Trurl Posted September 9, 2018 Share Posted September 9, 2018 I think the approach is brilliant and maybe simple enough to work. But I am still not convinced. You take 2 to a power and multiply by Prime factors. I agree that you can build any number this way, but I don’t believe multiplying by 2 here is the same as dividing an even number by 2 in the conjecture. The order of operations. And I cannot test this algorithm in a computer since we cannot factor the numbers in this series. Again I’m probably wrong but this is not a linear series. My understanding is that you are building a pattern linearly and the conjecture is not a one-to-one function. This is just my understanding but if you can prove your method does this I will buy multiple copies when you get it published. Link to comment Share on other sites More sharing options...

Zolar V Posted September 9, 2018 Author Share Posted September 9, 2018 (edited) 1 hour ago, Trurl said: I think the approach is brilliant and maybe simple enough to work. But I am still not convinced. You take 2 to a power and multiply by Prime factors. I agree that you can build any number this way, but I don’t believe multiplying by 2 here is the same as dividing an even number by 2 in the conjecture. The order of operations. And I cannot test this algorithm in a computer since we cannot factor the numbers in this series. Again I’m probably wrong but this is not a linear series. My understanding is that you are building a pattern linearly and the conjecture is not a one-to-one function. This is just my understanding but if you can prove your method does this I will buy multiple copies when you get it published. Where do you think i'm multiplying by a power of two? I'm doing nothing of the sort, I'm representing any natural number as a product of primes then to a sum of a prime. The iterative approach in conjunction with the inequality property of a prime plus 1 , is how I create a power of 2 number. take the number \(15 \) and the number \(18 \) then for \( 15 \): \[ 15 = 2^0 * 3*5 = \sum_{1}^{1*3}{5} = 5+5+5 \] let \(w = 1*3 \) then \[ \sum_{1}^{1*3 }{5} + w = (1 + 1 + 1) +5+5+5 = (5+1) + (5+1) + (5+1) \] Note: each \( (5+1) < 2*5 \) therefore the all primes in \( 5+1 \) are less than \(5 \) Notice the \(w \) is the result of the iterative approach to adding 1 to each prime, and is allowed via the conjectures wording. "For any natural number \( n \) there exists an \( i \) such that \( f^i (n) = 1 \)" . basically there is an \(i\)th iteration of the piece wise defined collatz function such that \( f^i( n) = 1 \) \(18 \)is solved similarly. for \(18\): \[ 18 = 2^1 * 3*3 = \sum_{1}^{2^1 * 3}{3} = 3+3+3+ 3+3+3 \] let \(w = 2*3 \) then \[ \sum_{1}^{2^1 * 3}{3} = 3+3+3+ 3+3+3 + w = (1 + 1 + 1 + 1 + 1 + 1 ) +3+3+3 +3+3+3= (3+1) + (3+1) + (3+1) + (3+1) + (3+1) + (3+1) \] clearly each \( (3+1) < 2*3 \) and again therefore all primes in \( (3+1) \) are less than \(3 \) What happened to me was this: While looking at the problem and breaking it into products of primes for each iteration, I began to wonder what the effect of adding 1 was. I made an intuitive leap to looking at what happens when I add 1 to a single prime, and therein lay the answer to the whole conjecture. the behavior of "adding 1 to a single prime" is embedded in the problem. The iterative nature implies we can add any number of 1's , which means we can "fully saturate" a sum of a prime. The multiplication by 3 in the collatz conjecture is irrelevant to the behavior, in fact we can look at a generalization and find a general formula to the collatz function. for any prime \(a)\ the general collatz is as follows Where \( b , c , ... ,(a-1) \) are primes less than \(a\) . Note \(a-1 \) is not literally \( a -1 \), it is the previous prime. 2 and 3 have a slightly special relationship as primes since they are right next to each other. So equations that do what the collatz does are simple to write, primes further away result in a less simply defined "collatz" function. Edited September 9, 2018 by Zolar V Link to comment Share on other sites More sharing options...

Trurl Posted September 13, 2018 Share Posted September 13, 2018 I think I have an understanding of what you are doing with factoring. You just want it in that form. I would like to see an updated proof, since this thread has so many explanations. This is the first time I saw this conjecture, so I hope some experts on this forum take interest. The problem is worth working through. Right or wrong it is a fresh approach. Just because the conjecture is unsolved does not mean the idea isn’t valid. Link to comment Share on other sites More sharing options...

Zolar V Posted September 13, 2018 Author Share Posted September 13, 2018 1 hour ago, Trurl said: I think I have an understanding of what you are doing with factoring. You just want it in that form. I would like to see an updated proof, since this thread has so many explanations. This is the first time I saw this conjecture, so I hope some experts on this forum take interest. The problem is worth working through. Right or wrong it is a fresh approach. Just because the conjecture is unsolved does not mean the idea isn’t valid. I agree. I really need to write the proof with the new information that has been garnered here. That will have to wait since I am moving. I really hope some experts take interest since I cannot find anything wrong with it. The method seems right, now I just have the daunting task of writing it well enough that other people can see what I see. Thank you for your interest in the problem! Link to comment Share on other sites More sharing options...

## Recommended Posts

## Create an account or sign in to comment

You need to be a member in order to leave a comment

## Create an account

Sign up for a new account in our community. It's easy!

Register a new account## Sign in

Already have an account? Sign in here.

Sign In Now