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Zolar V

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Posts posted by Zolar V

  1. Hello my old community.

    I'm going to be publishing my proof soon,  but I've solved the conjecture about 7-8 years ago.   Ive been sitting on the solution for a long while.  Since the introduction of LLM's ive taken the opportunity to test the consistency of my answer.  

     

    Following the solution, ive found a couple interesting patterns in primes reciently as well.   It seems to have a connection to the reinmann hypothesis. Namely the 1/2 part. 

    -

    The density of primes is contained in the first half of all numbers.   For m, m composite,  m-1 =2n.    The primes that constitute 2n are all from 0 to n and none from n+1 to 2n.   Meaning the relative density of prime numbers has a sort of upper bound where primes are sparse at higher than n.

     

    Take p,q consecutive primes and m distance between them.   Then m is a cyclical function, this relates fairly directly to the zetta function by the fact that the complex function is a description of rotation, and all solutions are within 0 and 1 where the real part is 1/2. 

     

    It means primes are predictable and there is a cyclical function in linear algebra that predicts primes. 

     

  2. 58 minutes ago, Ghideon said:

     

    I have some experience from similar situations. Nice to see you fixed the issues rather promptly!

    Anyway, I didn't know how much I appreciate the site (and it's members) until scienceforums.net suddenly was offline.

     

     

    Same thought exactly.   I went to go write a post and was immediately confused and saddened by the apparent loss of the community.

  3. I'm trying to understand what it means to take a number \(n \) to the power of \( i \).

    \[ F(x) = x^i     = x^{\sqrt{-1}} \]

     

    I'm trying to understand visually what taking a number to a squareroot really does to it. Examine the behavior.   I want to be able to understand it visually as well as numerically.

    I suppose it would be useful to also understand what a number \( n \) to a root \( \sqrt{a} \)  also means.
     

    \[ G(y) = n ^ { \sqrt{y}}  ,  y \in \mathbb{Z}  \] 

     

  4. My goal is to publish on arxiv, but of course I need to be endorsed.   In order to be endorsed, i need to be recognized by someone who does publish in the field.  However, I doubt many would actually take me and my proof seriously considering the conjecture that I indeed did prove.   Especially due to the lack of math publishing and non math major degree status.   I have my bachelors in science, with minors in both math and biology.  But that in itself wouldn't be enough to be noticed.

    I have emailed both my old undergraduate faculty and Dr. Lagarias, as he has written a lot about the history of the conjecture and I believe he would be the most knowledgeable and interested in my work. 

    uhh, sorry for the delay.  My office got invaded by a wood wasp. I promptly got up and closed the door and , not so promptly, got a vacuum, sucked him up, and transported him to the outdoors.

  5. I've had a few of my math major friends look, they're also confident it is true.   How would i publish to a math journal?  I dont even know where to start

    I do know it would be in Number Theory

  6. 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! 

  7. 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

    5b949bcfb865e_generalcollatz.PNG.0280b19c6c35b2238b16c11393eab6ad.PNG

    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.  

  8. 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:
    \[ G^w ( \sum_{1}^{ 2^r *P_1*P_2*...*P_{k-1}} {P_k}) = G(P_k)_1 +  G(P_k)_2 + G(P_k)_3+... + G(P_k)_w  \]

  9. 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
    \[ \sum_{1}^{14}{7}  + (7+1)  \]

     

  10. 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 \( 7+1 = 8 = 2*2*2 \)

  11. 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. 
     

  12. 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
    \( F_0 = 3^3 + 2^0 *  3^2 \)
    \( F_1 = F_0 + 2^{2*1} * 3^{1} \)
    \( F_2 = F_1 + 2^4 *3^0 = 2^{2*3} \)

    \( 3^3 + 3^2   + 2^{2*1} * 3^{1}  + 2^4 *3^0 = 2^{2*3}  \)

  13. 1 hour ago, taeto said:

    Did you mean to write that for any prime a and any natural number n you have

     

    an+an1=22an1?

     

    If + refers to usual addition of integers and means ordinary multiplication, then after cancellation of an1 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 \( 2^1  ->  2^4   ;   2^3 -> 2 ^6 \)

     

  14. 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 

    \[ \sum_{1}^{2^r*P_1 *P_2*...* P_{i-1}}{P} + i =  a^n + a^{n-1} = 2^2 * a^{n-1}   = 2r  \]

  15. 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.

  16. 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.

