Zolar V

I do recall the notation; unfortunately, I only briefly used the notation in a few of my classes. So I am familiar but probably not much past a familiarity. "I note that your paper is marked up nicely, which is good." - Thank you, i wrote it over the course of a few weeks as I learned how to use LaTex. I posted it here because I re-found the forum and remembered that this was a great place to have a good discussion. I believe a long time ago these forums were also integrated into LaTex or some other online math notation editor. I also don't really know where I would have posted this anyways. I did show my paper to a few math major friends of mine and we did have some discussion about it. They didn't seem to have the problems we are having here, so I wonder if I did some explanation to them verbally that I just entirely missed. "Are you saying that each of 3 and 5 eventually reduce to a power of 2 this way? Is that the idea?" - yes, every prime will eventually reduce to a power of 2. I am going to try to really define our function, or functions here so that it makes sense.

Thank you! good to see old members are still active here!

I'm american, though the problem with my grammar is likely the product of not talking very much to anyone else.

A couple notes: I stumbled on this solution, that i hope we can both work through so you can see it too, while sitting at work. I had a sudden flash of insight. It might be noteworthy that my school career has been anything but standard. I was home schooled till half way through 9th grade and then attended public school from 9th to 12th, during my time being home schooled i had a smattering of public attendance as well. 4 months here, 2 months there. After high school I joined the airforce and after that I attended college. In college I earned my B.S. in science with two supporting minors; mathematics and biology. In mathematics I took a number of classes, but what really piqued my interest was abstract algebra. Though I technically had the prerequisite (311W, which teaches you proofing and logic) for it, the class was supposed to be the capstone math course. I had to drop it since my proofing skills were subpar and I didnt want to risk failing. Typically a math major would take that course after a series of 300 -400 level advanced math classes that focused on proofing. My teacher noted that my proofs were basically written backwards. Here is one math research project I did: https://drive.google.com/drive/folders/0B_B_IgMx0w_3cG5LZjBQaUZZYkU?usp=sharing I developed a general formula (instead of stirlings approximation) for a certain set of factorial divided by factorial problems. There were some limitations that I hadn't quite explored by the end of the research project. Anyways I digress, my point is that I may not have a clear understanding of a lot of math concepts simply because my schooling was frequently interrupted. I believe I understand a function, but I may not know how math would define a recurrence relation. I do see what you're saying now about defining G(p), I think the definition you're looking for is a higher order recurrence relation. It seems I am defining one function on primes and then not defining the recurrence relation of that function on all integers. " G^n (p) = 2^m? " - well you see, in my mind im saying that after n iterations of G(p) the final result will be some number that is a power of 2. The number of iterations that cause that aren't related to the power of 2. " Yes perfectly well agreed, once we get to a power of 2 we're done. Yet what you originally wrote was wrong as stated. You need to explicitly call out this case so that we know you're paying attention to your own work " - Sure sure, I can totally treat both cases explicitly. " our part, have you considered rewriting the paper from scratch in the light of some of these comments from me and others? ... ... submitted to number theory " - Yes, I do entirely want to rewrite it to be bulletproof, if the proof actually stands up. That is the purpose of wanting to work with a real mathematician. I didn't submit it to number theory, I added that because I was following a template and also needed to remind myself that I think number theory is my target audience. Basically it was a mental note " What I saw was that you established your facts about the non-2 prime factors of p+1, then waved your hands that since each factor is prime you can recurse your way down to your conclusion. But I did not see a proof, just a vague intution that there's a recursion to be had. Without a clear definintion of G, probably a recursive one, you won't have a proof. " - Now that I think i understand what definition of G you're looking for we can make some headway and I would agree. "... So after 3 iterations we reach a 2^n number and we are done with G(p). So 2 < 3 <5 I still don't get it. Let's nail down what G is, maybe we can figure out the proper recursive definition needed to make the rest of this work. .... " - So I thought I was defining G as a function in just the same way that the original collatz conjecture had defined their odd function contained within the piecewise function. So G(p) = p + 1 , let p = 5. then G(5) = 5 +1. So iteration 1 is G(5) = 6. Iteration 2 is taking the result of iteration 1, reworking the result such that you have a prime number and then adding 1 to that prime. So iteration 2 is G(6) = 3 +(3 +1 ) Iteration 3 is adding 1 to the next prime number from iteration 2. G(3+ 4) =(3 +1) + 4 Iteration of G(p) is taking the prior result and inputting that as p. A better defined recursive relation would be Hk = Hk-1 +1 . As i said before, I was trying to follow the original notation and thought I could define my recursive function in just the same way. "> of course if I add 1 enough times I will encounter an integer that is a power of 2. But this result is why the collatz conjecture itself converges to 1. Well that's a handwavy argument that doesn't convince me." Yes it is handwavy but true. Just think about this. Take the largest number that is a power of 2 and add 1 however many times it takes to reach the next power of 2. Since integers are infinite you would always reach another one. think about it this way. let 2k be the kth power of two. then 2k - 2k-1 = n . n is the number of times you need to add 1 to 2k-1 to reach 2k Example: 2^4 - 2 ^3 = 8 , 2 ^3 + 8(1) is 2^4 multiplication is repeated addition of course. so 8 is the number of times you need to add 1 to go from the lower power of two, to the next power of two. " Good to know. Now I don't feel so bad being so critical. I'm curious to understand your idea and your exposition is in the way so I'm trying to help you clarify the exposition. " - be critical. I'm trying to craft the best proof I can. I did get irritated earlier since i read the first comments as being condescending. They have since changed tone and thank you for that.

