Zolar V

I understand you're frustration. It is my hope to show you I am right, though it may take some time in making it a proof given your commentary here. So lets get some terminology right: Convergence: I want to say a function converges to 2, if for any input the output is 2 after some amount of iterations. Definition: " property (exhibited by certain infinite series and functions) of approaching a limit more and more closely as an argument (variable) of the function increases or decreases or as the number of terms of the series increases" Would it be better to define convergence like this: For M iterations a function converges to 2 ? Theorem: A function that we are proving Lemma: A supporting argument for the theorem Corollary: An unintended result

Suppose the prime reduction function is true Take any odd natural number, It is either prime or composite. If it is prime, then adding 1 would result in all its primes being less than itself. (we've already agreed on this) if its a composite, then adding 1 would still result in at least 1 of its primes being less than the primes in the composite. (this needs to be shown. This will be the result of step 2) ----- if you split 3n+1 into 2n + (n+1), then you have an odd integer being added to 1. which fits our function. the drill down part of our function is the same iteration in the collatz function. Collatz has everything to do with the adding 1 mechanism. I believe the (3) part in 3n+1 is irrelevant. Since if this works it works for An+1 where A is any odd natural number.

yea i know, i just didnt want to say definitively... i didnt want to look like an idiot yeah so here is how its supposed to work 1: For P prime, show P+1 = 2^r *p_1 *p_2 *p_3... p_i . show 2 < p_i <P 2: Show how any odd composite number can be rewritten such that a function from step 1 can be applied to it 3: Since any odd composite number can have the function from step 1 applied to it, it also has the same prime reduction mechanism happening to it 4: Show how the odd collatz conjecture can be rewritten such that it looks like our function. 5: Since our function converges to 2^n , so does the odd collatz function 6: since the even collatz function divides 2^n, then after n times it converges to 1 7: and thats it.

Hmm, thats a tough question. I want to say it isnt going to matter for the proof, but I think it will be injective, i doubt it would be surjective. "So, onward. How does this prove Collatz?" - I believe our function is embedded in the odd function of the collatz conjecture. That is to say we can manipulate 3x+1 to be 2x + x+1 where x is our p. then 2x + (x+1) and (x+1) is our function. (yes i know, its not defined here) since G^n(p)= 2^n then 2x^n + 2^n -> 2^n(x+1) -> 2^n(G(x+1)) -> 2^n (2^m) then apply the even collatz function n*m times. 2^{n*m} / 2^{n*m} = 1 So I know there is a lot missing/wrong. But the underlaying mechanism to the collatz function is the prime reduction done by adding 1 to a prime. the odd collatz function will always converge at some point to a 2^n number, and as soon as it does the even function will reduce it to 1

all the terminal nodes are multiplied together to give a final result of 2^n and youre done

Possibly both operations, but the final result is going to be a 2^n number. (line 34 in my poorly written pdf) If all we need to do is multiply then that's all we need to do. Once we have that result, n will equal the number of steps/iterations we needed to get to a 2^n number. if we also divide by 2^n, then 2*n will equal the number of steps needed to get to 1.

Edit: I think it is sufficient. *

I think it is sufficient to show how the idea works and I think it gives a good indication of what mechanisms are in play that cause the collatz to converge. I am unsure if it is going to be sufficiently rigorous or if it will give us the right number of steps. For our purposes here, it may be enough to continue. If we reach a terminal branch and are stuck again, then it might be worth going back and looking at a different way to go about applying the Prime + 1 concept. E.G the addition idea

Oh thats really neat, i didnt know that. I thought we only had expressions for special primes.

I added a little extra to my last post that i thought of. Lets accept and work with the tree method first. If this gets nowhere then we can explore the adding idea. Though even if the tree method does work, I think it's also worth exploring the adding idea. I'm not surprised, being inductive would indicate you can systematically build primes. I mean technically you can, but this would be for all primes not special primes. Here is an odd thought: Suppose this idea was true. Some inverse equation/algorithm would build only primes. Each equation we have now that creates some subset of primes would somehow be a part of or equivalent to that equation/algorithm

Alright, I can work with that. I didn't know that is what you wanted. You are exactly right then for step 3. all you are doing is adding 1 to each prime, one prime at a time per iteration. in programming it would look something like this: P = prime While( P =/= 2^r ){ P +1 = 2^r P_1 *P_2 ... foreach (Prime in (P+1)) { Call (while P =/= 2^r) } } it's a nested loop. -very true that it does. but can you think of any number that is even, prime, and greater than 2? the largest prime in an even number must be half the size of the even number, Example: P is our first prime, if P+1 = 2^r * k then \[ \frac {P+1} {k} = 2^r \] Now suppose there is a prime factor in k that is larger than P, then if we were divide k we would not get a positive even number. I believe we also know that r >= 0 yeah, the powers of 2 we collect at the end were a very good indication of the number steps we needed to take.

ok then, I see the error. I will renew my efforts to clearly define a set of functions and the rules by which to use them. Cheers and thanks for all the help! Probably just be G( x+ h) = x + (h +1) ; where x is a natural number and h is prime. That would define the function well enough without losing the original idea behind a prime +1. Theorem 1 needs a complete rewrite.