  17. 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
    \[ 2 \leq q_i  < {P_1}  i = 1,2,3,4,...,n \]


     

  18. 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 \[ m*G(\sum{P_i}) \]
     

  19. 6 minutes ago, wtf said:

    It makes no sense at all to me. You've reverted to the "spinning out of control" narrative that I thought I talked you down from the past couple of days. I was mistaken. I did no good at all. I can't do anything else here. 

    I don't see your perspective.   Maybe it is a difference in thinking styles.  I think in big picture and then I work on the details.  So when I see something,  I see all these connected dots, though I might not yet fully understand or know what the dots are. Sometimes my big picture is off, so when I work on the dots they are off too.  But usually if my dots are off too then I get a good indication of what is wrong and fix my big picture.

    Maybe your idea of "spinning out of control" is my expression of the different ideas that I get while working through something.  I let my brain connect the dots and explore new dots, even if it has little to do with what I am doing.  I'm essentially letting my brain be creative and intuitive, supposedly these flashes of creativity are how our brain works on problems in the background.  

    What I did was show a set of lemmas that begin to outline how I am going to prove steps 1-3 from:

    On 8/23/2018 at 5:21 PM, Zolar V said:

    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


    I am going to design a theorem supported by the lemmas.  The Theorem will state: For any natural number N, there exists a function G such that any N converges to a power of 2 over M iterations.
    Right now I have a bunch of hypotheses and no proof.  The set of hypotheses (lemmas) are the framework in which we will show Theorem 1 is true.   The proof of the lemmas are some of the dots. 
    Imagine it like a couple nested big pictures with nested dots. 
    Big picture 1:  The collatz conjecture converges to 1 via this prime reduction method
                    Dots_1:  Show the collatz looks like the prime reduction function
                    Dots_2:  Show the prime reduction method works 
                                Big Picture 2:   Show the prime reduction method works
                                                  Dots_2_1:  Show lemma 1
                                                  Dots_2_2:  Show lemma 2
                                                  Dots_2_3:  Show lemma 3
                                                  Dots_2_4:  Show lemma 4
                                                  Dots_2_5:  Show Theorem 1

  20. These are proof of concept right now:

    Lemma 1:  2 < p_i < P   
    For a natural number P that is prime and P >2 , P+1 =  2^r * p_1 *p_2 * p_3 ... p_i 
    for every p_n, n = 1,2,3,.. ,i  
    2 < p_n <P

    Lemma 2: Any natural composite number can be rewritten as a sum of prime(s)   {might need to add about odd}
    Let k be a composite number
     \[  k = 2^r *p^{1}_1*p^{1}_2*p^{1}_3*p^{1}_14...*p^{1}_n   =  \sum_{i=1}^ { 2^r *p^{1}_1*p^{1}_2*p^{1}_3*p^{1}_4*...*p^{1}_(n-1)  }  p_n  \]

    Lemma 3:  Add 1 for each prime in a sum of primes.     {might need to say only for odd, not sure yet}
    adding 1 to each prime in a sum is equivalent to adding the upper index to the prime.


     

    Lemma 4:  do the adding part until a power of 2 is reached

    example:  3*7 =   7+7+7                             ( potentially 21 may be expressed as   2^3 + 5,    then applying lemma 3 to (5) :  5+ 1 = 6  :  6+2 = 8 : 16+8 = 24 :  24+8 = 32 )
    applying lemma 3:   7+7+7(+3)= 24                            
                                        24 = 2*2*2 *3= 8*3
    Applying lemma 3:   24+8 =32 = 2^5

    Example: 5*3*7 = 7+7+7+7+7+7+7+7+7...
    Applying Lemma 3:  105+15 =120
    Prime factorization:    12 = 2*2*2*3*5   : 2*2*2*3 = 24
    Applying lemma 3:      120+24=144
    Prime factorization:    2*2*2*2*3*3   :  2*2*2*2*3 = 48
    Applying lemma 3:     144+48 = 192
    Prime factorization:    2^6 * 3
    Applying lemma 3:     192+2^6 = 256  = 2^8
    Done


    NOTE:  All I am saying with " (potentially 21.... )"  is that there may be more than 1 way to solve the problem,  and each way may be more or less efficient.   If we were to measure efficiency by how many steps it takes to solve.                       

    Also, suppose we were to divide out the 2^r numbers each step, that would also shrink the number before we add 1 to it. 
    Again the purpose of keeping the 2^r numbers is that inside them they hold the key to how many steps are taken. 

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