Great critique! I didn't design the proof with any sort of programing language in mind. However, I typically think in that sort of nested fashion. So lets get started: " On line 21 you have Lemma 1, an extremely trivial result for which you supply a proof. " -Though the result is obvious and trivial, I think it is worth mentioning since it is the underlying concept to why adding 1 to a prime causes the resulting primes contained in the composite number to be less. " On line 28 immediately following your end-of-proof square, it says, "Proof" and then begins a long involved proof ... of something, we cannot tell what. " - This is the proof of the G(p) = p +1, but I believe you figured that out. " Also please clearly define what G is. You say G(p) = p + 1 and then you want to iterate it, but clearly you CAN'T ITERATE THIS because p + 1 is not prime in general. " -I think a note here that explains why we can still apply the function to a composite number would work. The reason why we can iterate G(p) on any integer is because every integer is either prime or a composite of primes. The latter needs a little bit of work, but we can extract a prime out such that we have a number + a prime. Like stated above, E.G. 7*3 = (3-1)*7+ 7 " You've explained this by saying that G is NOT actually the function p+1, but is rather your "prime reduction function" or some such, which you never actually seem to define. " -G(p) = p+1 is what I called the prime reducing function. "I'll add that G^n(p) = p + 1 hardly solves the problem, since the right hand side is entirely independent of n. " -I suppose a better way to write that would be G^n (p) = 2^m , since after n iterations you get a 2^m number. I mean you're really only adding 1 a number of times until you reach a 2^n. its rather trivial if you think about it like that. "On line 29 you say that if p is prime, then p+1 can be factored as 2^q x product((P_i)^n_i). But if p + 1 happens to be a power of 2, such as 7 + 1 = 8 = 2^3, then none of the other primes P_1, P_2, etc., exist. I'm not sure if this is relevant to your argument but you should definitely handle this case. " -Fair enough, but if p+1 happens to be a power of 2 then the problem is done. We're only trying to create a string of 2's, so then we can divide by 2 however many times to get 1. We only do this because that is part of the collatz cojecture. " On line 30 you have a flat out error though. You say that (P_k)^n_k < P. This is of course false. You have indeed proved that P_k < P, but that inequality does not hold for some arbitrary integer POWER of P_k. " -You are both correct and wrong. I shouldn't have written the notation with arbitrary powers of P_k, I should have written it as a fully expanded string of powers. 3^3 should be 3*3*3... We are attempting to handle individual primes, one at a time. Not some grouping of primes in a shorthand notation. " You compound the error on lines 31-32 by saying that (P_k)^n_k is prime. In fact you say it is "prime by definition." Of course for n_i > 1 it is not prime, it's a power of P_k. Right? " - I believe the prior answer to your prior question also solves this question. by definition of the fundamental theorem of arithmetic each P_k is prime. " ppps -- No, you don't have an argument left. On line 32 you apply the "prime reducing function" to a list of POWERS of primes, VERY FEW OF WHICH are prime in the general case. Your proof is busted right there. " - I believe rewriting the composite number as a series of primes with a power of 1 solves your question here. Very similar to the above two questions. " I'll also mention that line 33 doesn't follow from anything that came before EVEN IF we allow that prime powers are prime (which of course they are not)! And what is the 'x' on the right side of line 33? " - Wrong notation here. In my mind "x" was the first prime we punched into the problem. so it should have been written as "p" not "x" Line 33 is supposed to show that for each iteration of G(p), the resulting primes are less than p and also less than the last iteration of G(p). E.G for the first iteration, let our p = 5 . G(5) = 6 rewrite 6 as 2*3, then rewrite it as (2-1)*3+3. (again the purpose of the rewrite is to show that we can get a prime number to apply G(p) to. So G(5) = 3 + 3 Then the next iteration is G(3) = 4 . so G(5) = G(3) + G(3) (there are two iterations here, since there are two 3's we're reducing) So after 3 iterations we reach a 2^n number and we are done with G(p). So 2 < 3 <5 Note: I don't believe your questions yet disprove that this isn't the solution to the collatz conjecture. They do help enormously to clarify the logical jumps I made. In a simple sense, adding 1 to a number an arbitrary number of times such that you get a 2^m number is exceptionally trivial and intuitive. of course if I add 1 enough times I will encounter an integer that is a power of 2. But this result is why the collatz conjecture itself converges to 1.

