Zolar V Posted August 26, 2018 Author Share Posted August 26, 2018 (edited) Lemmas: Lemma 1: Let P be prime and P >2. The smallest natural number N > P to contain P as prime is 2P. Lemma 2: Any natural number N can be written as a sum of one of its primes. Case 1: N is prime \[N = \sum_{1}^{1} N \] Case 2: N is a composite \[ N = 2^q *P_1 *P_2 *..*P_i \] \[ N = \sum_{1}^{2^q *P_1 *P_2 *..*P_{i-1}}{P_i} \] Notice if q = 0 then N is odd, else N is even. Lemma 3: For a prime P >2 G(P) = P+1 and \[ P+1 = 2^r *q_1 *q_2 *..*q_n \] where \[ 2 \leq q_i < P \] for i = 1,2,3 ...n Lemma 4: For any natural number P >2 written as a sum of one of its primes P_i , then \[ m*G(\sum{P_i}) = G_1(P_i) + G_2(P_i) + G_3(P_i)+...+ G_m(P_i) \] notice each \[G_m(P_i) = P_i +1 \] . Since \[P_i +1 < 2p \] all of its primes are \[2 \leq q_i < P \] for i = 1,2,3..n Lemma 5: For any natural number P >2, a function \[H^k(m*G(\sum{P_i})) = 2^r \] Where k is the number of iterations of \[ m*G(\sum{P_i}) \] Edited August 26, 2018 by Zolar V Link to comment Share on other sites More sharing options...

wtf Posted August 26, 2018 Share Posted August 26, 2018 (edited) 1 hour ago, Zolar V said: Lemmas ... @Zolar, Did you understand @taeo's excellent point? Let f(n) = 2 for all natural numbers 2. Then f satisfies your condition on G. It always outputs a power of 2, namely 2^1. Now given some number n, let's apply Collatz. If it's even, then eventually it either hits ("converges to" in your terminology") 1; or else it hits an odd number. If it's odd, then f(n) = 2. So f is "inside the Collatz function" in your concept. Yet I hope you can see that f is just some arbitrary function that has nothing to do with Collatz. It doesn't prove anything. Do you understand this example? Edited August 27, 2018 by wtf Link to comment Share on other sites More sharing options...

Zolar V Posted August 27, 2018 Author Share Posted August 27, 2018 6 minutes ago, wtf said: Did you understand @taeto's excellent point? Let f(n) = 2 for all natural numbers 2. Then f satisfies your condition on G. It always outputs a power of 2, namely 2^1. Now given some number n, let's apply Collatz. If it's even, then eventually it either hits ("converges to" in your terminology") 1; or else it hits an odd number. If it's odd, then f(n) = 2. So f is "inside the Collatz function" in your concept. Yet I hope you can see that f is just some arbitrary function that has nothing to do with Collatz. It doesn't prove anything. Do you understand this example? Taeto's and your point here are both irrelevant. Of course there exists some function f(x) that satisfies the conditions on G, stating such a fact is mostly irrelevant to the topic at hand. Since we assume the function already exists, our purpose here is to elucidate that function and show how the Collatz function is a natural byproduct of it. Also it isn't "my" terminology, it is the terminology used by Collatz himself. If you notice Lemma 4, m*G(sum P_i) = 2k or written differently: Let K > 2 be an arbitrary natural number with P_i primes not necessarily distinct, and P be the highest element of the set P_i then K + i = 2r where each of 2r's primes are 2 \leq q_i < P Proof: \[ \sum_{1}^{2^r * P_1 *P_2 *..P_{i-1}} {P} + i = 2r \] \[ {P_1}_1 + {P_1}_2 +{P_1}_3 +... + {P_1}_i + 1_1 + 1_2 + 1_3 +... +1_i = 2r \] notice each P_1 has a corresponding 1, then \[ ({P_1}_1 + 1) + ({P_1}_2 + 1) + ({P_1}_3 + 1) + ({P_1}_4 + 1) + ... + ({P_1}_i + 1) = 2r \] since each P_1 is prime, it can also be written as 2k+1, so \[ (2k+1 + 1) + (2k+1 + 1) + (2k+1 + 1) + (2k+1 + 1) + ... + (2k+1+ 1) = 2r \] Clearly each element is 2k+2 = 2(k+1) = 2m , then \[ 2m + 2m + 2m +...+ 2m = 2r \] Notice each \[ {P_1}_i + 1 < 2p \] then \[ {P_1}_i +1 = 2^r * q_1 *q_2 *... *q_n \] where \[ 2 \leq q_i < {P_1} i = 1,2,3,4,...,n \] Link to comment Share on other sites More sharing options...

