Catch David on the Numberphile podcast: kzbin.info/www/bejne/b6qUc3qso7msh6M
@raydarable5 жыл бұрын
Why does the book show 1 going back to 2? Shouldn't it go to 4?
@ashkanledu22825 жыл бұрын
@@raydarable Cause as he explained in the video, an odd number multiplied by 3+1 is always even so he goes 1 step further and divides it by 2 So the formula would be : if it's even : n/2 if it's odd (3n+1)/2
@ashkanledu22825 жыл бұрын
@@tejaspathak375 Check my comment below, I get a bit deeper into this question
@raydarable5 жыл бұрын
@@ashkanledu2282 Thanks, didn't realize the cover was skipping that step.
@foxleo67295 жыл бұрын
Isnt this just a fractal if you layer all these solutions?
@scottanderson81676 жыл бұрын
“Seven seems to be an odd number.” That’s why he’s a maths professor.
@user-sx2lc4jt1r5 жыл бұрын
What d u mean
@HN-kr1nf5 жыл бұрын
@@user-sx2lc4jt1r he is clearly extremely skilled and educated in the field of mathematics
@debashisbarik48055 жыл бұрын
😆 😆
@sohee75975 жыл бұрын
>_>
@powerplay.5565 жыл бұрын
He's certainly no English professor. Love when people say "prehaps" (0:08) and then think they're smart.
@jalepenopvp23006 жыл бұрын
This guy has legit the most calming and comforting voice I’ve ever heard
@Nerine983 жыл бұрын
He reminds me of my university professor so it actually makes me nervous as it reminds me about the work I have to do damn
@Asofia202 жыл бұрын
it’s like mathematics with Bob Ross
@EliCreed Жыл бұрын
So frickin accurate 😤@@Asofia20
@severussnapeytp715Ай бұрын
I was ready to fall asleep, but the topic was too interesting lol
@finchisneat5 жыл бұрын
My friend in high school told me about this nearly 20 years ago, so cool to see a video about it!
@megauser85123 жыл бұрын
I notice that if you start with 0, you don't get to 1, but you keep looping back to 0 over and over again, since 0 is even. Also, at the end when they ask "Why not 3*n - 1?", I thought "Hey, if we do the 3*n + 1 problem with negative numbers, then if we let n = -m, where m is a positive odd number, then we get 3n + 1 = 3*(-m) + 1 = -3*m - (-1) = - (3*m - 1), and if m is even, then we get 2*n = 2*(-m) = -(2*m), so the negative cases reduce to the 3*n - 1 problem for positive n (if we take the absolute value)!"
@JohnSmith-xb4ux3 жыл бұрын
It's amazing how many teachers shouldn't be teachers , after just paying a few minutes of attention while I was eating I just posted the solution in the comments section.
@asiamies91533 жыл бұрын
@@megauser8512 0 not natural
@sahiba22973 жыл бұрын
@@megauser8512 The Conjecture is defined for Positive Integers ≥ 2
@Eleni_E7 жыл бұрын
I'm reminded of the wikipedia phenomenon where, if you click the first link in almost any article that isn't in parenthesis or italics, and keep doing that long enough, you will eventually end up at philosophy.
@traso567 жыл бұрын
tried nuclear power plant ended in philosophy tried C language, i'm looping when i reach math they have read this :o
@typo6917 жыл бұрын
And your comment reminds me of the six degrees of separation.
@DeathBringer7696 жыл бұрын
The Wikipedia degrees of separation from Philosophy... it usually gets you there within 5-10 clicks from my experience, lol.
@ThatJerryCamacho6 жыл бұрын
That's because philosophy is the source. Without philosophy, we could not be as advanced as we are. This is common knowledge to philosophers.
@I.amthatrealJuan6 жыл бұрын
Also try that with Hitler
@Croxmata8 жыл бұрын
And as we learned, Brady was most likely to choose 7 out of all numbers from 1 to 10.
@SatanicBunny6668 жыл бұрын
Actually, some studies have been done on this and turns out, if you ask people to pick a number between 1 and ten, 7 and 3 are the most common answer.
@Halo3Matalix8 жыл бұрын
But if there is a study showing that people pick 7 or 3 as a common answer, doing the test again would result in different numbers or the same numbers. we don't know if the data is correct simply because we can't tell if people are really just randomly picking those numbers or picking them on purpose because the information they got from the initial test. i guess what i'm trying to say is the moment we "record" something is the moment that bit of information becomes irrelevant.
@MugenJinSan8 жыл бұрын
There's actually a video about this on this channel, that's why Vexatos said that.
@furbyfubar8 жыл бұрын
Also, he obviously *knew* the subject he was asking about, so he was not very likely to pick 1, 2, 4 or 8 given how short a chain that would be... 5 would also be sort of short. It just made for the best example here.
@MadTiger200018 жыл бұрын
People often pick 7, sometimes 2 or 3. People like primes and 5 is too central, it seems too obvious.
@badhombre49424 жыл бұрын
This must be the formula that banks use, to calculate fees, that reduce my account to 1.
@mitodrumisra89723 жыл бұрын
🤣🤣🤣🤣
@mr.ditkovich63793 жыл бұрын
🤣
@simonreinsperger7188 жыл бұрын
"Wow... 16, very even number!"
@sarahvan38265 жыл бұрын
Because it's a power of 2!
@nohaylamujer5 жыл бұрын
That's right: it's quadruple even.
@arunpathak98514 жыл бұрын
Shut up
@thatoneguy89663 жыл бұрын
Square root, fourth root, divide by two, divide by four
@ibite1003 жыл бұрын
Your problem??
@andlabs8 жыл бұрын
My first introduction to the problem was one of the example in former Bell Labs Unix co-pioneer Jon Bentley's book "Programming Pearls", which is a collection of columns on the theory and practice of software design he wrote for seminal comp sci journal Communications of the ACM over the years, expanded and with exercises. One of the exercises in one of the early chapters was this conjecture (in the Second Edition it's Column 4 Question 5). In the back of the book is a collection of hints. Here is the hint for this question. It has stuck with me ever since I first read it: "If you solve this problem, run to the nearest mathematics department and ask for a Ph.D."
@elrak02196 жыл бұрын
Is there any reason that “n” can’t be negative? if you do it with a negative variable, it just ends up as -1 and not 1.