Our function G(p) = p+1. p = 2^n x 7 x 7 x 7 x 101 x 101 x 103 x 103 x 103 x 103 Which also equals 103 added together 2^n x 7 x 7 x 7 x 101 x 101 x 103 x 103 x 103 times. if we use the sum, we can apply G to it. Resulting in G(p) = .... + (103+1)

"As I asked earlier, what do you do with 2^n x 7 x 7 x 7 x 101 x 101 x 103 x 103 x 103 x 103? How do you recurse down this configuration of primes?" Lets select 103. then we have 2^n x 7 x 7 x 7 x 101 x 101 x 103 x 103 copies of 103 added together. G(103+103+103+103+103+103+103+103+.......+103) = 103 +103+103 .... + (103+1) \[ 2^n * 7 * 7 * 7 * 101 * 101 * 103 * 103 * 103 * 103 = \sum_1^{ 2^n * 7 * 7 * 7 * 101 * 101 * 103 * 103} 103 /]

"No prob, take your time. I doubt anyone will beat you to Collatz and claim credit. You have a few more days at least, I'm sure." - I would hope so, Ive been sitting on this for almost a year. I am attempting to explain how I handled 5 and 7 in the computation of G(139). We are following the paper, line 40 to be exact. "What do you mean "extracting" 7 20 times? This really is very unmotivated, it's not possible for me to follow your logic. As I asked earlier, what do you do with 2^n x 7 x 7 x 7 x 101 x 101 x 103 x 103 x 103 x 103? How do you recurse down this configuration of primes?" extract, to take out one piece of a whole. Think about a number as an entire collection of collections of objects that are equivalent. If we only consider the additive operation then the number of equivalent objects within a number is how many different numbers can add up to equal that number. E.G. 7. 1+6 = 2+5 = 3+4 ... = 2+2+2+1 = ... 7 A whole is 7, a piece would be take out a 2 or a 4 or some other part of 7. In our case our object is 2^2*5*7 which means you can extract a 2,5, or 7 and there are 2*5*7 , 2^2 *7, or 2^2*5 copies respectively. If we extract 7, then we are extracting all 2^2*5 copies of 7. So 7 +7+7+7+7+7+7+7+7+7+7+7+7+7+....+7 And we iterate through that sum of primes. "As I asked earlier, what do you do with 2^n x 7 x 7 x 7 x 101 x 101 x 103 x 103 x 103 x 103? How do you recurse down this configuration of primes?" - Currently we select a prime and reconfigure the product of primes into a sum of that prime. Then iterating through that on each prime. The core concept of adding 1 to a prime and reducing the primes within it is the primary mechanism we want to chase. On a very rudimentary level, the mathematics that I know, only allow you to do something like add to a particular value only if you can group it together. Obviously you cannot just select random primes within a product to add 1, but you can do just that over a sum of primes.

"As an example we have G(139) which is computed as 139 -> 140 -> 2^2 x 5 x 7. " - the next step is to extract either 7 or 5. if we chose 7 then we need to extract it 2^2 *5 times, (hence why i started to talk about sums.) I'm sorry i desperately looked for my notebooks and work has been real busy.

