This is the longest episode I've ever made. Let me show you an interesting pattern I recently found, and all of the cool mathematical connections I've been able to deduce about it.... Timestamps of different chapters: 0:00 - My Explorations Into Primeglitching 3:29 - Explanation of the Main Pattern 9:18 - Discoveries Among the First 50 Numbers 14:36 - Discoveries Among Larger Numbers 19:57 - Finding the Other Research About This 27:38 - What I Could Deduce About the General Pattern 32:30 - What I Could Deduce About Fixed Points 42:13 - What I Could Deduce About Two-Number Cycles 50:47 - Ideas for Larger Cycles and General Proofs 55:12 - Extending to Fractions and Square Roots 1:05:22 - Other Forms of Potential Primeglitching (See the video description for more information)
@ComboClass Great video, as often. If that's of any help, a function such as the one mapping p^e to p+e (component-wise) is said to be "multiplicative". Equivalently, it satisfies f(a*b) = f(a)*f(b) whenever gcd(a,b)=1. And this gives another reason to say that f1)=1.
What is this referring to lol, I haven't watched them for ages because I prefer this channel and a few other ones where it's one person presenting maths problems in a more personal way. I find numberphile too big screen type .I like the smaller scale channels like this.
That's what's so cool about math. Anything you learn about has hundreds of variations, it's like if every spot on earth had hundreds of different places a short walk away. with a little wandering around you can easily reach a spot no one else has ever been to.
holy fucking shit, so THIS is why you havent uploaded for a while.. wow. you fucking DISCOVERED SHIT THAT NOBODY HAS FOUND BEFORE??? AND THEN MADE AN HOUR LONG VIDEO ABOUT IT??? just... i dont even have words dude. thats insane
36:30 Noting that 4/3 < (6/5)^2 = 36/25, and that any odd prime higher than 5 contributes a factor less than 6/5 to growth, we see that the maximum additional growing factor that can be added to compensate for one power of 2 is 6/5. From 2^a to 2^(a+1), the additional shrinking factor added is [(a+2)/(2^(a+1))]/[(a+1)/(2^a)] = (a+2)/(2a+2), which for a >= 1 is less than or equal to 3/4, and therefore less than 5/6. For a = 4, we see already that (6/5)^4 < (3/2)^2 = 9/4 < 8/3 and therefore odd primes cannot compensate for the shrinking factor induced by 2; and for any higher a, the shrinking factor decreases faster than we can multiply 6/5s to compensate. Similarly at 39:50, 41:00, and 41:50, the combined factors (2^1)(3^3), (2^3)(5^2), and (2^3)(3^2)(5^1) shrink by (3/2)(6/27) = 1/3 < 5/6, (5/8)(7/25) = 7/40 < 5/6, and (5/8)(5/9)(6/5) = 5/12 < 5/6, respectively. Any higher power of 3 or 5 shrinks further without allowing more compensation. 39:00 Alternatively, note that a fixed point with 2 raised to the power of exactly 2 requires that the odd part also be a fixed point, which was just shown to be impossible in the non-trivial case. 42:00 Note no additional odd primes can be added since 3^1 and 5^1 together already create 3 powers of 2. 48:50 Counterbalancing the shrinking factor of 5/8, constrained by the factor of 2^3 within m, requires either m have odd part (3^1)(5^1), which just becomes the fixed point of 120, or can grow in compensation by at most (6/5)(8/7)^2 < (6/5)(4/3) = 8/5. Neither allows n to be growing. 48:30 The shrinking factor in (2^3)(3^2) within m is (5/8)(5/9) = 25/72. The most n can grow (aside from the fixed point of 120) given the 2^3 within m is (3/2)(6/5)(8/7)^2, and the most m can grow in compensation given the 2^1 within n this creates is 6/5; (25/72)(6/5) = 5/12, while (3/2)(6/5)(8/7)^2 < (3/2)(6/5)(4/3) = 12/5, so n cannot grow enough if it has a power of 2^1, and if it has any higher power of 2, its growth will be reduced by a factor greater than 6/5, as explained in the first paragraph. 49:15 Note no additional odd primes apart from 5 itself can be added since 5^1 and 2^2 together already create 3 powers of 2, and n cannot have a 5^2 within it since it can be compensated for by at most one other odd prime, and 7/25 < 5/6. No higher power of 5 allows more odd primes in n. 50:00 If n had a power of 2^1, it would induce a power of 3 within m, which either is greater than 1, reducing to 48:30, or is 1, which would necessitate that n had a power of at least 2^2, which is contradictory. Therefore, n has a power of 2 that is at least 2, and therefore constrained by the 2^4 within m, can grow at most by (4/3)(6/5)^2 < (4/3)(3/2) = 2, while m must shrink by at least (3/8)(6/5) = 9/20 < 1/2. Any higher power of 2 within n or m reduces growth by a factor greater than can be compensated as at 48:30.
This is so cool. I love how so many problems in math can be stated simply, but result in complex and even ground breaking insights. I love biology and chemistry too, but there are big barriers to entry with expensive equipment and education... base requirements to make any progress most likely. Nice work fellas!
I love when you get a simple concept and start developing it and applying it to bigger and bigger contexts (it's what made me subscribe with the tetration+ and unusual bases videos). I was almost disappointed that we wouldn't go that far, and then you pulled the fractions and square roots and everything was good again.