@danishqureshi85836 жыл бұрын
elrak0 2 I think it's because of the "+1", adding a positive to a positive is different than adding a positive to a negative, since 5+1=6 but -5+1=-4. But I'm no math genius, so idk. For example: -5 -5(3)+1=-14 -14/2=-7 -7(3)+1=-20 -20/2=-10 -10/2=-5 Cycle repeats.
@elrak02196 жыл бұрын
Danish Qureshi That’s exactly how it works, I believe there are four total loops fot starting at a negative(if you do a quick google search for 3n+1, you should find a Wikipedia page that has more than enough information on variation, loops, mathematics, etc.)
@alexandertownsend32914 жыл бұрын
@@elrak0219 n could be negative, but then it wouldn't be the Collatz Conjecture. A conjecture is an educated guess in math. The collatz conjecture was an educated guess by the mathematician Lothar Collatz. The conjecture is something like, "I think that every positive whole number n goes to one under this process". While these are probably not his exact words you get the idea. He said nothing about n being negative. You could ask about negative numbers, but that is its own separate, but still interesting problem. I hope that clears things up.
@tommyencrapera16293 жыл бұрын
What is the formula trying to decipher? Like why did they choose those numbers 3x+1 and not 4x+1 or 3x-1 and so on , what is the end game ? Is it the Only formula that always falls back to 1 and they dont know why? I'm trying to figure out what it would prove if there is a number that goes onto infinity or completely forms a separate loop , then what would that determine? Sorry for not understanding and seeing what I'm missing .
@nimets1233 жыл бұрын
it is interesting to look at the problem with binary numbers. dividing by 2 is just binary shifting to the right, because lowest bit is 0 for even number. 3n+1 is shifting the initial number to the left, adding the original number and then add 1. the middle path is reached when the highest bit is 1 and the rest is 0. they then can be reduced to "1" by shifting to the right.
@JimCullen8 жыл бұрын
Man I chose 4 initially. Worst possible number…
@blasttrash5 жыл бұрын
i chose 1. :P
@AlexKing-tg9hl5 жыл бұрын
Jim Cullen 2
@thefamousarthur5 жыл бұрын
4 goes to 2. 2 goes to 1.
@roylavecchia14365 жыл бұрын
I chose 0.
@WordoftheElderGods5 жыл бұрын
I picked 8.
@esotericVideos7 жыл бұрын
The first part of every numberphile video gets me excited to try to solve a problem, the second part convinces me that there's no point in trying because tons of people have already tried. :/
@numberboxgamer7 жыл бұрын
This is literally just "4 is cosmic" in math form lol I appreciate that very much.
@blu4able3608 жыл бұрын
Thank you for making this video!
@numberphile8 жыл бұрын
+Jake S (blu4) thank you for watching it. Please show your friends. :)
@eugenebayak68078 жыл бұрын
+Numberphile Ok (no)
@rafa3lico8 жыл бұрын
wat
@CoolJoe3308 жыл бұрын
Nissan > Suzuki
@1951split8 жыл бұрын
+Numberphile Can you explain to me why sequences that follow any number that solely consists of a lot of ones (like 1111111111111111111111111111111111111111111111111111111111111111111111 ) always follow the next general rule: {meaningless number}25{a lot of zeroes}4 {meaningless number}25{a lot of zeroes}2 {meaningless number}25{a lot of zeroes}1 {meaningless number}75{a lot of zeroes}4 {meaningless number}75{a lot of zeroes}2 {meaningless number}75{a lot of zeroes}1 and so on... the {meaningless number} gets bigger and the number of zeroes gets lower, until there are no more zeroes left and then the sequence turns ""normal"" like any other sequence...
@Faisal7102 жыл бұрын
I am working on Prime numbers and I am amazed to see your expression they are extremely useful in my work Thank you man
@simonmultiverse63493 жыл бұрын
The cover of the book is taking a short cut. It has an arrow going from 1 to 2 so it's showing two steps n-> (3n+1)/2 (combining the multiplication and the division by 2).
@Rock48968 жыл бұрын
what about the 360 noscope conjecture tho
@Xeverous8 жыл бұрын
link pls, I want to see this dank conjecture
@calholli8 жыл бұрын
360j -n +s = 360nc j= jump... n= n0sc0pe... s= sh0t... c= c0njecture I leave it t0 y0u t0 s0lve... if y0u can write this int0 a sentence, y0u win
@YoHoOMirster8 жыл бұрын
The 360 noscope conjecture equals a number between 360 and Conjecture.
@Biliklok8 жыл бұрын
MLG pro only
@xygomorphic448 жыл бұрын
it always goes off to 420
@maki62034 жыл бұрын
today was my first day at uni and my professor mentioned this conjecture. i was intrigued so i decided to look into it more THANKS NUMBERPHILE FOR COVERING IT
@ljfaag8 жыл бұрын
I found a counterexample, but the comment section is too small to contain it :P
@justlikedirt96348 жыл бұрын
Then your lying, or else you would take every way possible to show everyone collaz is wrong. Stop faking smarts to get attention.
@IceMetalPunk8 жыл бұрын
+JustLikeDirt I think you missed Ijfa's reference...
@sworupadhikari73198 жыл бұрын
+JustLikeDirt. Its a joke. Fermat wrote that he had a an elegant solution to his last theorem but the margin in the book he was studying was too small to contain it.
@rohitg15298 жыл бұрын
+JustLikeDirt it's a joke. it's what Fermat did about his theorem
@philofblood8558 жыл бұрын
made my day
@garrett38838 жыл бұрын
I was messing around on my calculator and I think I found a similar problem. It has 3 rules. If even dived by 2, If divisible by 3 dived by 3, IF the number isn't divisible by 2 or 3 the multiply by 5 and add 1. do this and It always seems to get stuck at the loop 6, 3, 1, 6, 3, 1. Or depending on If you divided by 3 or 2 first 6 ,2, 1, 6, 2, 1. This works regardless of the number.
@TimStamper892 жыл бұрын
The Garrett guesstimate ?
@yashrawat94093 жыл бұрын
*" Don't Judge a b̶o̶o̶k̶ problem by its c̶o̶v̶e̶r̶ statement "* *- Collatz Conjecture*
@robinaylott12642 жыл бұрын
Genius!
@the10000thspoon3 жыл бұрын
When he said any fourth grader could understand it, I was like, that's probably an exaggeration. Then he explained it and sure enough I remembered reading about it in a children's math book called "Math For Smarty Pants" in elementary school.
@Xonatron3 жыл бұрын
Ha, awesome.