Trurl Posted August 27, 2018 Share Posted August 27, 2018 Are you saying that to add 1 to a number divisible by 2 will make it odd? And if you can determine if this odd number is a semi-Prime the Prime factors must be less than 2 to the power of the semi-Prime? My question is what way did you use the conjecture to solve the problem of factoring? This will increase in difficulty with larger numbers. Lemme 2, lines 48 & 49 lost me. Link to comment Share on other sites More sharing options...

wtf Posted August 27, 2018 Share Posted August 27, 2018 12 minutes ago, Zolar V said: Taeto's and your point here are both irrelevant. Of course there exists some function f(x) that satisfies the conditions on G, stating such a fact is mostly irrelevant to the topic at hand. Since we assume the function already exists, our purpose here is to elucidate that function and show how the Collatz function is a natural byproduct of it. Why isn't Collatz a "natural byproduct" of f(x) = 2? Can you explain the difference in a few words? Why is G related to Collatz and f isn't? Link to comment Share on other sites More sharing options...

Zolar V Posted August 27, 2018 Author Share Posted August 27, 2018 3 minutes ago, Trurl said: Are you saying that to add 1 to a number divisible by 2 will make it odd? And if you can determine if this odd number is a semi-Prime the Prime factors must be less than 2 to the power of the semi-Prime? My question is what way did you use the conjecture to solve the problem of factoring? This will increase in difficulty with larger numbers. Lemme 2, lines 48 & 49 lost me. "Lemme 2, lines 48 & 49 lost me." Are you referring to the original posted draft? (draft 2) The function H(x) = x/2 corresponds to the even collatz function. Since our function G^m(P) = 2^t over some m iterations, then taking H^t(G^m(p)) = 1 after t times. "Are you saying that to add 1 to a number divisible by 2 will make it odd?" No, i'm saying adding 1 to any odd integer P results in a number who's prime factors are less than the the highest prime factor in P. "And if you can determine if this odd number is a semi-Prime the Prime factors must be less than 2 to the power of the semi-Prime?" I haven't even explored the semi-primes as a distinct set, so I am unable to answer your question."My question is what way did you use the conjecture to solve the problem of factoring?" - I used the fundamental theorem of arithmetic to write an arbitrary number of prime factors to any natural number i needed. 6 minutes ago, wtf said: Why isn't Collatz a "natural byproduct" of f(x) = 2? Can you explain the difference in a few words? Why is G related to Collatz and f isn't? They're both related. Since we already assume there exists a function , f(x) gives no information and is useless. Where as G is shown to reduce all primes to 2 and is the underlying mechanism behind why the collatz function converges to 1. f(x) gives no useful information G gives useful information. Link to comment Share on other sites More sharing options...

wtf Posted August 27, 2018 Share Posted August 27, 2018 (edited) 32 minutes ago, Zolar V said: f(x) gives no useful information G gives useful information. If I failed to see the distinction, what simple phrase or couple of sentences could you say to me that would give me a glimmer of an idea as to why you see a difference between G and f? G isn't related to Collatz, it's just some function you made up. That's how it seems to me. Can you articulate why you think these are different? Edited August 27, 2018 by wtf Link to comment Share on other sites More sharing options...

Zolar V Posted August 27, 2018 Author Share Posted August 27, 2018 (edited) 4 hours ago, wtf said: If I failed to see the distinction, what simple phrase or couple of sentences could you say to me that would give me a glimmer of an idea as to why you see a difference between G and f? G isn't related to Collatz, it's just some function you made up. That's how it seems to me. Can you articulate why you think these are different? If you are unable to understand the literal pages of dialog on the very topic, then no I cannot re-articulate it in a way you may understand. Edited August 27, 2018 by Zolar V Link to comment Share on other sites More sharing options...

taeto Posted August 27, 2018 Share Posted August 27, 2018 4 hours ago, Zolar V said: If you are unable to understand the literal pages of dialog on the very topic, then no I cannot re-articulate it in a way you may understand. It appears the issue is with your phrase "there exists". What you stated, by accident means literally that there are some functions that reach powers of 2 after some iterations. What you intended to express was that *your* particular function G has this exact property (among others). I am not sure about what properties G is assumed to have exactly; at least if it follows from those assumed properties that G also has this property, then it has some similarity to the conjectured property of the Collatz function, that is granted. Link to comment Share on other sites More sharing options...

Zolar V Posted August 27, 2018 Author Share Posted August 27, 2018 (edited) 8 hours ago, taeto said: It appears the issue is with your phrase "there exists". What you stated, by accident means literally that there are some functions that reach powers of 2 after some iterations. What you intended to express was that *your* particular function G has this exact property (among others). I am not sure about what properties G is assumed to have exactly; at least if it follows from those assumed properties that G also has this property, then it has some similarity to the conjectured property of the Collatz function, that is granted. Ahh i believe i see, but the statement "there exists" only concludes that some functions exist. The existence of those functions doesn't negate the existence of my function, nor does it conclude the uniqueness to any such functions. I can change the wording if that pleases the masses. This seems to fall out from the lemmas : for any a^n , a is prime and n >1 \[ a^n + a^{n-1} = 2^2 * a^{n-1} \] so \[ \sum_{1}^{2^r*P_1 *P_2*...* P_{i-1}}{P} + i = a^n + a^{n-1} = 2^2 * a^{n-1} = 2r \] Edited August 27, 2018 by Zolar V Link to comment Share on other sites More sharing options...

taeto Posted August 27, 2018 Share Posted August 27, 2018 1 hour ago, Zolar V said: Ahh i believe i see, but the statement "there exists" only concludes that some functions exist. The existence of those functions doesn't negate the existence of my function The existence of such functions has no logical impact whatsoever, since it is a trivial fact. 1 Link to comment Share on other sites More sharing options...

Zolar V Posted August 27, 2018 Author Share Posted August 27, 2018 Just now, taeto said: The existence of such functions has no logical impact whatsoever, since it is a trivial fact. exactly, hence why it was irrelevant to the discussion. Link to comment Share on other sites More sharing options...

taeto Posted August 27, 2018 Share Posted August 27, 2018 (edited) 2 hours ago, Zolar V said: for any a^n , a is prime and n >1an+an−1=22∗an− Did you mean to write that for any prime \(a\) and any natural number \(n\) you have \[a^n + a^{n-1} = 2^2 \cdot a^{n-1}?\] If + refers to usual addition of integers and \(\cdot\) means ordinary multiplication, then after cancellation of \(a^{n-1}\) this becomes \(a + 1 = 4,\) so \(a = 3,\) and the problem then is that this is not true for every prime \(a.\) Did you mistype something? Edited August 27, 2018 by taeto Link to comment Share on other sites More sharing options...

Zolar V Posted August 27, 2018 Author Share Posted August 27, 2018 (edited) 1 hour ago, taeto said: Did you mean to write that for any prime a and any natural number n you have an+an−1=22⋅an−1? If + refers to usual addition of integers and ⋅ means ordinary multiplication, then after cancellation of an−1 this becomes a+1=4, so $a = 3,$ and the problem then is that this is not true for every prime a. Did you mistype something? I suppose I did, I was checking values for \( a > 2 \) and \( n>2 \). The operations are the usual addition and multiplication under the group of natural numbers. I think a clarification would be \( a >2 \) and \( n>2 \) , but i have not fully investigated it. just cursory work while i was also at work. take \( 3^3 \) then \( 3^3 + 3^2 = 2^2 * 3^2 \) then \( 2^2 *3^2 + (2^2 * 3) = 2^4 *3 \) then \( 2^4 *3 + 2^4 = 2^6 \) i also noticed that odd powers of 2 skipped the next even power.. so \( 2^1 -> 2^4 ; 2^3 -> 2 ^6 \) Edited August 27, 2018 by Zolar V Link to comment Share on other sites More sharing options...

Zolar V Posted August 27, 2018 Author Share Posted August 27, 2018 I retract my statement. I am missing something. Link to comment Share on other sites More sharing options...

Zolar V Posted August 28, 2018 Author Share Posted August 28, 2018 (edited) For any prime \( a \) and any natural number \( n \) you have \[ F_0 = a^n+ 2^{2(0)} * a^{n-1} \] \[ F_1 = F_0 + 2^{2(k+1)} * a^{n-2} \] ... \[ F_k = F_{k-1} + 2^{2(k+1)} * a^{n-(k+1)} \] if \(k+1=n \) then \[F_k = F_{k-1} + 2^{2(k+1)} * 1 = 2^{(k+1)+2} \] Note: \(k\) as a counter for steps is off by 1, since we start at step 0. Hence \( k+1 \) in each \( k \) term I'm having difficulty writing the function properly for the terms. it seems it is the prior result + the prior result where its highest power is decremented by 1. I apologize for my atrocious math, I am exploring unknown areas and am prone to error. Example: a,n = 3 \( F_0 = 3^3 + 2^0 * 3^2 \) \( F_1 = F_0 + 2^{2*1} * 3^{1} \) \( F_2 = F_1 + 2^4 *3^0 = 2^{2*3} \) \( 3^3 + 3^2 + 2^{2*1} * 3^{1} + 2^4 *3^0 = 2^{2*3} \) Edited August 28, 2018 by Zolar V Link to comment Share on other sites More sharing options...

Trurl Posted August 29, 2018 Share Posted August 29, 2018 I am also an amateur mathematician. I agree your last example equals. But with all these posts the thread is out of control. I have read 3pages but all these changes confuse the reader. I know it is for your benefit to improve your proof, but I can’t understand it. I know it is bad practice to work with only limited examples, but could you show from the beginning an example? Here is an excerpt from the book Primes and Programing “The fundamental theorem 1.1 has a simple look about it, and indeed it is nice to know that every number greater than 1 is prime..or factors into primes. But actually finding the factorization…(or primality) can be very hard.” I request a through example because I don’t know what is the given and what we are solving. I think an example will show what you are describing and what variables we know. Can you write in 2 sentences what is the goal of the Proof. (A short abstract) Again I am also an amateur mathematician but for your proof to work everyone must understand what you are attempting to show. I like to work with examples when trying to understand series. Link to comment Share on other sites More sharing options...

Zolar V Posted August 29, 2018 Author Share Posted August 29, 2018 (edited) 3 hours ago, Trurl said: I am also an amateur mathematician. I agree your last example equals. But with all these posts the thread is out of control. I have read 3pages but all these changes confuse the reader. I know it is for your benefit to improve your proof, but I can’t understand it. I know it is bad practice to work with only limited examples, but could you show from the beginning an example? Here is an excerpt from the book Primes and Programing “The fundamental theorem 1.1 has a simple look about it, and indeed it is nice to know that every number greater than 1 is prime..or factors into primes. But actually finding the factorization…(or primality) can be very hard.” I request a through example because I don’t know what is the given and what we are solving. I think an example will show what you are describing and what variables we know. Can you write in 2 sentences what is the goal of the Proof. (A short abstract) Again I am also an amateur mathematician but for your proof to work everyone must understand what you are attempting to show. I like to work with examples when trying to understand series. Don't worry I'm an amateur as well. Abstract: I am attempting to show a function G(p) = p+1 for primes, reduces the primes in it. Next, I am going to show how that function applies not just to primes, but to any natural number. Then as an extension I will show how repeated iterations of that function converge to a power of 2. Then I am attempting to show the Collatz conjecture's odd function is a natural byproduct of it, and of course the even Collatz function will naturally result in 1 after repeated iterations. Couple sentences longer, but those are the steps. Essentially the core concept is the addition of 1 to a prime number ultimately reduces the primes within it. That is the behavior I noticed in the odd Collatz function. I only need to prove it for primes since every other number that is not prime, is simply a product of primes. So by some rewrite or manipulation, it can be shown that the same concept still applies. Note: we are working with Natural numbers for the symbols and their powers. Prime reduction example: Primes: 5,7,11,13,17,97 5+1 = 6 ; 6 = 2*3 ; Where 2,3 <5 7+1 = 8; 8 = 2*2*2 ; Where 2 <7 11+1 =12; 12 = 2*2*3; Where 2,2,3 <11 ... 97+1= 98; 98= 2*7*7 ; where 2,7,7 <97 Let \( k \) be prime then \( k +1 = 2^r * P_1 * P_2 * P_3 *... * P_n \) Since \( k +1 < 2k \) , the prime product does not contain \( k \) and \( 2 \leq P_i < k \) for each \( i = 1,2,3,4...n-1, n \) Non Prime number extension example: Let \( j \) be a composite natural number then \( j = 2^r * q_1 * q_2 *q_3 *... * q_n \) primes not necessarily distinct, where \( q_n \geq q_{n-1} \geq ...\geq q_1 > 2 \) If \( j \) is odd then \( r=0 \). else \( j \) is even and \(r >0 \) So \[ j =2^r * q_1 * q_2 *q_3 *... * q_n = \sum_{1}^{ 2^r * q_1 * q_2 *q_3 *... * q_{n-1}} q_n \] then \[ j +1 =2^r * q_1 * q_2 *q_3 *... * q_n +1 = \sum_{1}^{ 2^r * q_1 * q_2 *q_3 *... * q_{n-1}}q_n +1 \] Notice \( q_n +1 < 2q_n\) then the primes in \(q_n +1 < q_n \) Let \( i = 2^r * q_1 * q_2 *q_3 *... * q_{n-1} \) then \[ j +i =2^r * q_1 * q_2 *q_3 *... * q_n + i = \sum_{1}^{ 2^r * q_1 * q_2 *q_3 *... * q_{n-1}} q_n +i = 2 S \] Where each prime in \( S \) is less than \( q_n \) By the first example, we show how adding 1 to a prime number creates a new number who's primes are less than the original prime. By the second example we show how we can still apply the same concept to any natural number. Then its just a matter of rewriting the odd Collatz function to look like our function and showing how they both do the same thing to the numbers. Edited August 29, 2018 by Zolar V Link to comment Share on other sites More sharing options...

uncool Posted August 30, 2018 Share Posted August 30, 2018 (edited) On 8/28/2018 at 11:28 PM, Zolar V said: j+1=2r∗q1∗q2∗q3∗...∗qn+1=∑12r∗q1∗q2∗q3∗...∗qn−1qn+1 I am pretty sure that you have"moved an addition into a sum" wrongly here, or at leat should provide a set of parentheses to make it clearer. Can you provide an example? Apologies for the bad LaTeX; the "quote this" function seems to not work well with it. Edited August 30, 2018 by uncool Link to comment Share on other sites More sharing options...

Zolar V Posted August 30, 2018 Author Share Posted August 30, 2018 (edited) 2 hours ago, uncool said: I am pretty sure that you have"moved an addition into a sum" wrongly here, or at leat should provide a set of parentheses to make it clearer. Can you provide an example? Apologies for the bad LaTeX; the "quote this" function seems to not work well with it. Ahh yes, I see what you see. There should be a set of parenthesis. I believe this is what you're referring to: \[ (j )+1 =(2^r * q_1 * q_2 *q_3 *... * q_n) +1 = ( \sum_{1}^{ 2^r * q_1 * q_2 *q_3 *... * q_{n-1}}q_n ) +1 \] Example: let \( j = 3*5*7 \) then \ ( j+1 = (3*5*7) + 1 =( \sum_{1}^{3*5}{7}) +1 \) so \[ j+1 = (3*5*7) + 1 = 7+7+7+7+...+7+7+1 \] and of course notice that \( 7+1 < 2*7 \) so the primes \( z_n \) in \(7+1 \) are \(2 \leq z_n < 7 \) and of course they are since \( 7+1 = 8 = 2*2*2 \) Edited August 30, 2018 by Zolar V Link to comment Share on other sites More sharing options...

Trurl Posted August 31, 2018 Share Posted August 31, 2018 Ok here is what I don’t yet understand. Try a series for 85. In your example you would multiply by 3 and add 1. This would find Prime factors for the new number and not 85. I believe that would make it hard to factor Primes or semi-Primes. But then again I don’t have a full understanding of the problem. You are the expert on this problem. This is your problem. I am just making you defend and explain it more clearly. Also I took interest in you explaining your idea. I don’t know if you are right or wrong. A mathematician must decide for himself if a problem is worth pursing. But if it doesn’t work on first explanation don’t give up. If x is even divide by 2 If x is odd multiple 3 add 1 gives you even so divide by 2 still even divide by 2 till equals 1 The conjecture shows a relationship in factors but does not show those factors. Let me know what you think. I am problem wrong in the understanding of this problem. But that is ok. It just leads to more discussion. 1 Link to comment Share on other sites More sharing options...

uncool Posted August 31, 2018 Share Posted August 31, 2018 But in that case - how does 7+1=8 matter to 3*5*7+1 = 106, which decomposes to 2*53 (the latter of which is a larger prime than any appearing in 3*5*7)? Link to comment Share on other sites More sharing options...

Zolar V Posted August 31, 2018 Author Share Posted August 31, 2018 (edited) 35 minutes ago, Trurl said: Ok here is what I don’t yet understand. Try a series for 85. In your example you would multiply by 3 and add 1. This would find Prime factors for the new number and not 85. I believe that would make it hard to factor Primes or semi-Primes. But then again I don’t have a full understanding of the problem. You are the expert on this problem. This is your problem. I am just making you defend and explain it more clearly. Also I took interest in you explaining your idea. I don’t know if you are right or wrong. A mathematician must decide for himself if a problem is worth pursing. But if it doesn’t work on first explanation don’t give up. If x is even divide by 2 If x is odd multiple 3 add 1 gives you even so divide by 2 still even divide by 2 till equals 1 The conjecture shows a relationship in factors but does not show those factors. Let me know what you think. I am problem wrong in the understanding of this problem. But that is ok. It just leads to more discussion. I think you're right on track. But, I'm not trying to find out what the primes actually are. Instead I am trying to show how adding 1 to any prime results in a new number whose primes are less than the original prime. Applying that relationship to composite numbers is the tricky part. The relationship to composite numbers is where the iteration part of the collatz conjecture comes into play. since the iteration lets you add more than just 1 to a composite number. infact if you were to add that upper index of the sum to a prime number then you have added one to every prime number in that series. and each of those has the prime P +1 property 33 minutes ago, uncool said: But in that case - how does 7+1=8 matter to 3*5*7+1 = 106, which decomposes to 2*53 (the latter of which is a larger prime than any appearing in 3*5*7)? You are exactly correct in wondering and asserting this, but that is where the last equation comes into play. if you add 3*5 to 3*5*7 then each 7 has a corresponding 1. then the property of P+1 applies. Granted adding 3*5 hardly seems the most efficient route, but we don't particularly care about the most efficient route yet. We only care to show that every number will converge to 2 after some M iteration. note: 106 = 14*7 + 8 or if we maintain our notation \[ \sum_{1}^{14}{7} + (7+1) \] Edited August 31, 2018 by Zolar V Link to comment Share on other sites More sharing options...

Trurl Posted September 3, 2018 Share Posted September 3, 2018 Ok, I’m listening. Where does this go from here? This is how I understand you: Add 1 to a Prime number and it can be written as a power of 2. Since all Composite numbers factors are Prime numbers, those Prime factors added to 1 reduce to a power of 2. This will reduce to 1 and prove the conjecture. Your job is to prove this is possible. My question is how do you factor the Primes and apply the conjecture without changing the value of original number. By this I mean you factor the number into two Prime numbers and add 2 (+1 each Prime) to coverage to 1 and prove the conjecture. But at the same time you did not apply 3x +1 to the original value breaking the series. I am probably wrong again. I just don’t follow the modifications to the conjecture series. I’m am just making clear if I am following what you are doing. But if you explain this, I’m onboard. I am still looking at the original proof to make sure my question made sense. I am not completely sure of all the functions affecting x. But if the above question is confusing I will simply ask this: Are you breaking the rules of the conjecture by apply functions? Link to comment Share on other sites More sharing options...

Zolar V Posted September 7, 2018 Author Share Posted September 7, 2018 (edited) On 9/3/2018 at 5:21 PM, Trurl said: ... ... I am probably wrong again. I just don’t follow the modifications to the conjecture series. I’m am just making clear if I am following what you are doing. But if you explain this, I’m onboard. I am still looking at the original proof to make sure my question made sense. I am not completely sure of all the functions affecting x. But if the above question is confusing I will simply ask this: Are you breaking the rules of the conjecture by apply functions? I don't believe I am. What I believe I have done is break apart the conjecture into its pieces that cause it to function the way that it does. There seems to be a lot of lee-way in what you can do simply because of its iterative approach and that it is not given some limit on the iterations for the convergence. Quote "Add 1 to a Prime number and it can be written as a power of 2. Since all Composite numbers factors are Prime numbers, those Prime factors added to 1 reduce to a power of 2. This will reduce to 1 and prove the conjecture." - Exactly, you get what I am trying to do. Quote "My question is how do you factor the Primes and apply the conjecture without changing the value of original number. By this I mean you factor the number into two Prime numbers and add 2 (+1 each Prime) to coverage to 1 and prove the conjecture. But at the same time you did not apply 3x +1 to the original value breaking the series." - Here is where it gets a little murky. In the original proof, I did not flesh out the idea enough and it is clearly lacking in detail. Subsequent posts with WTF have allowed me to elucidate this part with greater clarity. Factoring the primes is a tricky way to do it, it seems. Something that should be explored in more detail. However let us think about a product of primes in a different way. Let us think about a product of primes be equivalent to a sum of one of its primes. Then we know how many 1's it takes to add 1 to each prime. Indeed we can add 1 simply by iterating the "Prime Reduction Function" on that sum of primes. That iteration is equivalent to the iterations of the Collatz function and since there isn't a limit on the number of iterations we are allowed to do then we can iterate or add any number of 1's it takes to every "fully saturated" sum of primes. Lets define "fully saturated": I mean a sum of primes where each prime within that sum has a 1 added to it. Or worded differently: every prime within the sum of primes has had the "Prime Reduction Function" applied to it 1 times. Let us use math symbols: Let \( N\) be a composite natural number and \( 2^r *P_1*P_2*...*P_k \) be its product of primes. where \( P_k \) is the largest prime, though not necessarily distinct. then \[ N = 2^r *P_1*P_2*...*P_k = \sum_{1}^{ 2^r *P_1*P_2*...*P_{k-1}} {P_k} \] let \(w = 2^r *P_1*P_2*...*P_{k-1} \) and \( H\) be an arbitrary function that applies our prime reduction function \(G(P) = P+1 \) exactly \( w \) times. for a fully saturated sum: \[ H( \sum_{1}^{ 2^r *P_1*P_2*...*P_{k-1}} {P_k}) = (P_k + 1)_1 + (P_k + 1)_2 +(P_k + 1)_3+ ...+(P_k + 1)_w = \sum_{1}^{ 2^r *P_1*P_2*...*P_{k-1}} {P_k} + w \] or if we use the second wording style: \[ G^w ( \sum_{1}^{ 2^r *P_1*P_2*...*P_{k-1}} {P_k}) = G(P_k)_1 + G(P_k)_2 + G(P_k)_3+... + G(P_k)_w \] Edited September 7, 2018 by Zolar V Link to comment Share on other sites More sharing options...

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