I've got a very strong skeleton of a proof, isn't formal but I could make it so. Let’s define some functions. Glitch(n) is the process outlined in this video - namely, glitch(p^n) = p+n and glitch(p^n * q^m * …) = (p+n) * (q+m) * …. I’ll also define primesum(n), which is the sum of the prime factors with multiplicity. So primesum(p^n) = p*n, primesum(p^n * q^m) = p*n + q*m + … Now I want to consider how glitching impacts primesum. First off, let’s show that primesum(giltch(a * b)) = primesum(glitch(a)) + primesum(glitch(b)) for relatively prime a, b. This is fairly elementary - primesum(a*b) = primesum(a) + primesum(b) and glitch(a*b) = glitch(a) * glitch(b) for relatively prime a, b (proof of those left to the reader). Next I want to see how primesum(glitch(p^n)) compares to primesum(p^n). Well, glitch(p^n) = p+n, and primesum(p^n) = p*n. Furthemore, we’re looking at primesum(p+n) and comparing that to p*n, and primesum never makes a number bigger. I want to particularly ask “when is primesum(glitch(p^n)) within 2 of primesum(p^n)”. If n ≥ 2, then we have primesum(p + ≥ 2) compared to p * ≥ 2. If p > 3, then this is never within 2, if p = 3 and n = 2 then we have 5 compared to 6, and if p = 2 and n = 2 we have 4 and 4. If n = 1, then primesum(p+1) is being compared to p, and only for p = 2,3,5,7 is it within 2. To be specific: p(g(2^1)) = p(2^1) + 1 p(g(3^1)) = p(3^1) + 1 p(g(5^1)) = p(5^1) p(g(7^1)) = p(7^1) - 1 p(g(2^2)) = p(2^2) p(g(3^2)) = p(3^2) - 1. p(g(2^3)) = p(2^3) - 1. For all other p, n we have p(g(p^n)) < p(p^n) - 2. Now, consider p(g(n)). We showed that this is equal to the sum of p(g(components)), and since components decrease by at least 2 unless they’re on that list, it means that if there is any component not on that list then p(g(n)) < p(n). Since p(n) is strictly positive, this means that each number has a finite amount of time moving around until it reaches a point where p(g(n)) stops decreasing and thus it has to eventually become made out of only those components.
This idea is definitely a strong enough idea. I felt like sharing it before I got it all out formalized but hopefully reading it you see why this is true
You have a typo. Your function should be glitch(p^n * q^m * …) = (p+n) * (q+m) * …. rather than glitch(p^n + q^m + …) = (p+n) * (q+m) * …. to do what is given in the video.
I spent this afternoon filling out the video's proof of 2-cycles by giving rigorous estimates for the growing and shrinking, but now I think that was pretty much unnecessary... With your argument of decreasing primesums, we can show that every cycle must contain a number ≤ 420, and those numbers are easily checked to produce no more cycles than the ones we know. And all that without needing to considering cycle length!
Fun observation: each number has at least about *n/logn* parents because there's about *n/logn* primes below a number n, and if you have any such prime p, the number p^(n-p) is a parent of n
i appreciate your videos so much. You provide a distinctive insight in number theory. Primes and prime factorizations are my obssessions so this is quite a treat for me. And i also love your taste for humor. ❤
You've inspired me to analyze what I call the "concatenated glitch primes", where instead of taking the sum of p and n for p^n, you concatenate p^n as pn (where an exponent of 1 would, for example, turn 7^1 into 71). Unlike the addition case, where numbers go up and down fairly evenly, concatenation causes even small numbers to explode in size. Consider 30: its prime factorization is 2^1*3^1*5^1, and this becomes (21)*(31)*(51)=33,201. Thus, I quickly realized that, unless this is behaving like some sort of Goodstein sequence where the numbers get absurdly large but then come down, that iterating the process isn't too interesting. What is more intriguing is what numbers go down: these are, at first, purely powers of 2 or 3, but eventually includes numbers like 864=2^5*3^3, and 25*33=825 is less than it. There are obviously infinitely many such numbers, since powers of small primes grow far quicker than the digit in the exponent (take 2^100 vs. 2100). But are there any fixed points in this process? I think it's possible, if not likely. I've found "twins" in the 4000s (4374 and 4375 both decrease), and one number that decreases 2 times in a row (1152->864->825). The battle between increasing numbers of prime factors vs. the increasing size of the numbers is an interesting battle (once we get past that 33,201 barrier, we can consider products of 3 primes, for example). Lots to ponder!
1) Excellent example of using cases to rule out cycles. 2) Your lower bound for the percent of increasing numbers is based on square free numbers. What if as we grow more and more numbers with square factors also grow? Could this percent grow towards 1? 3) How fun would it be if almost all numbers grow, but when they shrink they fall so fast all numbers converge to your cycles?
It's certainly possible the natural density tends towards 1. I'm not sure where it lies between 6/(pi^2) and 1. And yeah it's interesting that more than half of integers grow but the shrinking seems to cause them all to drop. Similarly, I've wondered about if there's a limit to how many steps a number can grow before it shrinks (and how each shrinking amount would compare to the total growth before it)
I played around with large semiprimes for a while trying different algorithms and changing bases and such. Sadly, I am terrible at math and therefore am not yet rich and famous for a fast factorization algorithm. Good luck in your endeavors.
Haha wow about how long ago? (To anybody wondering if this is true, I'm sure it is because I grew up in Berkeley and used to hang out near there sometimes!)
Fun fact: when you're factorizing a number x to check if it's prime and the square of prime you are about to devide x by is larger than x. You can conclude that x is prime; proof omitted
19:11 Don't forget four times odd squarefrees, which also grows. This means the total is at least 7/pi^2 (4/pi^2 odd squarefrees, 2/pi^2 even squarefrees, 1/pi^2 4*squarefrees)
I'm not entirely sure if extending this to rationals and radicals is well-defined. I remember reading about Gaussian integers -- numbers of the form a+bi, a and b as integers -- and how they can be factored into primes. But there's a twist; prime factorization is not *unique* in this field; instead, composite numbers are factored into two or more primes, each of which may optionally be multiplied by 1, i, -1, or -i (which constitute the "units" of this field). Maybe this isn't actually a concern. Maybe it is. I don't know, I haven't thought too hard about it. The point is; a transformation that requires you to find the prime factorization of a number is not well-defined if there is more than one valid way to factor that number. With that said, you could update the definition to select only one prime factorization from all that are valid. To use the Gaussian integers, you could take only factors of the form a+bi where a is positive and b is zero or positive, then multiply by exactly one of 1, i, -1, or -i as necessary. ...actually, what's gonna happen if we primeglitch Gaussian integers? (I am SO sorry for the nerdsnipe.) (That's a lie I'm not sorry.)
In your "reminder of terms" at 2:03, you say "Each whole number larger than 1 is composed of a specific multiset of prime numbers that multiply into that number" but it's actually true for 1 as well, with the multiset [ ].
This is interesting, essentially the larger the number the smaller the factorization value it would be neat to see the use of "primegliching" to a ratio of all whole ones of each between them. As a ratio. Though this can't be counted it will tend to a number.