I don't think Theorem 1 is wrong yet, but I think there is an important step missing between the application of Theorem 1 on primes to the application of Theorem 1 on all natural numbers. I really wish I could remember, or find, why I deduced that if Theorem 1 was true for primes it was true for all natural numbers. I believe my logic went something like this: Step 1: Theorem 1 is true for Primes. Step 2: Any number can be written as a product of primes. Step 3: Any product of primes can be rewritten to be a number (a) + a prime (b)... a +b. Step 4: Since I have a prime (b) that stands alone, I can apply Theorem 1, Then that means Theorem 1 is true for all natural numbers. Note: The result on Theorem 1 is it reduces value of the primes in the output to be less than the input prime ( 2 < p_i < p) . I think you are correct that the 3rd step isnt fully developed and it indeed cooks the goose. If we can flesh out Step 3 then we can get back on track. I also think at a minimum, this Prime Reducing function is ultimately why the collatz conjecture converges. It's just a matter of proving step 3, and then showing how the collatz conjecture is simply an extension of the result. Namely the odd function can be rewritten to be (2x) + (x+1). and x+1 is our prime reducing function. OK. onward to fixing step 3! " What is the above symbology supposed to be about? What's your claim, what are you proving, what is your conclusion? " - I believe a sum of primes is what i tried to state on line 40, though I did not state it well. - I claim any Natural number can be rewritten as at least a sum of a prime. There are many sums for natural numbers and that number of sums is related to how many primes there are in the composite number, though we only need to look at just one sum at the moment. - If we can prove that a sum of a prime is equivalent to a product of primes for any given natural number then we can apply our theorem 1 to that number. - I conclude that the number of times we need to apply G to a natural number is equal to the product of primes of that natural number divided by the prime we are summing. example: 5 * 7 * 47 * 47 * 113 * 113 * 113 = p \[ \sum_{i=1}^{5 * 7 *47 *47*113} 113 = p = 5*7*47*47*113*113*113 \] \[ n = \frac {5*7*47*47*113*113*113} {113} \] \[ G^n (p) = 2^m \] I believe my original assessment was applying G to the product of primes, which is not a valid move. So I suggest we look at rewriting the product of primes as a sum of a prime. We will need to later look at which sum is the right sum for the least number of times we need to apply G. Edit: I hope you dont think im an idiot.

I'm sorry WTF, I can't find all of my notebooks. The best I can discern is this was one of the places I needed some good thought and collaboration on. I think I naively applied G on none prime numbers. Though later I did add the Remark at line 37. It appears that what I did here was essentially rewrite a number as a sum of a prime number, and then that allowed G to be applied. Let us take writing any natural number as a sum of primes. \[ H(k) = \sum_{i=1}^{2^r * p^1 *p^2 *... p^{i-1} } p_{i} \] I think here there exists a sum that has the least steps needed for G(p) to equal a power of 2 Suppose there are at least i number of sums for any natural number then these are our sums \[ \sum_{i=1}^{2^r * p^1 *p^2 *... p^{i-1} } p_{i} = \sum_{i=1}^{2^r * p^1 *p^2 *... p^{i} } p_{i-1} = ... =\sum_{i=1}^{ p^1 *p^2 *... p^{i} } 2^r \] and \[ G_1^{2^r * p^1 *p^2 *... p^{i-1} } ( \sum_{i=1}^{2^r * p^1 *p^2 *... p^{i-1} } p_{i}) = 2^{m} \] \[ G_2^{2^r * p^1 *p^2 *... p^{i} } (\sum_{i=1}^{2^r * p^1 *p^2 *... p^{i} } p_{i-1}) = 2^{m} \] \[ G_i^ {p^1 *p^2 *... p^{i} } ( \sum_{i=1}^{ p^1 *p^2 *... p^{i} } 2^r ) = 2^{m} \] Where \[ G_1 \leq G_2 \leq ... \leq G_i \]

I think as we work through this i'm having new ideas about it, here is one of them: maybe it would make sense to first state that any natural number can be written as a non distinct sum of primes. E.G 3*7 = 3 +3+3... +3 or 7+7+7 The proof if we need to show it could probably be a straight forward one. Suppose there exists a natural number that is not a sum of primes. by the fundamental theorem of arithmetic, n is either prime or a product of primes. Case 1: n is prime if n is prime then n = ∑ n ; lower index = 1, upper index = 1 Case 2: n is a product of primes if n is a product of primes then n = 2^r * p_1.. *p_i = ∑ p_i ; lower index=1, upper index = 2^r *p_1 *p_2 *...p_(i-1) Since n is either a sum of itself or a sum of primes given by the prime factorization then there does not exist a natural number that is not a sum of primes and all natural numbers greater than 2 are sums of primes. I suppose this also says that any number is either a sum of itself or has a i number of equivalent sums of primes. The purpose in stating that any natural number can be written as a sum of primes is to validate our G function on all natural numbers. G is still G(p) = p + 1 ; p is an element of Natural numbers and is prime. Here i am saying you need 2^r *p_1 *p_2 *...p_(i-1) iterations of G on H(k) in order to add 1 to each prime p_i. you would be left with each (p_i + 1) = 2^m * q . so you would have 2^r *p_1 *p_2 *...p_(i-1) many (2^m * q) Note: q does not divide (p_i + 1), in fact q < p_i I think the next step would be to rewrite (2^m *q) to a new sum of primes. and then apply G to it again. and keep on rewriting and applying until you drill down to 2's

So here is where it gets a little tricky. You cant recurse 5,7 separately since they're multiplied together, but you can select which you want to recurse through. Once you select which prime you want to recurse, you manipulate the product such that you have a copy of that prime you selected. Suppose you select 7 then 5*7 = (5-1)7 + 7 and you can recurse through the first copy of 7 you have extracted. Then recurse through all 5 copies of 7. Unfortunately my gut is saying im missing something here or something isnt quite right. I need to go find and look at all my notes again. Something is a little off, but its also very late. I'm gonna think about it and see if I can't remember whatever it is that's lurking in the shadows of my mind.