@Albeit_Jordan5 жыл бұрын
I think the key to this conjecture takes the form of yet another such - if we find the probability of *n* either being odd or even in the simple expression *x∙y=n* (where x and y are odd and even whole numbers) then I reckon it'd just be a game of permutations from there.
@air9music5 жыл бұрын
"I bet the person who found it thought maybe he was on to something..." I'm willing to bet the guy who found the tree for 63,728,127 wasn't sitting down and doing it 😂
@CineGeeks0014 жыл бұрын
Hold my java program for finding the steps
@haoyu79375 жыл бұрын
I have a truely marvelous proof to the collaz conjecture whitch this planet is too small to contain
@АбдаллахМуслим5 жыл бұрын
)))))actually, math is going to end in this way for every single new proof someday)))))))
@arpitdas42635 жыл бұрын
Outstanding move
@bambolere5 жыл бұрын
Fermat is everywhere
@air9music5 жыл бұрын
Is that you, Fermat?
@alexandertownsend32914 жыл бұрын
Aaahhh! Zombie Fermat!
@TimJSwan8 жыл бұрын
You did it!! The Collatz! Thank you! Now, I am about to be away from electronics for 9 months. At least I'll know that you guys made a video on the conjecture like I asked. I tried to prove it as an exercise. Going to watch the extra footage, of course. :)
@kevingil18174 жыл бұрын
0:16 He chose seven on purpose! He knows it's the least random number! Bold strategy.
@doodlefox98378 жыл бұрын
I like how it started like the fibonacci sequence; 1, 1, 2, 3, 5. I wonder if it follows that pattern even longer with a bigger tree..
@travispetit24108 жыл бұрын
Finally!! A video about the 3x+1 problem :) Liked
@davidhahnert29143 жыл бұрын
I didn't know that mathematics had a Bob Ross. Homeboy mentions trees, clouds, and visual patterns in his explanation.
@BurakBagdatli8 жыл бұрын
Thanks for ruining my life. I've been absolutely obsessed with this since the video came out and cannot function as a productive member of the society anymore.
@manijoy64376 жыл бұрын
i feel your anguish
@tgrizz213t5 жыл бұрын
so does he still find a pattern?
@thesheq50235 жыл бұрын
Burak Bağdatlı zero will not result in one
@zer0python5 жыл бұрын
How far did you get? I went as far as translating the function over 4x. But stopped working on it. I'm sure if I analyzed the x sequences, I'd end up just as lost as everyone else. :-) -- I prefer spending my time on the factoring problem though. It's just such a fun challenge. For anyone else who wants to play with it (yanked from my notes): Lets recap, originally we defined the function as follows: f(2n) -> n/2 f(2n+1) -> 3n+1 but with our new found knowledge we can define it further as: f(4n+0) -> n/2 f(4n+1) -> 4 * (3 * (n/4) + 1) f(4n+2) -> 4 * floor(n/8) + (2 + -1^((n/4)+1)) f(4n+3) -> 4 * (3 * (n/4) + 2) + 2
@esotericVideos5 жыл бұрын
Don't solve it before me! lol I've been working hard on it. I keep thinking I have a thread on a solution.
@jess47508 жыл бұрын
I love watching your videos! They always give me something new to think about, as a future mathematician it's great to get a glimpse of what I could be working on!
@apollion8886 жыл бұрын
Great speaking voice, I'm sure you are a terrific teacher
@stephen28768 жыл бұрын
73 takes 73 steps to reach 1
@numberphile8 жыл бұрын
+Stephen H I've not checked but of so, that's awesome. How many numbers have that property?
@rafa3lico8 жыл бұрын
so many questions can be made.. damn
@stephen28768 жыл бұрын
My bad. Some poorly written code on my part, I forgot to set up the count properly (oops). 73 actually has 115 steps. The only number in the first 100 numbers is 5, although the number 91 comes close with 92 steps.
@stephenkamenar8 жыл бұрын
73 takes 73 steps if you use the shortcut he mentioned to do (n*3+1)/2 as one step
I feel that this problem may be easy enough to find a counter example for if you could search for loops without testing any number. Maybe try plugging in 3(2x)+1=x, or other equations that show the process that might occur to any number, and if you find that a number that would, following the rules of the conjecture, normally be able to follow such a pattern, you would find a loop that hopefully does not end in 1.
@JHashcroft Жыл бұрын
Hey it’s been 5 years, did you manage to solve the Collatz conjecture?
@subscheme Жыл бұрын
lol@@JHashcroft
@KirosanaPerkele8 ай бұрын
Surely 6 years was sufficient to find your counter example. After all, you said it's easy enough. Let's hear it.
@dragoda2 жыл бұрын
Veritasium also made a great video about this conjuncture. I love math and your channel. Please never change.
@reziik69048 жыл бұрын
I just learned to make simple stuff in python yesterday and my second program was making a loop to take a random number between a set range and do this.
@davidwiatr84028 жыл бұрын
Thanks for the book reference. I ordered it!
@lukeblankenberg73716 жыл бұрын
It seems to me (although this has probably been pointed out many times) that just because (3n+1)/2 > n we do not have to show that there are more n/2's in order to reach 1. What we have to do is show that it hits some power of 2 (as then it will immediately go to 1)(the last 1/2 should be fine as hitting a power of 2 or the next power of 2 is equivalent for the purposes of going to 1). The issue then is whether this is the limiting case (all cases with more n/2's in them before a power of 2, so that there aren't equal amounts of them, will all go to a power of 2 if this one does/they will eventually goes on a track that goes to this alternating between even and odd sequence that we might be able to show goes to a power of 2, or if some do not necessarily do so and so go to infinity). However, I have a nasty feeling that actually showing this is pretty hard. But I thought I might as well throw this out there.
@joaovitormatos81477 жыл бұрын
This conjecture was in my maths test, it asked how many non-repeating primes was in the Collatz Conjecture. People failed because they thought 1 was prime
@logan26698 жыл бұрын
I actually bet my friend 20$ that he couldn't do the 3n + 1 without getting to one I won the bet
@cheesywiz94435 жыл бұрын
xD
@leobitencourt47197 жыл бұрын
I think it has to do with an odd number multiplied by 3 adding 1 always being an even number, but an even number divided by two could give ya an odd number or another even number, so it kinda eventually forces it into powers of 2 and we all know it's downhill from there
@kb98802 жыл бұрын
Yeah that's the intuition behind the "conjecture" but we can't prove that eventually it will reach powers of two.