Now that the Fermat's Last Theorem was proven, we were in need of a Conjecture made informally that would baffle mathematicians for centuries... Who knew it would come from Combo Class?? I want historians and mathematicians in the far future seeing your videos and to see their reaction
Moral of the story: never underestimate a hobo. FWIW, a computer search up to one million produces only the cycles [1], [4], [90], [120], [20, 24] and [6, 12, 16].
You also could increase your background music's volume, make it basically foreground, and then you could even shout louder to enhance the general user experience.
So for powers of 2 We know ×1 ½ the time ×1.5 ¼ the time × 1 ⅛ the time × ⅝ ¹/¹⁶ the time . . . Then do that for every prime to find the expected change (log). For collatz we know the expected change is √3 ÷ 2 < 1 Pretty sure its < 1 here
Dude, no! Non-squarefree numbers do NOT always shrink. Consider (3^2) (5)(7)(11)...(p_N), this decreases by a factor of 5/9 but increases by a factor of \Product_{2
When did he say that? His very first example is "nonsquarefree". He phrases it ambiguously, but he says that numbers where ALL the prime factors have higher exponents than 1 always shrink (except for 4) Mixed numbers where some prime number exponents are 1 but not all of them will vary, as some components shrink and other grow.
dude i am messing up with some rather interesting combinotorics right now, as my sort of independed mathomusicological study, and i`ve got to a point where i am evaluating summ over recoursive function involving summ of devisors. It shows "invariant" behaviour at prime arguments, and has multiplicative rules! Thanks for your wid, i will definetely watch it again! Hit me up in asnwer if you want to chat about what i am "researching", because i am at the start of preparing material for my first UA-cam video 🥳. You are such a cool creator and probably one of the guys who inpire me to actually make something both content and mathematics wise! And i love your locations / actions going on in the scenes non stop 🥵
Note that every integer has an infinite product of p^0 and in the algorithm, that is not altered so even though you are adding exponents to bases, you are not doing so for the zero powers. Therefore, the consistent thing to do is to define 1 to be a fixed point.
Is that the common way of viewing it? Can't you just define 6 as (2^1)(3^1)? Do you have to also have (5^0)(7^0)... And so on? Results would be the same.
@@hugoperhammer you’re right but the point is, I’m considering the alternate viewpoint in order to make sense of how to apply the rule to the number 1.
im plotting these relationships in an organic grapher, some interesting patterns emerge! apparently a number's parents are related to their partitions, specifically two-part partitions that start with prime numbers, for instance, the number 7 has 6 two-part partitions: (1+6 is discarded), (2+5 means 32 goes to 7), (3+4 means 81 goes to 7), (4+3 is discarded), (5+2 means 25 goes to 7) and (6+1 is discarded).
There ia a mapping of positive integers onto the odd numbers by taking n -> 2n-1. There is also a mapping of odd numbers back onto positive integers by demoting the prime factors. So 35 which is is 5 x 7, turns into 3 x 5; 99 which is 3x3x11 becomes 2x2x7 which is 28. [Note that because these mappings are isomorphic, you can use the second step to prove that only one third of positive integers are even.] If you do those two maps; 1,2 and 3 map onto themselves, then 4 and 5 swap places, then we get a long cycle 6->7->11->10->17->14->8->6. I think I found another long cycle but I forget. Most of the numbers I don't know if they are on a finite cycle or not.
I can't stop thinking about how any number would have prime factorisation of zeroth powers like a⁰b⁰c⁰d⁰ which would contribute terms like (a+0)(b+0)(c+0)(d+0) which (since prime numbers are infinitely many) would go to infinity each time. 26:00 or that
About the breaking poiny of 3+ cycles: I feel like the basic rules of shrinking, evenness and odd primes to the first power could be given to a computer to explore (and give automated proofs of their inexistence, if one further invests the time to make that translation happen)
The OEIS sequence that corresponds to the primeglitch rule is A008473 In that OEIS entry you can see some implementations of the rule in a few languages If none of those are suitable, I put a Python implementation on pastebin: ZpqE02f5
This is really cool, i love patterns. Could someone help me understand why adding exponent to prime factor isn’t arbitrary? What is the meaning behind the idea that incrementing a prime factor’s exponent results in incrementing the prime factor number itself in this framework? Thanks
@@saifleonardotariqquraishi5188oh I was wondering if I was missing something. Still really cool and if arbitrary probably practically applicable one day!
It's arbitrary in some sense, but not in another. Altering the operations performed while leaving the input numbers unchanged, for the components in prime factorizations, will tell you something about what happens to the set of elements under a particular function and it will have a set of base-independant properties. It's not fundamentally utilitarian or practical (having no known applications), but oftentimes mathematics don't need to be constrained by that property of being instantly useful in some way, and indeed occasionally things that seem arbitrary for some indeterminate period of time can someday find themself being useful. Not implying that's the case here, just that it does happen. One thing that makes it really interesting is that it's an evidently relatively unexplored concept. I personally found a lot of really interesting things about the sequences generated by iterating the function. It might be somewhat more arbitrary than finding the sequence of squares, for example, but only by some arbitrarily large degree 😂.
@@user-rb8zl2me1jabout half an hour in he connects it to another operation you can do with the factors of a number, but that one isnt really any less arbitrary. In any case, seeing how numbers change when we futz around with their primes, no matter how arbitrarily, will surely tell us at least something about the structure of primes and how numbers depend on them- and its worth exploring to see if anything interesting or novel comes of it- or if the relationships are essentially random and no structure can be decoded, which would even still at least tells us that the structure of primes is essentially unrelated to this operation
it is arbitrary, it's just something arbitrary that results in really neat stuff which describes basically all of pure mathematics once you get to things disconnected from the real world pure math isn't about applications (even if sometimes they come up after the fact), it's about doing different things arbitrarily and seeing if anything interesting comes of it
This is an exercise in Chaos Theory along the same lines of "3x+1" because of the iteration of the calculation either converging or not (ie. a fractal). Once you understand that the results are AKA "strange attractors" and in the same class as any pseudo random number generator. Call it Number Theory if you like, but I see it as Chaos Theory.