As stated awkwardly: For any prime p > 2 the function G(p) = p + 1 (converges) to 2 after n iterations. So G(7) = 8 and 8 = 2^3. so each prime in 8 is 2. A better example i think: G(11) = 12 = 2*2*3 = (2*2 -1) *3 + 3 substep: G(3) = 4 = 2*2 So G(G(11)) = (2*2 -1) *3 + G(3) G(G(11)) = (2*2 -1) *3 + 2*2 The next step is to take out another prime and apply the function again (2*2 -1) *3 = (2*2 -1) + (2*2 -1) + (2*2 -1) or ((3-1)(2*2 -1)) + (2*2 -1) = ((3-1)(2*2 -1)) + 3 = 2*3 + 3 G(G(G(11)) = 2*3 + 3 + 2*2 = 2*3 + G(3) + 2*2 = 2*3 + 2*2 +2*2 = 2*3 + 2*2*2 you can then rewrite 2*3 as (2-1)*3 + 3 G(G(G(G(11))) = (2-1)*3 + 3 + 2*2*2 = (2-1)*3 + G(3) + 2*2*2 and of course (2-1)*3 = (1)*3 = 3 G(G(G(G(G(11)))) = 3 + 2*2 + 2*2*2 = G(3) + 2*2 + 2*2*2 = 2*2 + 2*2 + 2*2*2 so, 5 iterations cause G(11) to be a series of primes where each prime is 2. The purpose of G(p) = p + 1 is to reduce each prime within the composite number to 2. once we have our 2's then the application of n iterations of the second function H(x) = x/2 , results in 1. after n iterations of course. What this paper is trying to state is that G(p) = p + 1 is embedded in the odd function of the collatz conjecture. Since G(p) over some m iterations results in a 2^k number, it's clear that dividing by 2^k equals 1. thus the collatz conjecture always results in 1 after some m iterations.

That would be all from the first equation. the next part is to take the even equation from the collatz conjecture and apply it however many times it takes to reduce each 2 into a 1. The purpose of G(p) = p + 1 is to reduce each prime to 2.