@thej3799 Жыл бұрын
@@kb9880 I don't think you can ever prove by doing it the hard way from here the upper limit of a diverging infinite set of " " prime values when he talks about records he's making a very important point because what is the biggest number in the list of records how can you define that as being the biggest without a sense of something greater it means nothing without the idea that there is yet another number. I was thinking this morning about this powers of two thing it's weird to see it show up in my KZbin feed because I wasn't saying much of it out loud. I was thinking of the difference between square roots cube roots and different numerals versus decimal integers so I found the whole thing fascinating to think about but part of it was because you're dealing with recursion once you start adding a decimal and going beyond and so you're back into these recursive trees when thinking about decimal roots it's weird that this would show up I like this video.
@seadub49444 жыл бұрын
2^950 Now i just need a universe large enough. Maybe ill find a lever large enough there too. 🎣🎣🎣
@parsec53663 жыл бұрын
You can: 2^950 = 9516908214257811601907599988159363584840065290620124537956939899622020205826587990689077212775400643774711832257235027522909345571487396529861315719055325605011013378863743193233193022939505515969530853007049198118833591724018432564205433218231411731277088674906521042072098232413978624 Steps 950
@ー-ーー7 ай бұрын
Here are some axioms I could come up with General: 1. Odd number x Odd number is always results in an Odd number. 2. Adding 1 to an Odd number always makes it an Even number. 3. All Even numbers repeatedly divided by two will eventually turn into a Odd Number. Specific to 3x + 1: 1. In a 3x + 1 sequence the amount Odd numbers will always be less then the amount of Even numbers. 2. There is never two Odd Numbers in a row during a 3x + 1 sequence. 3. As the Even Numbers size increases the amount of divisions by 2 in the sequence increase.
@enlongchiou6 жыл бұрын
Reciprocal of (3m+1)/2^n rule of Collatz conjecture is (2^n*m-1)/3, can use (2^n*m-1)/3=m for both, it's only solution is m=1.
@Transyst5 жыл бұрын
This only shows that it doesn't return to the same number after just one 3m+1 step, no matter how many consecutive /2 steps afterwards, except for m=1. But it doesn't exclude cycles with multiple 3m+1 steps, or the possibility that it doesn't end.
@HL-iw1du3 жыл бұрын
Please post more videos Enlong
@drenzine3 жыл бұрын
Here is something I figured out: If the numbers will eventually end in a loop other than the 1-4-2 loop, it cant be in the pattern odd-even-odd-even... and so on, ending with an even and go back to the same odd number at the start. This might help!
@dmtc6913 Жыл бұрын
I did some extremely precise calculations and all and here are the results. There's an infinite quantity of powers of 2 and of numbers that lead to a power of 2. Every single step you take with any number is literally just playing with fire and in the end you will get burnt.
@drenzine Жыл бұрын
@@dmtc6913 so you solved it or somethin?
@dmtc6913 Жыл бұрын
@@drenzine My previous post included the full proof. There's no way to avoid the infinite number of death traps forever. Unless you could end up in a loop other than 1 4 2 1. Surely there's no such thing. My maths is too precise for that.
@drenzine Жыл бұрын
@@dmtc6913 wym by death traps? all will eventually lead to 1? just because there are an infinite amount of numbers that go to a power of two, doesn't mean other numbers will eventually branch into them. it's kind of like the that one orchard problem that numberphile covered (Tree Gaps and Orchard Problems) : in an infinite lattice grid, there is a line goes infinitely far but will miss all the lattice points if its gradient is irrational. maybe a special number is that line, and all the points are powers of two.
@dmtc6913 Жыл бұрын
@@drenzine I must admit that my comments were mostly tongue in cheek, sorry. However, I can assure you with the utmost confidence that all numbers are doomed from the start. I didn't have to do any maths, I just come equipped with that knowledge. And give it out for free.
@theroadie75376 жыл бұрын
I have been running this project on BOINC for so long i forget when (2 years?) and now i know what it is doing.
@Verrisin7 жыл бұрын
I see why it tends towards 1: 2>1.5+c (for all big numbers and all small are checked) but I wonder what keeps it from ever entering a loop...
@kinyutaka2 жыл бұрын
The numbers in the paths are on separate moduli. Take the number 7, and think of it as 128x+7 or 7 mod 128. Multiply by 3, add 1 and divide by 2 to get 192x+11 or 11 mod 192. Do that again, and you get 288x+17, then 216x+13, and 81x+5. Furthermore, you can cut out parts of these mods to show that the paths are part of even smaller moduli: 11 mod 192 is part of 11 mod 64. 17 mod 288 is part of 17 mod 32. And 13 mod 216 is part of 5 mod 8, leaving 5 mod 81 to be 2 mod 3 Because of this, we can clearly show the direction of the path along the branching tree of Collatz, and show that the number 7 (which we already knew) goes all the way to one. Generalizing it, we can show that any number above the point where we can verify has a matching pattern, a matching path, that lies within the numbers we can verify, and thus must also drop below itself to a number that has been verified, thus proving that all numbers, without exception, drop down to 1.
@lopata_of_death68945 жыл бұрын
"I've become more powerful than any jedi."
@moonman____4 жыл бұрын
Twice the pride double the fall
@zrebbesh11 ай бұрын
Here's another function that generates a different 'hailstone sequence': if n is divisible by 3, n/3 if n+1 is divisible by 3, 2n if n+2 is divisible by 3, 2n-1 Every positive integer eventually goes to 1.
@kgc060910 ай бұрын
That isn't true, since any number when entering phase of n~1(mod 3) cannot escape since 2n-1~n (mod 3) in that case, and it diverges. Perhaps you interchaned the cases of remainder being 1/2.
@zrebbesh10 ай бұрын
@@kgc0609 The remainder is never 1/2. The only division is by 3, and that only happens when it is divisible (with no remainder) by 3. There are quite a number of functions like this that generate various sequences. Collatz is the simplest I'm aware of. This function family has some other kind of deep connection that I have never been able to figure out. I once thought that if I could figure it out I could fully understand what causes these sequences to converge. But I never could. There's no way I know of to predict whether a function has the property or not, but I wasted quite a bit of time trying to figure it out.
@TheHereticAnthem207 жыл бұрын
5:43 and then eVENTUALLY
@accounttest16604 жыл бұрын
eV *ENT* U A ll y
@morgiewthelord86484 жыл бұрын
Your point?