Have you explored polynomial iteration modulo p? What cycles develop? What are the coefficients of the polynomials produced by iteration? Let P( n) = 1 + 2*n + 3*n*n mod 7. Then P( 0) = 1, P( P( 0)) = 6 , etc., giving the following cycle: 0, 1, 6, 2, 3, 6, ... Let's calculate iteration coefficients. Let Q( n, a, b, c, d, e, f, ...) = a + b*n + c*n*n + ... Then Q( Q( n, a, b, c)) = a + b*n + c*n*n + b*a + b*b*n + b*c*n*n + c*a*a + 2*c*a*b*n + ( 2*c*a*c + b*b) * n*n + 2*c*b*c*n*n*n + c*c*c*n*n*n*n = Q( n, a+b*a+c*a*a, b+b*b*2*c*a*b, c+b*c+2*c*a*c+b*b, 2*c*b*c, c*c*c) or something like that? Have you started to notice a pattern? I haven't! -- but there must be one!
I checked for cycles of *any* length containing numbers without any prime factors or exponents above 18, and I found no new ones... Just to illustrate, the familiar ones (4, 6-12-16, 20-24, 90, and 120) don't contain any prime factors above 5 and no exponents above 4, so the fact that nothing new pops up should go quite far.
I loved this video. One note though: the application to rational numbers feels a bit contrived, since, while all positive integers have exactly one prime factoring, a rational numbers has an infinite number of fractional representations that lead to different primeglitched results, and selecting the simplified fractional form of a given rational number seems a bit artificial to me.
Thanks, and I do see your point, but if you follow a few simple rules you can create a single, unambiguous prime factorization for each non-zero rational number. For example, the prime factorization (2^2)(3^(-2)) describes the quantity 4/9 (whether you write it as 4/9 or 8/18 or .444 repeating or whatever) in a way that no other prime factorization does, assuming you follow some simple rules. It's not necessarily a common method, but it does work, and I cover the concept in my earlier episode here: ua-cam.com/video/3eT0uQoAwms/v-deo.html
For the proof you provided, one way to prove the "narrowing the range" arguments you kept reiterating is to find an upper bound to the product for all primes p of 1+1/p, which gives an upper bound gor the expansion factors possible for all possible numbers. It may not be as powerful as you desire, but it will certainly reduce the search space to be finite and relatively small if complex. EDIT: or that product could grow to infinity. I believe it's larger than the sum of all reciprocals of primes, which, if I remember correctly, diverges :(
For both 4k+1 and 4k+3 primes the growth has to end as these will contribute to earlier primes higher power while hitting 4k+4, therefore shrinking is inevitable.
So this is basically Multiplicative version of Aliquot Sums, that's cool. This for sure carries weight for the notion that any squared Prime greater than 3 are exactly one greater than a multiple of 24
Hello Folks :) an hour long episode - i need 6 days to accomplish it :) - maybe there are many picturesque landscapes then i need to set up FHD download :)
even easier to show that no other fix point have 2^2, be A and B two number, if they do not share prime factor, the child of A*B is the child of A times the child of B, if A is an odd number B time a multiple of 4, it means that A=4*B where B do not share any prime factor with 4, so the child of A is the child of B time 4, if A is a fix point, so need to be B without being a multiple of 2, but that is impossible (because if we suppose the existence of 1, it would only contain strictly decreasing component), so there are no other fix point that are of the type 2^2
I did that prime factorization stuff in calculus 3 because I was trying to eliminate errors from my calculations I was having trouble what triple integration and calculation error control
Humans were not supposed to find that sequence. Order to destroy finder has just been sent. When I see your office, I see that Aliens just missed the target, OMG you are safe !
Addendum: I realize that the integers are disjoint pairwise as regards factorization. And I think Eulers proof relies on this fact. But if we can find a counter example for P! +2, then it's not as rigorous as it should be. Also I meant to use 2^3 en toto in my example
This is the longest episode I've ever made. Let me show you an interesting pattern I recently found, and all of the cool mathematical connections I've been able to deduce about it.... Timestamps of different chapters:
0:00 - My Explorations Into Primeglitching
3:29 - Explanation of the Main Pattern
9:18 - Discoveries Among the First 50 Numbers
14:36 - Discoveries Among Larger Numbers
19:57 - Finding the Other Research About This
27:38 - What I Could Deduce About the General Pattern
32:30 - What I Could Deduce About Fixed Points
42:13 - What I Could Deduce About Two-Number Cycles
50:47 - Ideas for Larger Cycles and General Proofs
55:12 - Extending to Fractions and Square Roots
1:05:22 - Other Forms of Potential Primeglitching
(See the video description for more information)
Wonderful video! You might want to pin this comment 😊
@ComboClass Great video, as often. If that's of any help, a function such as the one mapping p^e to p+e (component-wise) is said to be "multiplicative". Equivalently, it satisfies f(a*b) = f(a)*f(b) whenever gcd(a,b)=1. And this gives another reason to say that f1)=1.
Numberphile has gotten weird ever since the apocalypse.
lol
What is this referring to lol, I haven't watched them for ages because I prefer this channel and a few other ones where it's one person presenting maths problems in a more personal way. I find numberphile too big screen type .I like the smaller scale channels like this.
I love the implication that after the apocalypse we still have UA-cam, but clothes and houses get a different vibe
@@AntimatterBeam8954op is implying that this channel is post-apocalyptic numberphile
🤣🤣🤣 WELCOME BACK TO COMBO CLASSSS!!!
that thumbnail gave me a collatz ptsd attack
I only spent three days down that rabbit hole and it was enough to give me ptsd
only Domotro would look at Collatz and think, "not difficult enough"
I spent so many boring math lessons in high school just throwing everything I could think of at Collatz 😂
imagine if this is actually related to collatz... mindblown
Same here,
19:57 cat
54:27 cat
1:01:02 cat
thank you for the long episode
Merci _purr_ le cat 😸 alogue de toutes les apparitions du chat
@@OghamTheBold 3 Quantum cat states.
@@Danny_Fantom we love the kitties! 😻
6:45 spider
Domotro breaking actual new ground in prime numbers is crazy
He breaks so many other things, he was bound to break new ground eventually.
That's what's so cool about math. Anything you learn about has hundreds of variations, it's like if every spot on earth had hundreds of different places a short walk away. with a little wandering around you can easily reach a spot no one else has ever been to.
One of this days Domotro might end up publishing an article in a number theory journal.
I would be surprised if he doesn’t. I assume he is a math graduate student no?
9:21 this is the first time I think I’ve seen this man indoors.
It's not like people on vacation are using their houses to do any number theory anyway.
One hour of this guy? He'll burn down three major cities
lmao real
The Worlds first Pyro-Mathematician.
Interesting stuff. Love it!
holy fucking shit, so THIS is why you havent uploaded for a while.. wow. you fucking DISCOVERED SHIT THAT NOBODY HAS FOUND BEFORE??? AND THEN MADE AN HOUR LONG VIDEO ABOUT IT??? just... i dont even have words dude. thats insane
36:30
Noting that 4/3 < (6/5)^2 = 36/25, and that any odd prime higher than 5 contributes a factor less than 6/5 to growth, we see that the maximum additional growing factor that can be added to compensate for one power of 2 is 6/5. From 2^a to 2^(a+1), the additional shrinking factor added is [(a+2)/(2^(a+1))]/[(a+1)/(2^a)] = (a+2)/(2a+2), which for a >= 1 is less than or equal to 3/4, and therefore less than 5/6.
For a = 4, we see already that (6/5)^4 < (3/2)^2 = 9/4 < 8/3 and therefore odd primes cannot compensate for the shrinking factor induced by 2; and for any higher a, the shrinking factor decreases faster than we can multiply 6/5s to compensate.
Similarly at 39:50, 41:00, and 41:50, the combined factors (2^1)(3^3), (2^3)(5^2), and (2^3)(3^2)(5^1) shrink by (3/2)(6/27) = 1/3 < 5/6, (5/8)(7/25) = 7/40 < 5/6, and (5/8)(5/9)(6/5) = 5/12 < 5/6, respectively. Any higher power of 3 or 5 shrinks further without allowing more compensation.
39:00
Alternatively, note that a fixed point with 2 raised to the power of exactly 2 requires that the odd part also be a fixed point, which was just shown to be impossible in the non-trivial case.
42:00
Note no additional odd primes can be added since 3^1 and 5^1 together already create 3 powers of 2.
48:50
Counterbalancing the shrinking factor of 5/8, constrained by the factor of 2^3 within m, requires either m have odd part (3^1)(5^1), which just becomes the fixed point of 120, or can grow in compensation by at most (6/5)(8/7)^2 < (6/5)(4/3) = 8/5. Neither allows n to be growing.
48:30
The shrinking factor in (2^3)(3^2) within m is (5/8)(5/9) = 25/72. The most n can grow (aside from the fixed point of 120) given the 2^3 within m is (3/2)(6/5)(8/7)^2, and the most m can grow in compensation given the 2^1 within n this creates is 6/5; (25/72)(6/5) = 5/12, while (3/2)(6/5)(8/7)^2 < (3/2)(6/5)(4/3) = 12/5, so n cannot grow enough if it has a power of 2^1, and if it has any higher power of 2, its growth will be reduced by a factor greater than 6/5, as explained in the first paragraph.
49:15
Note no additional odd primes apart from 5 itself can be added since 5^1 and 2^2 together already create 3 powers of 2, and n cannot have a 5^2 within it since it can be compensated for by at most one other odd prime, and 7/25 < 5/6. No higher power of 5 allows more odd primes in n.
50:00
If n had a power of 2^1, it would induce a power of 3 within m, which either is greater than 1, reducing to 48:30, or is 1, which would necessitate that n had a power of at least 2^2, which is contradictory. Therefore, n has a power of 2 that is at least 2, and therefore constrained by the 2^4 within m, can grow at most by (4/3)(6/5)^2 < (4/3)(3/2) = 2, while m must shrink by at least (3/8)(6/5) = 9/20 < 1/2. Any higher power of 2 within n or m reduces growth by a factor greater than can be compensated as at 48:30.
If you close your eyes, you can hear Dr. Evil teaching math.
Or Dr. Horrible even.
This is so cool. I love how so many problems in math can be stated simply, but result in complex and even ground breaking insights. I love biology and chemistry too, but there are big barriers to entry with expensive equipment and education... base requirements to make any progress most likely. Nice work fellas!
Hello. It's more fun than the Collatz's conjecture! Thanks!
I love when you get a simple concept and start developing it and applying it to bigger and bigger contexts (it's what made me subscribe with the tetration+ and unusual bases videos). I was almost disappointed that we wouldn't go that far, and then you pulled the fractions and square roots and everything was good again.
I've got a very strong skeleton of a proof, isn't formal but I could make it so.
Let’s define some functions. Glitch(n) is the process outlined in this video - namely, glitch(p^n) = p+n and glitch(p^n * q^m * …) = (p+n) * (q+m) * …. I’ll also define primesum(n), which is the sum of the prime factors with multiplicity. So primesum(p^n) = p*n, primesum(p^n * q^m) = p*n + q*m + …
Now I want to consider how glitching impacts primesum. First off, let’s show that primesum(giltch(a * b)) = primesum(glitch(a)) + primesum(glitch(b)) for relatively prime a, b. This is fairly elementary - primesum(a*b) = primesum(a) + primesum(b) and glitch(a*b) = glitch(a) * glitch(b) for relatively prime a, b (proof of those left to the reader).
Next I want to see how primesum(glitch(p^n)) compares to primesum(p^n). Well, glitch(p^n) = p+n, and primesum(p^n) = p*n. Furthemore, we’re looking at primesum(p+n) and comparing that to p*n, and primesum never makes a number bigger. I want to particularly ask “when is primesum(glitch(p^n)) within 2 of primesum(p^n)”. If n ≥ 2, then we have primesum(p + ≥ 2) compared to p * ≥ 2. If p > 3, then this is never within 2, if p = 3 and n = 2 then we have 5 compared to 6, and if p = 2 and n = 2 we have 4 and 4. If n = 1, then primesum(p+1) is being compared to p, and only for p = 2,3,5,7 is it within 2. To be specific:
p(g(2^1)) = p(2^1) + 1
p(g(3^1)) = p(3^1) + 1
p(g(5^1)) = p(5^1)
p(g(7^1)) = p(7^1) - 1
p(g(2^2)) = p(2^2)
p(g(3^2)) = p(3^2) - 1.
p(g(2^3)) = p(2^3) - 1.
For all other p, n we have p(g(p^n)) < p(p^n) - 2. Now, consider p(g(n)). We showed that this is equal to the sum of p(g(components)), and since components decrease by at least 2 unless they’re on that list, it means that if there is any component not on that list then p(g(n)) < p(n). Since p(n) is strictly positive, this means that each number has a finite amount of time moving around until it reaches a point where p(g(n)) stops decreasing and thus it has to eventually become made out of only those components.
