

Brandon Sams, how about a program?

working on it, don't worry

Albert Zhang, I was here first!

Tyler Matthew Harris I'd say he looks more like Michael Sheen.

his ugly , nerdy twin then.

+Antony Jones , damn! You're right.

No one has ever seen Andy Serkis. Even the interviews he does are CGI. For all we know, he's a hyper-intelligent goldfish in a bowl, connected to a brainwave-interpreting computer through electrodes.

The ratio between odd/even numbera in sequences is exactly 0.66(counted in program), so it appears twice as frequently as odd number do.

how far did you go ??

Not "created by", just "described by". Mathematics is mankind's way of giving linguistic adjectives to nature, describing things and events in the abstract and extrapolating outwards to gain ever-more precise logical understanding of those things and events. I'm sure you already knew that though, since you did a thesis on it. I just wanted to make it clear for others who might read this and mistakenly think that mathematics creates things. If people like Stephen Hawking can make such a silly blunder, then anybody can.

Yeah thanks! That's why I put the "created by" in quotation marks :)

That's what I figured. Like I said, just wanted to clarify it for others.

It's in reverse order because the conjecture is that all branches die in 1, rather than start at 1 (which would just cycle 1-4-2-1).

Mage of Void is right imo. Connecting 5 to 8 makes as much sense as connecting 8 to 2: it's skipping a step. Sure, 5 will go to 8 as 8 will go to 2 but 5 gets to 8 through 16. Same with 13 to 20. It goes through 40 so should connect to it. The diagram is consistent in combining an odd step with its subsequent even step so no technically wrong as a schematic but I don't see how that is helpful.

Mage of Void... the tree splits at those nodes. Even to the right, odd to the left.

...I see what you're saying now. The line should go through the bigger number only.

*3×3+1=10 , 10÷2=5*

Are you a Medusa? O_o

yoou mean a gorgon. Medusa is an individual.

Mebbe

I mean, your avatar is a Trapinch, which has Arena Trap, so just like Medusa, if you look at it, you can't get away :3

You need some apply some collagen? Get it?... Because, collagen... Collatz

Oh and btw *3+1 is related to *2+2 which is much easier to understand.

How do you know _EVERY_ starting number will go from even to odd until it finds a power of two?

OrangeCreeper217 Because every odd number will have 1 added to making that odd number even.

OmegaMike 108 Thank you for the oh so insightful commentary on how the conjecture works. ( ͡° ͜ʖ ͡°)

Seriously, though, not all even numbers are a power of 2. How would you prove that it goes to a power of 2 at one point? How do you know if there isn't a number that doesn't go to a power of 2 at all?

OrangeCreeper217 Beats me, I didnt think I would get this far.

Too add what I was saying, primes have a relation to 3. 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199 What number or prime, less one divided by three would give a prime? Because ending on a prime would force a triple +1. Is there a number that less one of a prime that can be divided by three and is odd. I guess more or less, if it were to three times three. to keep growing it would work. Alternatively, one could look at how many odds could be hit. first few primes. Less one. Example. 1 2 4 6 10 12 16 18 22 28 30 36 40 42 46 52 58 60 66 70 72 78 82 88 96 100 102 106 108 112 126 130 136 138 148 150 156 162 166 172 178 180 190 192 196 198 divided by 3. / / / 2 / 4/ 6 / / 10 12 / 14 / / 20 22 / 24 26 / / 32 / 34 / 36 /42 / 46 / 50 52 54 / / 60 / 64 66 Perhaps an interesting thought to the conjecture is to ask? Well what times three gives me an odd when adding one? 0, 4, 7, 10, 13, 16, 19, 22, 25, 28, 31. what within that cycle can produce another 3n+1off within that sequence. Perhaps primes have nothing to do with the relation. It would seem, to hit a number that would be odd at atleast two intervals would be pretty rare. So 3n+1 and 9n+4 would have to have the same source number, and so on. 27n+13. Where 3n+1. What is the question asking. I could try to write it as 3n+1 = 9n+4 which is something or 3n+1 = 9n+4 = 27n+13 Where n is the same source number but is always an odd product. Which is not in any mathematical formula to force an odd number as a product as we often do not start from an answer and work back, but if we did, it would be something x = /2* condition.(odd check) Which sounds more like computer code, lol to add a some sort of condition. The above = is not wanting the same product as the product is different but a series of odds from n number that is odd. 3n+1 = x 9n+4 = x (solve for n) one could add more and more if it works in the first case of odd numbers that meet the formula and condition, into a series of odds that must be multiplied in succession by the formula. I don't know lol. So I want all the products to meet the odd check, which means it must fail to divide by two. N has to be the same in both equations. Which would give a number that would increase. The more I think about it along this root, I think the more I would get lost, as it gets away from math as a philosophical relation of numbers as a language of use, and measurement. SO what did this tell me. What would a proof really say or prove? I mean what does a solution give? If it has no means of practical application? THE SOFA question last week, has limits one can add, and has an answer or maybe even several answers that contain the same area. So there would be a maximum area. Of course the material of the sofa is considered solid, not malleable or bendy.. lol. Pretty uncomfy sofa. if n =10 31/94/283

cheers

I found it useful to look at 3x+n functions such as 3x+3, 3x+5, and 3x+7, however there are some limitations to rule modification that either complicate the study or are so simple it hurts to find answers. (For example, 7x+1 wanders to infinity and 1x+1 has almost everything loop or terminate instantly). The second problem is creating a function that is guaranteed to go to infinity (ex. 1x+4, 3x+8, 6x+3, etc.). The third problem is finding a multiple of a pre-existing function (ex. 2x+4 is really 1x+2, 6x+2 and 12x+4 are really 3x+1, etc.) I tried to experiment with dividing by three instead of 2 and I fried my brain. I would love to go back and look into that, however I would like to better understand the Collatz sequences first. If I am stuck with were I am, as a last resort I can experiment with dividing by different numbers. Experimenting with the rules is always fun. Just be careful not to get stuck into a loop of 70 something steps! (That was a long day...)

Sander Huisman Wow. Thanks!

Yeah, I just spent over an hour just now based on this comment trying to use the 5n+1 function for the number 5. 1 through 4 are in a loop, but it was 5 that threw me for a loop. I didn't get to finish, maybe someone else can finish what I started: 5 26 13 66 33 166 84 42 21 106 53 266 133 666 333 1666 833 4166 2083 10416 5208 2604 1302 651 3256 1628 814 407 2036 1018 509 2546 1273 6366 3183 15916 7958 3979 19896 9948 4974 2487 12436 6218 3109 15546 7773 38866 19433 97166 48583 242916 121458 60729 303646 151823 759116 379558 189779 948896 474448 237224 118612 59306 29653 148226 74133 370666 185333 926666 463333 2316666 1158333 5791666 2895833 14479166 7239583 36197916 18098958 9049479 45247396 22623698 11311849 56559246 28279623 141398116 70699058 35349529 176747646 88373823 441869116 220934558 110467279 552336396 276168198 138084099 690420496 345210248 172605124 86302562 43151281 215756406 107878203 539391016 269695508 134847754 67423877 337119386 168559693 842798466 421399233 2106996166 1053498083 5267490416 2633745208 1316872604 658436302 329218151 1646090756 823045378 411522689 2057613446 1028806723 5144033616 2572016808 1286008404 643004202 321502101 1607510506 803755253 4018776266 2009388133 10046940666 5023470333 25117351666 12558675833 62793379166 31396689583 155983447916 78491723958 39245861979 196229309896 98114654948 49057327474 24528663737 122643318686 61321659343 306608296716 153304148358 76652074179 383260370896 191630185448 95185092724 47907546362 23953773181 119768865906 59884432953 299422164766 149711082383 748555411916 374277705958 187138852979 953694264896 467847132448 233923566224 116961783112 58480891556 29240445778 14620222889 73101114446 36550557223 182752786116 91376393058 45688196529 228440982646 114220491323 571102456616 285551228308 142775614154 71387807077 356939035386 178469517693 892347588466 446173794233 2230868971166 1115434485583 5577172427916 2788586213958 1394293106979 6971465534896 3485732767448 1742866383724 871433191862 435716595931 2178582979656 1089291489828 544645744914 272322872457 1361614362286 680807181143 3404035905716 1702017952858 851008976429 4255044882146 2127522441073 10637612205366 5318806102683 26594030513416 13297015256708 6648507628354 3324253814177 16621269070886 8310634535443 41553172677216 20776586338608 10388293169304 5194146584652 2597073292326 1298536646163 6492683230816 3246341615408 1623170807704 811585403852 405792701926 202896350963 1014481754816 507249877408 253620438704 126810219352 63405109676 31702554838 15851277419 79256387096 39628193548 19814096774 