@AaronHollander3146 жыл бұрын
Rule #1 has a lowering effect on any number and outputs an even number a certain percentage of the time. While rule #2 increases the number, it always produces an even number which is reduced further. Seems obvious that the number would tend downward. One step forward and two steps back. The misdirection is that you're tripling the number half the time, that's simply not the case.
@rajeevbagra52762 жыл бұрын
Agree completely.
@legendgames1282 жыл бұрын
But if there was a number where the steps you take are like this: (3n+1)/2, (3n'+1)/2, (3n''+1)/2,... keep going like that forever, then the number would never reach 1.
@evanherk2 жыл бұрын
what a lovely teacher.
@masonhunter27484 жыл бұрын
Xkcd collatz conjecture: “if it’s odd, multiply by three and add one, if it’s even, divide by two, and eventually your friends will ask if you want to hang out”
@lakshsind84634 жыл бұрын
Lol nice one
@masonhunter27484 жыл бұрын
Laksh Sind just remember, xkcd came up with it
@nixtoshi4 жыл бұрын
your friends will stop* calling to see if you want to hang out lol
@juliusalbe20708 жыл бұрын
So every number is eventually gonna hit a power of 2 and is done than. Even if they tend towards infinity, they would allways somewhere hit a power of 2 and than go all the way down to 1.
@numberphile8 жыл бұрын
But what if it gets caught in its own loop BEFORE it hits a power of 2 and drops to the ground --- for example see the extra footage linked in the description... If you use 3n-1 instead of 3n+1 that happens to 7....
@buildasnowman46018 жыл бұрын
Why would they always hit a power of two?
@austinmitchell88468 жыл бұрын
Maybe. We assume it is true, and all our evidence so far points to it being true, but we can't prove that it is.
@dariusdurian19108 жыл бұрын
This is mathematics that involves powers of 2. Something math could do but not us humans
@YourMJK8 жыл бұрын
+BuildA Snowman Because the only way for a number to become smaller (to become 1) is by dividing by 2, so if all will eventually get to 1 they must hit a power of 2
@medexamtoolscom4 жыл бұрын
I found a collatz calculator online and right off the bat found 2 numbers that last a long time. 447 for 80-something and 447123 which lasts for 138 iterations.
@jeremybuchanan47597 жыл бұрын
"If they never get to earth, of course, they're not hailstones."
@michaeldeierhoi40963 жыл бұрын
😄😁😆
@lilcriz91876 жыл бұрын
And this is, why I love mathematics :)
@esaedromicroflora12478 ай бұрын
i'm gonna call all the powers of 2 "very even numbers" from now on, thanks professor
@DjVortex-w8 жыл бұрын
Couldn't you build the tree in the opposite direction, ie. from the bottom up? In other words, start from 1 and do the reverse operations, always branching out when you can do both operations to the number.
@treufuss-yt8 жыл бұрын
Yes you can. But it doesn't help.
@DjVortex-w8 жыл бұрын
Treufuß It helps building a complete tree up to a certain depth, which will then help you with further calculations.
@DjVortex-w8 жыл бұрын
***** Uh, no. You do the reverse operations, branching out every time you can do them both. In other words, you build the actual tree, but from bottom up.
@ITR8 жыл бұрын
I think I saw a program that did that when I searched for more videos on the subject on youtube earlier
@MaxRay168 жыл бұрын
+WarpRulez You would only be getting even numbers
@GlobalWarmingSkeptic7 жыл бұрын
Extremely interesting! Just going over it I can already kinda see why this is a problem. You would have to find a set of numbers either where 3n+1/2 appears more than n/2, or you'd have to find a set of numbers that creates its own tree, which would probably be an extraordinary large number considering that, as numbers increase, it's less likely that you'll get a cyclical sequence.
@alantyte33177 жыл бұрын
A cyclical sequence is inevitable as there are only six rules for finding the pivot of the 1-chain which effectively boil down to 3. The length of the cycle triples at every step up. The numbers do get huge but are consistent. Easier to talk directly.
@tommykarrick91305 жыл бұрын
I’ve watched this video so many times and this is so frustrating because it is so intuitive and clear that every number will obviously get to one and yet we can not prove it
@tommykarrick91305 жыл бұрын
awebmate the interesting thing about the problem is that it will always get to one and we don’t know why If it had a clear simple proof it probably wouldn’t be interesting
@TheReligiousAtheists7 жыл бұрын
1:26 I always knew no one uses multiplication tables!!😃
@wotsac4 жыл бұрын
No, he just broke it down into something that fit in the (US) multiplication tables.
@chizzicle8 жыл бұрын
I feel like there's been a video about this here before, but considering how many maths videos I watch, it's entirely possible that was someone else's video, especially since some of the people I watch occasionally appear here as well
@DrGerbils8 жыл бұрын
Yeah. I remember seeing this recently as well, but a search for "collatz" doesn't turn up any videos I've watched recently.
@NoriMori19928 жыл бұрын
Pretty confident they've never ever covered this.
@tamiratgebremariam89834 жыл бұрын
If one can arrive at a power of 2, or power of power of 2, etc, then one is sure to arrive at 1 eventually. Categorization of the evens based on their 'proximity' to the powers of 2 could help.
@fyfferguy5 жыл бұрын
To the OP: What is the source of the graph I see in the "collage" at 2:45? It's the one in the bottom center of the screen with the red dots on white background. This is exactly the data set that I have been playing with since I first heard of this problem in my college days over 25 years ago. I'd never seen the data presented that way before, and then I saw "my" graph in your video! Thanks in advance! (and hopefully someone will see this ... :( )
@russellgokemeijer42693 жыл бұрын
I have seen that graph on the Wikipedia page for collate conjecture and I am sure they list a source
@chonchjohnch4 жыл бұрын
I decided to try a graph approach to this, specifically digraphs, every vertex has two incoming edges, one representing 3n+1 for some n and one representing n/2 for some n. Additionally each vertex has one outgoing edge which goes to a vertex representing the collatz function being applied. Basically every number has two incoming edges and one outgoing edge
@Knurf3 жыл бұрын
The next number with more steps than 63,728,127 has 1 step more and is double of it. And even though that 63 million number has 949 steps, the lowest number with over or equal to 1000 steps is 1,412,987,847 with exactly 1000 steps and is way bigger than that. And the number with the most steps that is under 4 billion is 2,610,744,987 with 1050 steps.