This idea is definitely a strong enough idea. I felt like sharing it before I got it all out formalized but hopefully reading it you see why this is true
You have a typo. Your function should be glitch(p^n * q^m * …) = (p+n) * (q+m) * …. rather than glitch(p^n + q^m + …) = (p+n) * (q+m) * …. to do what is given in the video.
This is a very clever approach, thank you for sharing!
I spent this afternoon filling out the video's proof of 2-cycles by giving rigorous estimates for the growing and shrinking, but now I think that was pretty much unnecessary...
With your argument of decreasing primesums, we can show that every cycle must contain a number ≤ 420, and those numbers are easily checked to produce no more cycles than the ones we know.
And all that without needing to considering cycle length!
Your set design really helps me focus on your videos and I love prime patterns!
Fun observation: each number has at least about *n/logn* parents because there's about *n/logn* primes below a number n, and if you have any such prime p, the number p^(n-p) is a parent of n
its n/logn, not logn
@@felipegiglio2047 Thank you for the correction. I knew something was off! I'll fix it
1 hour of combo class?????? YES PLEASE
i appreciate your videos so much. You provide a distinctive insight in number theory. Primes and prime factorizations are my obssessions so this is quite a treat for me. And i also love your taste for humor. ❤
The thing I most relate to mathematicians about is when they do a seemingly-arbitrary thing just to see what happens
You've inspired me to analyze what I call the "concatenated glitch primes", where instead of taking the sum of p and n for p^n, you concatenate p^n as pn (where an exponent of 1 would, for example, turn 7^1 into 71). Unlike the addition case, where numbers go up and down fairly evenly, concatenation causes even small numbers to explode in size. Consider 30: its prime factorization is 2^1*3^1*5^1, and this becomes (21)*(31)*(51)=33,201.
Thus, I quickly realized that, unless this is behaving like some sort of Goodstein sequence where the numbers get absurdly large but then come down, that iterating the process isn't too interesting. What is more intriguing is what numbers go down: these are, at first, purely powers of 2 or 3, but eventually includes numbers like 864=2^5*3^3, and 25*33=825 is less than it. There are obviously infinitely many such numbers, since powers of small primes grow far quicker than the digit in the exponent (take 2^100 vs. 2100).
But are there any fixed points in this process? I think it's possible, if not likely. I've found "twins" in the 4000s (4374 and 4375 both decrease), and one number that decreases 2 times in a row (1152->864->825). The battle between increasing numbers of prime factors vs. the increasing size of the numbers is an interesting battle (once we get past that 33,201 barrier, we can consider products of 3 primes, for example). Lots to ponder!
Love when the Combo Cats make an appearance. ❤
1) Excellent example of using cases to rule out cycles.
2) Your lower bound for the percent of increasing numbers is based on square free numbers. What if as we grow more and more numbers with square factors also grow? Could this percent grow towards 1?
3) How fun would it be if almost all numbers grow, but when they shrink they fall so fast all numbers converge to your cycles?
It's certainly possible the natural density tends towards 1. I'm not sure where it lies between 6/(pi^2) and 1. And yeah it's interesting that more than half of integers grow but the shrinking seems to cause them all to drop. Similarly, I've wondered about if there's a limit to how many steps a number can grow before it shrinks (and how each shrinking amount would compare to the total growth before it)
ooh i was working with prime factorizations and graphs of sequences so this is a very pleasant surprise :) loved the video, great proofs!
I played around with large semiprimes for a while trying different algorithms and changing bases and such. Sadly, I am terrible at math and therefore am not yet rich and famous for a fast factorization algorithm. Good luck in your endeavors.
Also, I’m envious of your staghorn ferns.
Congratulations on the channel growth - halfway to 100K!
This is actually really cool
1 hour math video??? My night is about to be wild
being your cameraman has got to be such a hard job
four always goes back to itself man
I knew it would be a banger for the last main episode of grade -2
I used to see this guy on Telegraph Ave in Berkeley.
Haha wow about how long ago? (To anybody wondering if this is true, I'm sure it is because I grew up in Berkeley and used to hang out near there sometimes!)
Fun fact: when you're factorizing a number x to check if it's prime and the square of prime you are about to devide x by is larger than x. You can conclude that x is prime; proof omitted
I love the parody to other math channels. Awesome episode. Thank you.
19:11
Don't forget four times odd squarefrees, which also grows. This means the total is at least 7/pi^2 (4/pi^2 odd squarefrees, 2/pi^2 even squarefrees, 1/pi^2 4*squarefrees)
I'm not entirely sure if extending this to rationals and radicals is well-defined. I remember reading about Gaussian integers -- numbers of the form a+bi, a and b as integers -- and how they can be factored into primes. But there's a twist; prime factorization is not *unique* in this field; instead, composite numbers are factored into two or more primes, each of which may optionally be multiplied by 1, i, -1, or -i (which constitute the "units" of this field).
Maybe this isn't actually a concern. Maybe it is. I don't know, I haven't thought too hard about it. The point is; a transformation that requires you to find the prime factorization of a number is not well-defined if there is more than one valid way to factor that number. With that said, you could update the definition to select only one prime factorization from all that are valid. To use the Gaussian integers, you could take only factors of the form a+bi where a is positive and b is zero or positive, then multiply by exactly one of 1, i, -1, or -i as necessary.
...actually, what's gonna happen if we primeglitch Gaussian integers?
(I am SO sorry for the nerdsnipe.)
(That's a lie I'm not sorry.)
In your "reminder of terms" at 2:03, you say "Each whole number larger than 1 is composed of a specific multiset of prime numbers that multiply into that number" but it's actually true for 1 as well, with the multiset [ ].
1:50 YES!! thank you !
Not everything sounds like a livestock auction when played at 2× speed, but this video does.
This is interesting, essentially the larger the number the smaller the factorization value it would be neat to see the use of "primegliching" to a ratio of all whole ones of each between them. As a ratio. Though this can't be counted it will tend to a number.
Bro came up with a new Collatz Conjecture!!!
Now that the Fermat's Last Theorem was proven, we were in need of a Conjecture made informally that would baffle mathematicians for centuries...
Who knew it would come from Combo Class??
I want historians and mathematicians in the far future seeing your videos and to see their reaction
29:37 this blew my mind. The product of the growing/shrinking factors of the components is the growing/shrinking factor of the number itself!
I dare to believe him to be an absolutely sane and accountable individual 😅😅😅
When are you going to make some merch, I would totally buy something from you. More people need to learn about the combo class.
This looks just like my program to graph primey stuff I made a while ago along with Parker's Happy work
Wake up babe, new Collatz Conjecture just dropped
Professor Props! Nice video!
Moral of the story: never underestimate a hobo.
FWIW, a computer search up to one million produces only the cycles [1], [4], [90], [120], [20, 24] and [6, 12, 16].
You also could increase your background music's volume, make it basically foreground, and then you could even shout louder to enhance the general user experience.
So for powers of 2
We know
×1 ½ the time
×1.5 ¼ the time
× 1 ⅛ the time
× ⅝ ¹/¹⁶ the time
.
.
.
Then do that for every prime to find the expected change (log).
For collatz we know the expected change is √3 ÷ 2 < 1
Pretty sure its < 1 here
Dude, no! Non-squarefree numbers do NOT always shrink. Consider (3^2) (5)(7)(11)...(p_N), this decreases by a factor of 5/9 but increases by a factor of \Product_{2
When did he say that? His very first example is "nonsquarefree". He phrases it ambiguously, but he says that numbers where ALL the prime factors have higher exponents than 1 always shrink (except for 4)
Mixed numbers where some prime number exponents are 1 but not all of them will vary, as some components shrink and other grow.
Unexpectedly very long video... guess i'll sleep to this 💤
dude i am messing up with some rather interesting combinotorics right now, as my sort of independed mathomusicological study, and i`ve got to a point where i am evaluating summ over recoursive function involving summ of devisors. It shows "invariant" behaviour at prime arguments, and has multiplicative rules! Thanks for your wid, i will definetely watch it again! Hit me up in asnwer if you want to chat about what i am "researching", because i am at the start of preparing material for my first UA-cam video 🥳. You are such a cool creator and probably one of the guys who inpire me to actually make something both content and mathematics wise! And i love your locations / actions going on in the scenes non stop 🥵
Note that every integer has an infinite product of p^0 and in the algorithm, that is not altered so even though you are adding exponents to bases, you are not doing so for the zero powers. Therefore, the consistent thing to do is to define 1 to be a fixed point.
No
Is that the common way of viewing it? Can't you just define 6 as (2^1)(3^1)? Do you have to also have (5^0)(7^0)... And so on? Results would be the same.
@@hugoperhammer you’re right but the point is, I’m considering the alternate viewpoint in order to make sense of how to apply the rule to the number 1.
You crammed for this material for so long that every nearby clock died?
Some clocks were elsewhere temporarily but have now returned. Others broke or burnt
@@ComboClass Any flip clocks?
Every negative number goes to 0. Because each negative number has a factor of -1^1, and -1+1 is 0.
im plotting these relationships in an organic grapher, some interesting patterns emerge! apparently a number's parents are related to their partitions, specifically two-part partitions that start with prime numbers, for instance, the number 7 has 6 two-part partitions: (1+6 is discarded), (2+5 means 32 goes to 7), (3+4 means 81 goes to 7), (4+3 is discarded), (5+2 means 25 goes to 7) and (6+1 is discarded).
57:30 ok now that sounds kinda fun. I wonder if you could fit that in a game
There ia a mapping of positive integers onto the odd numbers by taking n -> 2n-1.
There is also a mapping of odd numbers back onto positive integers by demoting the prime factors. So 35 which is is 5 x 7, turns into 3 x 5; 99 which is 3x3x11 becomes 2x2x7 which is 28.
[Note that because these mappings are isomorphic, you can use the second step to prove that only one third of positive integers are even.]
If you do those two maps; 1,2 and 3 map onto themselves, then 4 and 5 swap places, then we get a long cycle 6->7->11->10->17->14->8->6.
I think I found another long cycle but I forget. Most of the numbers I don't know if they are on a finite cycle or not.
Correction: 1050 => 672, not 840. 3*4*7*8 = 672
True. Luckily the point still holds. f(1050) < 1050.
I can't stop thinking about how any number would have prime factorisation of zeroth powers like a⁰b⁰c⁰d⁰
which would contribute terms like (a+0)(b+0)(c+0)(d+0)
which (since prime numbers are infinitely many) would go to infinity each time.
26:00 or that
About the breaking poiny of 3+ cycles: I feel like the basic rules of shrinking, evenness and odd primes to the first power could be given to a computer to explore (and give automated proofs of their inexistence, if one further invests the time to make that translation happen)
Imcredibly interesting
Um, can I see a computer program that automates this? It's very interesting!
I might make a desmos graph, actually. Desmos is great for this kind of stuff!
@@simonwillover4175do it!!
The OEIS sequence that corresponds to the primeglitch rule is A008473
In that OEIS entry you can see some implementations of the rule in a few languages
If none of those are suitable, I put a Python implementation on pastebin: ZpqE02f5
"I have a concept of a proof", to paraphrase a certain politician. Meaning: I have no idea 🙂
This is really cool, i love patterns. Could someone help me understand why adding exponent to prime factor isn’t arbitrary? What is the meaning behind the idea that incrementing a prime factor’s exponent results in incrementing the prime factor number itself in this framework? Thanks
It is arbitrary
@@saifleonardotariqquraishi5188oh I was wondering if I was missing something. Still really cool and if arbitrary probably practically applicable one day!
It's arbitrary in some sense, but not in another. Altering the operations performed while leaving the input numbers unchanged, for the components in prime factorizations, will tell you something about what happens to the set of elements under a particular function and it will have a set of base-independant properties. It's not fundamentally utilitarian or practical (having no known applications), but oftentimes mathematics don't need to be constrained by that property of being instantly useful in some way, and indeed occasionally things that seem arbitrary for some indeterminate period of time can someday find themself being useful. Not implying that's the case here, just that it does happen. One thing that makes it really interesting is that it's an evidently relatively unexplored concept. I personally found a lot of really interesting things about the sequences generated by iterating the function. It might be somewhat more arbitrary than finding the sequence of squares, for example, but only by some arbitrarily large degree 😂.
@@user-rb8zl2me1jabout half an hour in he connects it to another operation you can do with the factors of a number, but that one isnt really any less arbitrary. In any case, seeing how numbers change when we futz around with their primes, no matter how arbitrarily, will surely tell us at least something about the structure of primes and how numbers depend on them- and its worth exploring to see if anything interesting or novel comes of it- or if the relationships are essentially random and no structure can be decoded, which would even still at least tells us that the structure of primes is essentially unrelated to this operation
it is arbitrary, it's just something arbitrary that results in really neat stuff
which describes basically all of pure mathematics once you get to things disconnected from the real world
pure math isn't about applications (even if sometimes they come up after the fact), it's about doing different things arbitrarily and seeing if anything interesting comes of it
This is an exercise in Chaos Theory along the same lines of "3x+1" because of the iteration of the calculation either converging or not (ie. a fractal). Once you understand that the results are AKA "strange attractors" and in the same class as any pseudo random number generator. Call it Number Theory if you like, but I see it as Chaos Theory.
38:31
The two levels of evenness are already made by the 2², sir
Trying to imagine what kind of life dudes like this lived a few hundred years ago
I am from India, I don't know how to understand English, still I see that I understand maths.❤❤❤🙏🏻🙏🏻🙏🏻🙏🏻
Vsauce4
Have you explored polynomial iteration modulo p? What cycles develop? What are the coefficients of the polynomials produced by iteration?
Let P( n) = 1 + 2*n + 3*n*n mod 7. Then P( 0) = 1, P( P( 0)) = 6 , etc., giving the following cycle: 0, 1, 6, 2, 3, 6, ...
Let's calculate iteration coefficients. Let Q( n, a, b, c, d, e, f, ...) = a + b*n + c*n*n + ...
Then Q( Q( n, a, b, c)) =
a + b*n + c*n*n +
b*a + b*b*n + b*c*n*n +
c*a*a + 2*c*a*b*n + ( 2*c*a*c + b*b) * n*n + 2*c*b*c*n*n*n + c*c*c*n*n*n*n =
Q( n, a+b*a+c*a*a, b+b*b*2*c*a*b, c+b*c+2*c*a*c+b*b, 2*c*b*c, c*c*c)
or something like that? Have you started to notice a pattern? I haven't! -- but there must be one!
I can't get over how much I love your chaos gremlin energy, never change that part please :3
36:20
To formalize this:
The product, for all odd primes p, of 1+1/p, is pi^2/8, less than the 8/3 we need.
Грибной сезон в разгаре
I checked for cycles of *any* length containing numbers without any prime factors or exponents above 18, and I found no new ones...
Just to illustrate, the familiar ones (4, 6-12-16, 20-24, 90, and 120) don't contain any prime factors above 5 and no exponents above 4, so the fact that nothing new pops up should go quite far.
The white rabbit took a weird potion
I loved this video. One note though: the application to rational numbers feels a bit contrived, since, while all positive integers have exactly one prime factoring, a rational numbers has an infinite number of fractional representations that lead to different primeglitched results, and selecting the simplified fractional form of a given rational number seems a bit artificial to me.
Thanks, and I do see your point, but if you follow a few simple rules you can create a single, unambiguous prime factorization for each non-zero rational number. For example, the prime factorization (2^2)(3^(-2)) describes the quantity 4/9 (whether you write it as 4/9 or 8/18 or .444 repeating or whatever) in a way that no other prime factorization does, assuming you follow some simple rules. It's not necessarily a common method, but it does work, and I cover the concept in my earlier episode here: ua-cam.com/video/3eT0uQoAwms/v-deo.html
@@ComboClass oh yes, of course, i missed that. thanks!
Fantastic, I love the Math. But can you sometime pull out a ukulele and give us a rendo of "Tip Toe Through The Tulips"?
Stig of the dump is a mathematical genius, who knew?
For the proof you provided, one way to prove the "narrowing the range" arguments you kept reiterating is to find an upper bound to the product for all primes p of 1+1/p, which gives an upper bound gor the expansion factors possible for all possible numbers. It may not be as powerful as you desire, but it will certainly reduce the search space to be finite and relatively small if complex.
EDIT: or that product could grow to infinity. I believe it's larger than the sum of all reciprocals of primes, which, if I remember correctly, diverges :(
if p and q are coprime then f(p)f(q)=f(pq). so if there's a fixed point with 2^2 then there would be a fixed point with 2^0 but there isnt
For both 4k+1 and 4k+3 primes the growth has to end as these will contribute to earlier primes higher power while hitting 4k+4, therefore shrinking is inevitable.
I feel like I need to take a shower 😂
So this is basically Multiplicative version of Aliquot Sums, that's cool.
This for sure carries weight for the notion that any squared Prime greater than 3 are exactly one greater than a multiple of 24
Domotro is America’s Perelman
Hello Folks :) an hour long episode - i need 6 days to accomplish it :) - maybe there are many picturesque landscapes then i need to set up FHD download :)
That's Collatz on steroids!
even easier to show that no other fix point have 2^2, be A and B two number, if they do not share prime factor, the child of A*B is the child of A times the child of B, if A is an odd number B time a multiple of 4, it means that A=4*B where B do not share any prime factor with 4, so the child of A is the child of B time 4, if A is a fix point, so need to be B without being a multiple of 2, but that is impossible (because if we suppose the existence of 1, it would only contain strictly decreasing component), so there are no other fix point that are of the type 2^2
"If I had X squared..eeeeeehhhhh" 🤣
You mentioned at the start you were looking for a pattern that also worked in other bases but have not gotten back to that yet
The main pattern is this episode is an example of that. It is base-independent and doesn't matter which base you write the digits in.
This must be how karma works. Shrinking and growing factors, fixed points are archetypes/gods
I did that prime factorization stuff in calculus 3 because I was trying to eliminate errors from my calculations I was having trouble what triple integration and calculation error control
I watched the whole thing
13:50 he says cubed but means (and writes) squared.
I only mention it because it really threw me off.
It's like Beakman's world with more meth.
Humans were not supposed to find that sequence. Order to destroy finder has just been sent. When I see your office, I see that Aliens just missed the target, OMG you are safe !
Addendum: I realize that the integers are disjoint pairwise as regards factorization. And I think Eulers proof relies on this fact. But if we can find a counter example for P! +2, then it's not as rigorous as it should be. Also I meant to use 2^3 en toto in my example