9907048387 49535241936 24767620968 12383810484 6191905242 3095952621 15479763106 7739881553 38699407766 19349703883 96748519416 48374259708 24187129854 12093564927 60467824636 30233912318 15116956159 75584780796 37792390398 18896195199 94480975996 47240487998 23620243999 118101219996 59050609998 29525304999 147626524996 73813262498 36906631249 184533156246 92266578123 461332890616 230666445308 115333222654 57666611327 288333056636 144166528318 72083264159 360416320796 180208160398 90104080199 450520400996 225260200498 112630100249 563150501246 281575250623 1407876253116 703938126558 451969063279 1759845316396 879922658198 439961329099 2199806645496 1099903322748 549951661374 274975830687 1374879153436 687439576718 343719788359 1718598941796 859299470898 429649735449 2148248677246 1074124338623 5370621693116 2685310846558 1342655423279 6713277116396 3356638558198 1678319279099 8391596395496 4195798197748 2097899098874 1048949549437 5244747747186 2622373873593 13111869367966 6555934683983 32779673419916 16389836709958 8194918354979 40974591774896 20487295887448 10243647943724 5121823971862 2560911985931 12804559929656 6402279964828 3201139982414 1600569991207 8002849956036 4001424978018 2000712489009 10003562445046 5001781222523 25008906112616 12504453056308 6252226528154 3126113264077 15630566320386 7815283160193 39076415800966 19538207900483 97691039502416 48845519751208 24422759875604 12211379937802 6105689968901 30528449844506 15264224922253 76321124611266 38160562305633 190802811528166 95401405764083 477007028820416 238503514410208 119251757205104 59625878602552 29812939301276 14906469650638 7453234825319 37266174126596 18633087063298 9316543531649 46582717658246 23291358829123 116456794145616 58228397072808 29114198536404 14557099268202 7278549634101 36392748170506 18196374085253

mvmlego1212 I have not checked it and i have no idea how they did it but i would guess by the value of each branch where they intersect. could just be random though.

Since a different person colored it, it's kinda hard to know.

When you have an odd number and multiply it by three, you will have an odd number. Add one and you get an even number. So then you divide it by two. For convenience, some people make (3n+1)/2 into one step. 5 goes directly to 8 instead of 16.

to skip a step, It is common to use (3x+1)/2 as the odd rule, as it is already known to be an even number. Makes it a tiny bit faster on a computer.

But it can go to infinite or after some steps go back to itself. So you don't preview this cases.

or coral ...

Isn't (x+y)^2 different from x^2+y^2?

x^2 + y^2 = (x+y)^2 - 2xy = A^2 + 2B

31 337 Wait... so every number that you use must be a square number?

every odd number 2k+1 equals (k+1)^2 - k^2 and every even number 2k equals [(2k+1)/2]^2 - [(2k-1)/2]^2

it is a very easy problem! observe that in the following equation: a = ({a - n^2}/2n)^2 + ({a + n^2}/2n)^2 equality holds for all real a and n (except n = 0, of course) so we can simply put any integer value for a and any integer(or rational) value for n , and you will obtain a as a difference of two rational squares

en.m.wikipedia.org/wiki/Waring's_problem I didn't really get what your question was but here is something similar.

woah i didn't expect these replies lol +Yalim O basically you have to get a sum of a number by only using square integers (that are rational, that is 2^2 or 1.5^2), but only using it once. also, i'll look up this waring's problem. thanks!

Opposite conclusion depending on sign. odd->3*odd-1 versus odd->3*odd+1. So there is something in it.