Avoid the Collatz Conjecture at All Costs!

You just need some mathematician's father to write a letter to his son telling him it's impossible and not to waste his time on it. Problem will be solved in no time.

I had some fun trying to make some progress on it by running in reverse, and it quickly showed how this problem depends on an infinitely-growing number of calculations, but while also showing that it is true for certain n mod x = 0. Still though, since prime numbers exist, we can know that there will always be more numbers which will appear within the gaps presented by these values of x where n mod x = 0. My method was to start at 1, and do the opposite operations, building branching trees of numbers. So instead of 3n+1 and n/2, you do 2n and (n + 1) / 3, taking only integer results of the latter. Each branch then consists of consecutive doublings of different starting numbers. E.g., one branch starts with 1 and immediately starts doubling, and it corresponds to all powers of 2. The next branch would be constructed by adding 1 and then dividing by 3, but that doesn't produce an integer result and so it is discarded. The next then would be to move on to the second item in the powers-of-two branch, and adding one, then dividing by three. That comes out to (2 + 1) / 3, and so is a result already within the tree, so it too can be discarded. The next operation would then be (4 + 1) / 3, which is also discarded as it's a non-integer. The next, however, gives us (8 + 1) / 3, which is 3, a new value. With this new value, we can construct 3(2^y), or the first branch of powers of two but with the multiplication of a factor of 3. This process can then be repeated to paint out classes of numbers which are covered by the conjecture, rather than having to work with individual numbers themselves. These classes take the form two to the power of all non-negative integers, times some collection of prime numbers. It establishes that, once a prime number is found to be within the conjecture, all numbers equal to a power of two times that number will also be within the conjecture. This reduces the problem from one of testing every number to testing only prime numbers, while providing the route to reach that prime from some number previously established to be within the conjecture. Now who knows whether that ends up being more calculations to do or not, because it requires a list of known primes, as well as mapping out the route from 1 to each of those primes. But, if we presume the conjecture to be true, we can then rely on naively finding the backwards route in reverse -- i.e. we can do the normal Collatz conjecture process to prime numbers to find their route to/from 1. So this should reduce the amount of calculations needed, once working with a dataset consisting of known prime numbers. This also, however, still falls short of coming anywhere close to proof of the conjecture, as it would require finding the path for all prime numbers, which would require there to be a finite amount of prime numbers. But yeah, this could give a method to establish the validity of the conjecture as to all numbers up to and including all known primes.

It's amazing how quickly this problem goes into the weeds -- these examples do a good job of explaining why exactly it is so hard. We try to come up with a simple heuristic about cycle length, and then the nature of the problem forces us into asking questions about the distribution of powers of primes.

Here are the most facts about that simple conjecture - thank you! Very thorough work!

Man, and all this time i've been vising my local Collatz Conjecture not knowing about the consequences of doing so!

super glad to stumble onto your channel! really cool talk!

Very entertaining talk. Thanks!

Great talk! Thank you 😊

Great video. Nice to have a compilation of what we do and do not know!

Thank God YouTube recommended me this great video, 47 minutes flew by really quickly

What a nice video! I'm very glad to stumble upon it, never knew where an attempt at solving the Collatz Conjecture can lead you, and it was very interesting to follow the history of it in mathematics

Very interesting watch, thank you!

This was an amazing video.

Can’t wait to see what type of curves we need to look at to solve this one. :)

Thanks for this it was very informative!

Just found your channel! New subscriber!🎉❤

great lecture❤

Very cool video! Have you considered making more long-form videos like these? I'd definitely watch :)

I love how the comments prove exactly how tempting this problem is to anyone with a basic understanding of maths. Yes, we've tried that. And that. And that. My favourite way to UNDERSTAND the problem is in binary (of course) since you can sum the 'even' rule by just truncating trailing zeros, aka only tracking significant digits. This is what the strange digital funnel images are based on. Then, make trees by working in reverse from 1. This let my brain actually fully viaualize the puzzle and realize it was approximately impossible.

Tried my hand at it once. Never again. Ate up too much time just after high school. I do think I’ll at least post my methodology and conclusions here quickly though, although mostly what I can remember. Given that the problem works on multiples of 3 and 2, I decided to split all numbers into 6 (3*2) groups, giving them letter names. A (1,7,…) is an odd number 1 above a multiple of 3 B (2,8,…) is an even number 1 below a multiple of 3 C (3,9,…) is an odd number that is a multiple of 3. D (4,10,…) is an even number that is 1 above a multiple of 3 E (5,11,…) is an odd number 1 below a multiple of 3 F (6,12,…) is an even number that is a multiple of 3. And with these you can start figuring out where numbers trend to, and which are more important and which aren’t. F will become a lower F or a C A,C,E will get multiplied by 3 to become a C, and then have 1 added to become a D. A D divides into either a B or an E A B divides into either an A or a D. With these rules in mind it’s pretty clear F and C don’t matter at all; they will eventually become a D, and there’s no path to reach either once that happens. These rules also demonstrate that a D will be reached no matter what. There is no path that avoids a D. In fact, you can use D (basically a 3n+1) as the start and ends of chained together functions, with 3 potential paths at each D encountered. 1: the E-function, goes D->E->D with an increase. 2: the B-function, goes D->B->D with a decrease 3: the A-function, goes D->B->A->D with a slight decrease except at D1 where it repeats (the 4-2-1 cycle). As such you just ignore every number that isn’t a D and figure out which function a D goes down, as they will go in the pattern A-E-B-E… for growing values of D. Also make sure to give each D a number, with D1 being 4 and D2 being 10. Finding all suitable candidates for going down a chain of functions isn’t too difficult as you just insert each function into the previous function’s “X” (or is it “n” here?) and solve algebraically. And yes, all 3 functions have a specific formula by how much they change the D when applied. No, I don’t remember them off the top of my head, but they don’t take long to figure out. Anyways, the first interesting thing of note is that since the formula for E-functions is just changing factors of 2 into 3 (ok, I do remember that one), you can’t infinitely string them together with a finite number to start with as you’d eventually run out of factors of 2 to convert into 3s, and be forced to undergo a B or A function. The other interesting thing is each time you stack A, B, or E functions with each other, the denominator will grow by a multiple of 2 or 4. For a path of finite length, this means each additional step is either halving or quartering the number of starting D values that will immediately go through it. This effectively debunks infinitely increasing repeating cycles, as the number of D values that could start such a cycle eventually decrease to 1/infinity with infinite steps (aka exactly 1 number on the number line that’ll result in this specific path) and with an infinite repeating growing cycle you could shift 1 down the cycle and find a second D value that gives the same result. Although thinking about it now there are some interesting applications of this method, as there can only exist 1 starting value for any unique infinitely long chain in the conjecture. Although I don’t think it helps much. Anyways, that’s about as far as I got. There was a pattern with the chaining of B and A functions in regards to the addition changes to the next D value, but that’s about it. Thank you for reading this.