You need to convert a non prime number into a composite of primes then extract a prime from that composite. p+1 is a composite number. p+1 = a*b*c*d*e.. where each of those are primes. What you need to do is then take (a*b*...*(d -1) ) *e + e E.G: http://www.wolframalpha.com/input/?i=(5*3*7)+%3D+(5*3+-1)*7+%2B+7 once you do that you can then utilize the function to reduce each prime E.G: G(5)= 5 +1 = 6 6 = 2*3 = (2-1)*3 +3 = 3 + 3 so 3+ 3 -> G(3+3) = G(3) + G(3) yes, x+1 is a composite number. you can then follow the rules of math to give you a expression that has a prime that you can use. any composite number can be rewritten as a product of primes. as such you can rewrite that expression of primes to be an even product of primes multiplied by a certain prime and added to that prime. E.G 486485 + 1 = 2*3*3*3*3*3*7*11*13 = (2*3^5*7*11-1)*13 + 13 Since you have a composite part and a prime part you can then apply the function on the prime part. Maybe you can see more of what I am trying to explain by taking a look at a few primes. G(p) = p + 1 where each prime in the composite number p+1 is less than p 7 -> 7+1 = 8 = 2*2*2 | 2,2,2 are < 7 11 -> 11+1 = 12 = 2*2*3 | 2,2,3 are <11 13 -> 13+1 = 14 = 2*7 | 2,7 are <13 17 -> 17+1 = 18 = 2*3*3 | 2,3,3 are < 17 19 -> 19+1 = 20 = 2*2*5 | 2,2,5 are <19 23- > 23+1 = 24 = 2*2*2*3 | 2,2,2,3 are < 23 ... and so on.

I edited it for -> instead of equals. to me "->" means a sort of action. G(p) = p + 1 is the function and G^n(p) = 2^n is the result of n iterations of the function. I tend to think in a very step oriented logical (if then) fashion. unfortunately I dont express myself very well.

Or differently stated: Consider a function f(x) = x + 1 where x is prime. then x+1 is a composite of primes where each individual prime is less than x. Suppose you applied f(x) to that resulting number for some m iterations. Then each resulting prime eventually tends to 2. That is to say the first iteration of f(x) produces a number whos primes are less than x, then each subsequent iteration produces a number who's primes are less than the first prime x and each subsequent prime that is plugged into the function. such that after a number of iterations the composite number is 2^m. of course after dividing 2^m by the even function m times that results in 1.

Thank you for your insight: Here is the prose version: Since the most important function within the collatz conjecture is F(x) = 3x + 1 , I really tried to understand what it was this function did to the problem. Two facts are present, any number is either prime or a composite of primes. Therefore I broke it down and looked at a different equation: G(y) = K + 1. For y prime. Notice the composite number K +1 is less than 2K, therefore all of the primes within the number are also less than K. Now since K + 1 is a composite number it can be rewritten as a multiplicative sets of primes. Moreso, you can reduce one set by 1 and extract a prime. E.G. 9 +1 = 10 = 2*5 → (2-1)5 + 5 Therefore you can extract a prime (E.G: 5) and plug it back into G(y). For each iteration you reduce the resulting composite number and after some M iteration you can reduce it to 2. Now given the original collatz function F(x) = 3x + 1 , you can rewrite this as F(x) = 2x + x+1 . Notice that we have an x+1 and that we can also rewrite any composite number in such a way as to extract a single prime to work with. Ergo F(x) = 2x + G(y). Since G(y) = K +1 for K prime and converges to 2 after M iterations. Then F(x) = 2x + G(y) → F(x) = 2 (x + 1) and again (x +1) is our G(y) so F(x) = 2(G(y)) → F(x) = 2*2 for some M,N iterations. Then of course you can apply the even collatz function and it converges naturally to 1. Side note: Any odd function of the form ax+1 also converges to 1 following the exact same logic, and the collatz odd function is just a case where a = 3. I also hope it’s well enough written that you can see my thoughts on it.

I would like you guys to take a serious look at the mathematics, instead of attacking the semantics. Is it really that difficult for you to comprehend what the paper is trying to state?