@michaelbauers88008 жыл бұрын
Is it just me, or is the quality of comments on this one weaker than usual? ;)
@oluwatoyinokwunwa49896 жыл бұрын
Michael Bauers Noticed that too. Thought I would see more educated comments.
@medexamtoolscom6 жыл бұрын
Roses are red, Violets are blue, 3n+1, Is eventually 2. Or maybe "Or divide it by 2".
@jameshuddle47125 жыл бұрын
(raises champagne glass) I found 21 to be amusing... But 27 simply did not know when to stop!
@adityakhanna1138 жыл бұрын
2:50 to the right, it's not a paper, It's an XKCD comic.
@YoHoOMirster8 жыл бұрын
it's comic on the collatz conjecture
@NoriMori19928 жыл бұрын
Well-spotted! Don't think I'm familiar with that page, I'll have to Google it later!
@rohanpandey20378 жыл бұрын
lol I saw that too
@edwardoliveira174811 ай бұрын
Dude has such a relaxing way of speaking!
@52.yusrilihsanadinatanegar794 жыл бұрын
2:36 I've done checking 10 ^ 100 + 1 on my pc and it still got 1. A googol plus one. Down to ONE.
@inx18194 жыл бұрын
how many steps?
@Quantris4 жыл бұрын
@@inx1819 It only takes 2173 steps. What's kinda interesting is that 10^100 + 2 and 10^100 + 3 also take this many steps
@inx18194 жыл бұрын
@@Quantris yeah I made a program for that as well, it was pretty fun. got the same results
@nutzeeer5 жыл бұрын
cant we roll this up from the back side by calculating numbers up from 1? like all the powers of 2, and deviations thereof. would guarantee completion i think
@lonestarr14903 жыл бұрын
That's basically what people have in mind when they're building up those trees. The problem is, how do you prove that every single natural number appears as a node eventually?
@kinyutaka2 жыл бұрын
@@lonestarr1490 Exactly. 1, 2, 3, 4, 5... they all come in very quickly. but 27 dooesn't appear until after the 60th ROW of numbers.
@Mat_Rix3 жыл бұрын
For any positive integer which can be reduced by the Collatz conjecture to 1, it can be proposed the special kind of equation. On the other hand If this equation can be created for the particular positive integer, this integer can be reduced by the Collatz conjecture to 1. (v1xr4 2105.0003) As another step it can be proven that such equation can be created for any particular positive integer.
@BucketCapacity8 жыл бұрын
I have studied the Collatz Conjecture in my spare time. The closest thing I got to anything useful was, when n =/= 24*2^x+1, x in the integers, n going to 1 in the collatz function is equivalent to there existing integers a and b such that 4a | n+b and b | n +4a. This does work for certain n in the form of 24*2^x+1, but not all of them (ex n = 193)
@thesheq50234 жыл бұрын
If you look at (3k+1)/2. It is consistent that after some trials you will find a number that fulfills the request of (2^n). So really you’re looking for a number that doesn’t satisfy the equation (3k+1)/2 = 2^n + 2^q + 2^..... so find a number that when put into (3k+1) isn’t a series of 2, however i don’t think one exists...
@ClaudioBrogliato5 жыл бұрын
The only pattern I see is that powers of two with even exponential, once you subtract one, become odd numbers that can be divided by 3.
@cuentafake1405 жыл бұрын
If you have odd powers of two and add one, that number is divisible by 3
@giridharlprabhu89395 жыл бұрын
This is a result in number theory which is more formally defined as 4^x-1|3
@official-obama3 жыл бұрын
@@cuentafake140 subtract* Also the only odd power of two is 1
@cuentafake1403 жыл бұрын
@Orion Hunter Oh, I meant to say that 1 + 2^(odd number) is always divisible by 3, for instance: - 2^5 + 1 = 33 - 2^7 + 1 = 129 - 2^15 + 1 = 32769 My english is pretty mediocre at best so... sorry z.z
@0megaPi7 жыл бұрын
I've seen some comments here, proposing 0 as a solution to the problem. Though, 0 is a whole number, the definition of Collatz Conjecture calls for positive integers, thus we can't use 0 as an example of a number that doesn't fall to 1, because 0 is neither positive or negative integer.
@maldova3 жыл бұрын
Isn't this another unintentional asmr video? Professor's voice is so calming
@ZolarV5 жыл бұрын
I have the solution. It's very simple and elegant, the collatz conjecture is a corollary of it. I'll publish it on arxiv once i fix the lettering. I tend to reuse symbols while writing the proof.
@АбдаллахМуслим5 жыл бұрын
are you kidding... or???
@ZolarV5 жыл бұрын
@@АбдаллахМуслим no. Seems weird that i would kid.
@usualunusualkid71494 жыл бұрын
@@ZolarV Have you finished it yet?
@ZolarV4 жыл бұрын
@@usualunusualkid7149, you know I have not. I still haven't been able to get a mathematician to take me seriously enough. I'm even thinking about straight up signing up for undergrad at UC Michigan just to get access to Dr. Lagarias. I only have a minors in mathematics, my major was general science. I was only 2 or 3 classes off from having a second BS in mathematics however.
@charlieangkor86495 жыл бұрын
proof: 1) we postulate that Collatz conjecture is valid as an axiom. There are axioms in mathematics, and noone can prescribe you what you can accept as am axiom 2) proof follows trivially. QED.
@ilprincipe80945 жыл бұрын
Harvard: You want a free scholarship?
@mosshotep5 жыл бұрын
the problem is not just proving it, the problem is proving it using the given axioms
@nohaylamujer5 жыл бұрын
But then you have to prove it IS an axiom and not a property derived and provable from other properties. Math works that way.
@alexeyvlasenko66224 жыл бұрын
Actually, you can't add axioms arbitrarily. They must be consistent and independent of each other. Adding the Collatz conjecture as an axiom would only make sense if it's independent of the existing axioms of real number arithmetic (i.e., undecidable), and we don't know if that's the case. For instance, if we ever find a counterexample, we'll know that the Collatz conjecture is false, and so then we certainly can't add it as an axiom. Or, if someone proves that the conjecture is true, then it isn't an axiom either, since it would follow from the others. Of course, if you can prove that the Collatz conjecture is undecidable, then you can add it as an axiom.
@alexeyvlasenko66224 жыл бұрын
@PotatoTornado You're right. If it's unprovable, then it's impossible to prove that it's unprovable, as this would imply that it is true. So, it can't be provably unprovable, but it could be unprovably unprovable. But we still can't safely stick it into arithmetic as another axiom, because this would be inconsistent if a counterexample does exist.
@WaitTillNextYearChi Жыл бұрын
The Collatz Conjecture states that for any positive integer n, repeated application of the Collatz function f(n) = n/2 (if n is even) or f(n) = 3n+1 (if n is odd) eventually produces the number 1. To prove this, we will first show that f is a well-defined function from the positive integers N to the eventual outputs {1,1,1,...} of the Collatz function. We will then show that f is injective, meaning that different inputs produce different outputs, and that f is surjective, meaning that every output is produced by some input. This will establish that f is a bijection and hence that the sets N and {1,1,1,...} have equal cardinality, proving the Collatz Conjecture. To show that f is well-defined, suppose that we have two different sequences of Collatz function outputs that start from the same input n and eventually reach different outputs. Then, by definition of the Collatz function, these sequences must differ at some point. However, this contradicts the fact that the Collatz function is deterministic - given an input, there is only one possible output. Therefore, the Collatz function is well-defined. To show that f is injective, suppose that there exist two different inputs n and m such that f^k(n) = f^k(m) for some k ≥ 0, where f^k denotes the k-th iterate of the function f. We will show that this leads to a contradiction. Suppose that n and m have the same parity (i.e. they are both odd or both even). Then, applying the Collatz function once to both inputs produces two new inputs, either both even or both odd, which are still different from each other. Since the parity of the inputs remains the same, this process can be repeated indefinitely, producing an infinite sequence of different inputs. Therefore, if n and m have the same parity, they cannot produce the same output after repeated application of the Collatz function. Now suppose that n and m have different parity. Then, without loss of generality, suppose that n is even and m is odd. Then f(n) is even and f(m) is odd, so f^k(n) and f^k(m) will always have different parity for any k ≥ 0. Therefore, if n and m have different parity, they cannot produce the same output after repeated application of the Collatz function. In either case, we have reached a contradiction. Therefore, if n ≠ m, then f^i(n) ≠ f^i(m) for all i ≥ 0, and hence f is injective. To show that f is surjective, fix k ≥ 1. We will explicitly construct an n such that f^i(n) = k for some i ≥ 0: Let n0 = k. If n0 is even, set n1 = n0/2; otherwise, set n1 = 3n0+1. Let n2 = f(n1) (apply f to n1). Continuing in this way, we construct a sequence ni decreasing to 1. Let n be the first ni that satisfies f(ni) = ni. Then f^i(n) = k where i is the number of steps taken. To see that n exists, we note that each ni satisfies ni ≥ 1, and so the sequence ni must eventually reach 1. Moreover, since f^i(n) is strictly decreasing, there can be at most one ni satisfying f(ni) = ni. Therefore, for all k ≥ 1 we can construct an n such that f^i(n) = k for some i ≥ 0, which shows that f is surjective. Since we have shown that f is injective and surjective, we have established that f is a bijection from the positive integers N to the eventual outputs {1,1,1,...} of the Collatz function. Therefore, the sets N and {1,1,1,...} have equal cardinality, which proves the Collatz Conjecture. QED
@lionbryce101016 жыл бұрын
4:58 1 goes to 2 on the cover?
@SystemOfATool5 жыл бұрын
they used the shortened version for that tree. So... not 3n + 1 but (3n + 1)/2
@SystemOfATool5 жыл бұрын
3:23
@MonsterPianoPlayer4 ай бұрын
Tons of great Videos on Numberphile and Numberphile2! : ) Keep up the great work!! As I commented earlier on another KZbin Channel... I just wanted to officially say that I Solved the Collatz Conjecture at 3:34 AM, September 26th, 2024. I also solved it in the easiest way possible, and simple enough for kids in school to understand! It is now Solved without any doubt. I put in that extra time that I needed to figure it out, and who would of ever thought that I would of solved it with just a 1 and a 2. : ) Enjoy your Day!! ☺ MonsterPianoPlayer 😊
@GuySperry8 жыл бұрын
Brady please! your videos are so quiet. I love the material, but it's so hard to hear. then the ad for the next video blows up my ears.
@numberphile8 жыл бұрын
it's pretty normal volume on my computer... I'm reluctant to start an arms war with advertisers and/or attention seekers who jack up their volume more and more... But I'll check it out.
@zanzlanz8 жыл бұрын
My issue is mostly how much bass the voices have. I actually have to turn down my subwoofer to understand what you guys are saying (in a lot of videos actually), haha. Maybe that's why it's quiet for some people - all the noise is in the bass :)
@TypIch8 жыл бұрын
Just tested. To hear what they are saying: TommyEdison video: speakers 14% This video: speakers 31%
@Hjerpower8 жыл бұрын
The video doesn't seem to be quiet to me but some of the ads can be really loud
@GuySperry8 жыл бұрын
+Numberphile thanks for taking it as constructive. thanks for taking a look at it. I really love the content.
@bengalbasi47534 жыл бұрын
The " collatz behavior " is first shown when converting an odd p into even using p+1 and repeating collatz process infinitely. This collatz behavior originates from 1(p)+1. This exact behavior is observed using 3p+1. But loses using 5p+1.
@matetoth22925 жыл бұрын
When our computers will have enough memory, we might be able to get into a loop, that also starts and ends between two powers of 2.
@TheMinecraftACMan5 жыл бұрын
It's potentially an infinite problem, so memory isn't really the issue. It's finding an efficient enough way to program it that you get a reliable answer in a reasonable amount of time...
@PickleRickkkkkkk4 жыл бұрын
L'homme Baguette im getting P vs NP vibes...
@Carcharoth3134 жыл бұрын
When there's a still not understood randomness underlying the Collatz conjecture, couldn't you use that conjecture for encryption purposes / for creating random numbers?
@cezarcatalin14063 жыл бұрын
It’s not exactly the most efficient way of screwing around with cryptanalysts.
@alexritchie45867 ай бұрын
It's amazing how this problem seems so intuitive. That there's just something about the rules that seems so obvious the numbers will tend back to 1, but it's like having the solution on the tip of your tongue. Intuitively it's true, but you can't quite see why, despite the simplicity. I can see why people become obsessed with it.
@piipecek49608 жыл бұрын
What if I take the record holder for the most amount of steps to get to 1, and multiply it by 2? Boom, new record holder!
@polyacov_yury4 жыл бұрын
Then it exceeds the upper bound. 62 million × 2 is 124 million, which is more than 100 million.
@wetbadger21744 жыл бұрын
Then for being twice as big it only has 1 extra step
@official-obama3 жыл бұрын
@1729 math_blog what if you get an even number by that?
@official-obama3 жыл бұрын
@1729 math_blog and what if it’s a fraction?
@KevinHosley8 жыл бұрын
I hope that 7,382,036,203,215,362,117 works! Can someone check? ;)
@simonsallen8 жыл бұрын
7,382,036,203,215,362,117 terminates after 492 iterations
@KevinHosley8 жыл бұрын
How in the world did you figure that out so quickly? Wow! I was just fooling around when I said that number.
@pablossjui8 жыл бұрын
maybe it programmed something to check it
@justaway_of_the_samurai8 жыл бұрын
He probably wrote a program to calculate it
@miticander8 жыл бұрын
Computers are really fast
@grabern6 жыл бұрын
The sequence of 2^n in the middle column is also connected to a 5, which is connected to a 3, and is then attached to a 7. This shows that all integers of the form 5(2^n), 3(2^n), 7(2^n) and 2^n are all going to collapse to one. If we can prove that at least all odd numbers end up in any of these numbers, we can prove that Collatz conjecture. The first thing is that every second number in the 5(2^n) pattern is of the form 3n + 1 where n is a positive integer. 10 reduces to 3, 40 reduces to 13, 160 reduces to 53, 640 reduces to 213. 2560 reduces to 853. They are all of the form (10(2^(2n)-1))/3. In the 7's, every second number is also of the form 3n + 1. 7 reduces to 2, 28 reduces to 9, 112 reduces to 37, 448 reduces to 149. These are all of the form (14(2^(2n)-1))/3. This shows that in x(2^n) where x is of either of the forms above, it will reduce down to one. This doesn't prove anything; feel free to mess around with it though.
@wakeupnawaf8 жыл бұрын
why do they always draw on brown toilet papers?
@leedaniel20028 жыл бұрын
That's clearly not toilet paper
@EGarrett018 жыл бұрын
Who poops on brown paper?
@DjSamvy8 жыл бұрын
Remind me to never use a bathroom where you live if that's what the toilet paper is like
@marianpalko25318 жыл бұрын
iiDioxide Brown toilet paper, to use your vocabulary, is the iconic symbol of Numberphile.
@@usualunusualkid7149 absolute legend Unless you programmed it then legend
@dougr.23986 жыл бұрын
I was David Eisenbud’s father’s Student, 1975-76 at SUSB. Professor Leonard Eisenbud was on my (exit Master’s Degree) oral exam committee along with Dr. Nandor Balazs & Dr. Harold Metcalf, my sponsor. Some years later, I briefly met Dr. L. Eisenbud’s doctoral advisor, Dr. Geno (Eugen) Wigner at a NY Academy of Sciences meeting at which I solved a puzzle posed in a lecture by a guest Japanese Physicist, thanks to discussion of the Aharanov-Bohm effect, taught to me by Dr. Max Dresden. The puzzle involved a seemingly mysterious (it isn’t, once you know how it works) jump in flux quanta in a SQUID magnetometer.
@DiegoTuzzolo8 жыл бұрын
dis anyone notice the fibonacci sequence on the 7's tree? (quantities of numbers going down) 1 1 2 3 5 then the sequence ends
@lizardbaron37278 жыл бұрын
Stop with all the "first" comments, it accomplishes nothing and wastes your time.
@lizardbaron37278 жыл бұрын
By the way why won't Brady start another podcast, or has he got too much work already?
@lizardbaron37278 жыл бұрын
+Sebastian Karlsson just had to get it out of my system, it frustrates me.
@brightface50058 жыл бұрын
reading your useless comment and replying to it was a bigger waste if time, thanks :/
@awesomedude60298 жыл бұрын
First
@thomas10552138 жыл бұрын
first
@ManticoreRO2 жыл бұрын
I think I may have found the solution. Not sure what to do with it. It's a constant: 6
@ManticoreRO2 жыл бұрын
I will state it here: 1: we do not care about the even numbers. They always go down. 2: all odd numbers, 1,3,5, etc., by using the formula provided 3x+1 have between them a difference of 6. With f(x) = 3x+1, f(3x+3)-f(3x+1) = 6... always. All odd numbers, if applied the formula 3x+1 have a difference of 6 between them. 1 is 4, 3 is 10, 5 is 16 etc. 3: to prove the conjecture, you need to prove that the result of the formula applied for number n at any step is lower than the odd value f(of that step + 1). Remember that (3x+1) always produces an even number. By my calculations, for a number to fail to reach 1, it needs to be lower that -13/3 (my math is bad so I might be wrong), thus, proved. Or so I think until a true mathematician hits me in the head with it.
@Grizzly012 жыл бұрын
@@ManticoreRO "my math is bad so I might be wrong" 🤣🤣
@KeystoneScience4 жыл бұрын
If a number is to not reach one, a constraint is that it must be congruent to 63 mod 96.
@LeventK4 жыл бұрын
Why?
@alexandertownsend32914 жыл бұрын
I don't believe you. Prove it to me.
@Xonatron3 жыл бұрын
??
@paulmccartney82933 жыл бұрын
?
@paulmccartney82933 жыл бұрын
@@official-obama yeah but we dont care about non-positive integers
@AndrewErwin737 жыл бұрын
Any counter example would have to avoid any exponent of 2. Doesn't seem possible.
@armelstsrt7 жыл бұрын
If you can prove it isn't possible, then it's interesting, otherwise it's just some "it seems that..." which isn't bringing much more about this problem than what people already know
@idontwannanamemychannel84087 жыл бұрын
It's possible with a loop, which is the main thing you have to prove. How do you prove no number except 1 ever returns to itself?
@awawpogi30366 жыл бұрын
I don't wanna name my channel i have a proof don't call me a liar
@MGmirkin6 жыл бұрын
Evidence or it never happened... "Lawyer, Lawyer!" :P
@NoriMori19925 жыл бұрын
That doesn't follow. There are countably infinitely many positive integers that aren't powers of 2, and the gap between powers of 2 doubles with every power. So avoiding powers of 2 doesn't even rule out a countably infinite sequence of integers that never leads back to 1, let alone a finite loop that doesn't contain 1.