Ohhh! im sorry I read it wrong. I see now what you're doing. You have defined 2^n as the largest power of 2 within p+1. You then move it over and leave yourself the non-distinct product of prime factors. I got mixed up on your step 4 and assumed that you were further along, forgetting entirely that we are only trying to define G. In that case, yes I think you are on the right path. Hey man! Glad to see you're still around here too!

Maybe I'm reading yours wrong, but it appears you're doing a modified form of the collatz conjecture. if it is odd then add 1 if it is even divide by 2 What I am trying to show is this: given any prime number ,P > 2, and adding 1 to it results in a new number who's primes are all less than the original prime. (placeholder detail step) Then doing that again to each subsequent prime number until you're left with 2's 2 <p_i < p * The placeholder detail step is a very important step. You decompose the product of primes so that you can add 1 to a particular prime without affecting the rest of the number(s). That way can add 1 to only that prime and not some odd number that we know little about. I first toyed with the idea that we didn't need to constrain the function to primes, but it was by primes that everything falls into place. Also given every number written as a product of primes, that number can also be written as a summation of primes. The summation part is the part we need in order to apply our p+1 to particular primes. It seemed to me that I only needed to show that primes worked and that any nonprime odd number can be rewritten as a summation of primes. E.G Let p,q,r be prime p*q*r + 1 doesn't seem to yield much information at all, other than being even. but (p*q-1)*r + (r +1) seems to yield more information since we can add 1 to our particular prime r. we know that (r+1) < 2r , which means the composite number (r+1) does not contain r in its prime factorization. maybe a summation would be a better visual tool: p*q*r = ∑r lower index = 1 upper index = p*q Here the summation tells us that we need to add 1 to each r a minimum of p*q times. p*q here is the number of iterations we add p+1 the subsequent (r+1) still yields the (r+1) < 2r information as well. So we know whatever number (r+1) is, that number doesnt contain r in its prime factorization, and each prime factor is also strictly less than r " Well then now I'm perfectly well dispirited " - My apologizes my friend; I'm as terrible as I am at relating my thoughts and ideas in person , I am even worse online. Let P be prime If P+1 then the prime factorization of P+1 = 2^m * K_1 *K_2*K_3... and each K_i < P

I actually was about to use LaTex, but decided to use MS Word's insert equation function instead. Its more clunky but a bit faster for a mockup. I would say that you're generally right. I don't divide mine by 2n for the sole purpose of saving them all for the end. I think the tricky detail you're missing is converting the product of primes into a summation of the primes. That conversion lets you work with one prime at a time and also gives you some detail about how many times you need to do step 3 Maybe a bit better of a way to think about it would be to convert all products back into basic addition. 5*7 = 7 + 7 + 7 + 7 + 7

"Anyway I just want to make sure I've got this. You input a prime and add 1. If it's a power of 2 you're done. Otherwise divide out the highest power of 2 from p+1 and factor it into a product of primes 2 < p_i < p, each prime written as many times as needed. Then you recursively apply G to each of the primes and eventually everything drills down to a power of 2. I have no idea what that proves or why it's important, but we'll get to that later. Do I have the basic algorithm right?" -Mostly, I don't divide out the 2's until the end. That way I collect all of the division that I need to do all at once. You input a prime and add 1, if its a power of 2 you're done. if not then you factor the composite number into a product of primes and manipulate it to give you a prime to work with. Then recursively apply G again to each of the primes. once you have everything drilled down to a power of two for that piece of the proof. Otherwise for the collatz part, you collect your two's and divide them all at once, giving you a single letter representation for how many steps it takes to go to 1 from whatever power of 2 number you drilled down to. maybe the definition you're trying to get at looks like this ignore H, I don't believe it gets us anywhere. Maybe the definition you're looking for is G(j(X)). So we have this prime factorized number that is manipulated to be a number + a prime. We then plug that into G who takes the only prime number we have available and adds 1 to it. We then go back and factorize, manipulate and replug into G. we do this for as many steps as it takes until we get a power of 2. the 2 < p_i < p inequality gives us a lot of information about the resulting primes. Interesting enough, a few weeks after I had written my idea and worked with it on this problem I stumbled on this: https://en.wikipedia.org/wiki/Bertrand's_postulate Which appears to be saying very similar things to me. " I never realized it at the time, but a lot of people get degrees in math without learning any math. " I suppose, but they were the ones that spent the time studying, going to office hours, and getting high marks.

All the people that used to be on these forums, relatively (to the lifespan of a fruit fly) of course.