"That's utter nonsense. If you claim a proof of an open problem you can't make up all your own terminology." - precisely why I need to work with a true mathematician. G(p) = p + 1 -> G^n(p) = 2^n G^n(p) / 2^n = 1 What I was trying to say was that after n iterations, G(p) = 2+ 2+ 2.... = 2^m AND G(p) / 2^m = 1 Here is a simple online calculator that shows G^m (p) = 1 else multiply by 3 <- replace 3 with 1. https://www.dcode.fr/collatz-conjecture

yes typo. It should read 3*2*11 +11 . The point of the rewrite is to show that any integer can be rewritten such that it is a number + a prime. Of course the whole purpose is that when you have a number + a prime you can then use the prime reduction theorem upon the (+ prime) part. "Oh I see you're using the word "convergence" in a funny way." - yes, my abilities to write proofs are limited and inexperienced.

I meant to say that using the prime reduction theorem, the odd collatz function can be rewritten such that it is a 2^m number and thus divided by the even function m times results in 1. What do you mean limiting? I didn't limit the collatz conjecture by any means. "ps -- Line 41 is wrong. 99 does not equal 33. " correct, the line should read 3 * 2 *11 + 11

Well here, let me just post my pdf. Im quite serious about my proof. Rip it to pieces and tell me im an idiot please. Draft2.pdf

Is anyone here interested in the Collatz Conjecture? If so I believe I have the solution, seriously, and I need to work with someone who knows how to write a formal proof better than I. There are some errors in my proof, but the underlying principle is right. I only have my minors in mathematics, so my proof writing skills are subpar.

Ahh, i remember this experiment and post! Forgive my inability to accomplish the experiment in question, I was preoccupied with serving in the air force. I picked an exponential function because at a 0 distance there would be a near maximum repelling force exerted, and at some distance (r) there would be a slight or nil force exerted. After (r) the rest of the distances would also have a very small force, whereas between 0 and (r) the forces would get a good bit stronger.

After a long hiatus, nearly 7 years, I have returned. I went and joined the air force, got a bachelors of science, a minors in math and a minors in bio; and currently work in IT/ software engineering. Any of the old members still active?

i don't seem to be able to stress this enough, I cant do it. it is a simple experiment, and im sure that the people who read this forum are likely to have access to/ or the supplies to accomplish it.

Like i said, i don't have the equipment anymore. Right now performing the experiment by myself just isn't feasible. I picked exponential (in my initial hypothesis), Subsequently the data i collected even though it was jittery still supported this hypothesis.

Hey fellas, I know its been a freaking long while since i posted or visited here last. But i was remembering an old experiment i did back in high school that i couldn't do right. I know that the data i was supposed to see was going to be a positive exponential increase. But my data was too jittery, mostly due to poor experiment setup and execution. Ok, this is what i was trying to find: If you have 2 permanent magnets ( i used 2 Neodymium barrel type magnets. 1 In Dia, 2 In height) How much force is one magnet being pushed away from the second magnet. My hypothesis: As the second magnet gets closer, the first magnet will be pushing away harder. Scaling exponentially, until there is no more space left between the two magnets. As distance between the two magnets decreases the force exerted by the first magnet to move away from the second one increases. My experiment design wasn't all that great as i had to basically jerryrig all of the devices together for unintended purposes. I used a grooved aluminum yardstick, (grove fit magnets pretty well and easily slid) and there was this electronic force scale that im pretty sure i had to turn on its side and attach to one end of the yard stick. Though, after calibrating, the only real issue was that i used my hand to slowly move the second magnet close. Several attempts were thwarted by jittery hand movements, and some were thwarted by the one magnet being pushed up instead of back. I set one magnet on the rail with the backside touching the force scale, while i placed the other magnet on the rail away from the first. i grasped the second magnet and slid it closer to the 1st. + = magnet - = Rail [ = force scale [+------------+ i moved the second magnet closer: [+----+ and so on. Also note, i used like sides as to repel one another. SO: now on to you guys. IF you would be so kind as to gather the data i was looking for. I don't have any of the equipment necessary, nor the magents either. I would really like it if you would also test several different sizes of magnets too. E.G: 1cm Diameter, 2cm long: 3cm diameter, 4cm long: 5cm diameter, 5cm long: 2cm diameter, 4cm long: 4cm diameter, 5cm long: 6cm diameter, 6cm long: 3cm diameter, 6cm long: 5cm diameter, 6cm long: 7cm diameter, 7cm long: