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Prime Number Distribution Solved?

  1. Sep 30, 2003 #1
    Hi everyone, I'm new here. I have an interesting conjecture I have been trying to prove for some time. I've tried it out on a couple of forums, but perhaps some of you can help. Please make whatever comments you can (good or bad) and all input is welcome.

    I will give this conjecture in algorithm form, along with an example, and then put it in a general theoretical form.

    Take the first (n-1) primes so that p(1)=2, p(2)=3, etc. Now create two numbers, A and B, such that each prime in the above set is a factor of exactly one of them, A or B. It doesn't matter what the exponent on each of them is, as long as it is an integer greater than zero. Notice, for example, that "3 is a factor of A" could mean that 3 to the 56th power is a factor of A.
    Good time to introduce the example:
    Say p(n-1) is 11. Then A could be 2^4 * 7^3 and B could be 3*5*11^2. Notice each prime <=11 is in A or B, and the exponents are arbitrary. In fact, I usually leave the exponents variable:
    A=2^a*7^b, B=3^c*5^d*11^e

    Now, check out abs( A +/- B ) = Q* where abs() is absolute value, +/- is plus or minus, and Q* is the set of solutions dependant upon the original exponents.
    Now, if q is in Q*, and q < [p(n)]^2 (in the example, 13^2=169), then q MUST be prime (by the distributive principle, q is relatively prime to 2,3,5,7,11 and since it's less than 169, it is prime) or the number 1.

    In practice, not all primes less than 169 can be found with the above equation. However, by repartitioning the set into A and B, i.e. make different choices of which primes are factors of A and which are factors of B, all solutions can be gotten this way. In fact, all solutions up to 1369 can be gotten the same way by using the primes up to 31 in A and B.
    As I said, every such solution - given the less-than condition - is either prime or 1. The only thing I haven't proven is that EVERY prime can be gotten in this manner.
    Notice that, if every prime does follow this pattern, we have a finite set of exponential equations that gets any finite range of primes given as being between p(n-1) and [p(n)]^2.
    Does anyone have any ideas on how I could begin to prove that EVERY prime follows this pattern? Thanks for taking the time to read this and in advance for any input.
    ==================================================

    ===============
    Theoretical approach:
    key:
    p(n)# ("read p(n) primorial") = 2*3*5*7*...*p(n)

    Given: 2,3,5,...p(n)
    If p(n-1+i) does not divide A*B for i an integer > 0 and,
    if p(n-1)# divides A*B then,
    abs( A +/- B ) = Q*
    and if q is an element of Q* and q < [p(n)]^2
    then q is prime.
    ==================================================

    =================
    Note: to get a "feel" for this stuff, try the following example, where I leave all of the exponents constant=1 except the exponent on 2:
    Abs(105 +/- 2^x)=q where q<121 (note 105=3*5*7)
    (note also that abs() takes care of negative primes)
    Thanks again,
    Aaron
     
  2. jcsd
  3. Oct 2, 2003 #2
    Take a Look

    I find it interesting that nobody has anything to say about this. It really only involves 4th grade math. Is it too boring, or too simple, or what? Would somebody tell me what they think? I've tried using logarithms and modular arithmetic and even straight algebra to prove this, but I can't help thinking I'm overlooking some basic technique that will push the proof through. Basically, the challenge is to show that every exponential equation of the proper form has enough solutions at integer exponents to produce all of the primes in the range from p(n-1) to [p(n)]^2. I'm not sure where to go from there. Are there any ideas out there? I appreciate all input, you never know what will help. Thanks to all of you that looked at this - 'bye for now.
    Aaron
     
  4. Oct 9, 2003 #3
    Correct me if I'm wrong...but are you trying to find an equation that will give all the prime numbers within a given range of a certain number?


    If this is the case, what you are trying to prove reminds almost of the &epsilon;-&delta; definition of a limit.


    Could you clarify your description of your theorem and explain what every part of refers to in complete detail.

    Could you describe what you mean by p(n-1)? What is the p, and what is the n? And finally what are the A, B, q, Q you use in your theorm. I think I understand the concept, I'm just having a hard time figuring out what your various variables represent, and how you are deriving these relationships between them.

    Perhaps you could attempt an explanation in straight algebra. If you can clarify some of this for me, I'd be interested in trying to help you prove it.

    Thanks,
    Reddhawk
     
  5. Oct 13, 2003 #4
    reply to q's and some good examples

    Thanks for your interest.
    p(n) means the nth prime - p(5) is the fifth prime, p(10) is the tenth prime, etc.
    Now if we take all the primes from p(1) up to, say, p(6) then we are talking about 2,3,5,7,11, and 13.
    The procedure I'm talking about uses all of the first x-1 primes to obtain all of the primes less than the square of the (x)th prime, so in this case it uses the 2,3,5,7, and 11 to get all of the primes up to the square of 13, 13^2 = 169.
    It obtains these primes by dividing the set 2,...,11 into two sets and performing a simple set of operations on them.
    Notice that ^ means "to the power", +/- means put plus in to get one solution, and then minus to get another solution, and abs() means take the absolute value of whatever you obtain - if anyone doesn't know, this simply means take any negative solutions and change them to positive.
    Q* was simply the set of all possible solutions, since you get one solution every time you specify a different bunch of numbers to the exponents in the equation. q was used to represent a single solution from the set of all possible solutions.
    In the case of the example, suppose you define the exponents like this: abs( 7^1 * 11^1 +/- 2^2 * 3^1 * 5^1).
    Then you have abs(77 +/- 60) and you get two solutions: 77+60=137 and 77-60=17. Since both of these are less than 169, the part of my theorem that I HAVE proven guarantees they are both prime. Suppose you increase the exponent on 7 so that you have abs(49*11 +/- 60). Then you have 539+60=599 and 539-60=479, and both are greater than 169 so the theorem only says that they don't have 2,3,5,7, or 11 as factors, but they COULD STILL HAVE 13,17,19, or 23 as factors so they may not be prime (you only need to check primes up to their square root to prove primality. The next prime after 23 is 29 and it's bigger than their square root). Notice if we increased the number of primes in our equation to include up to 23, say
    abs( 7^1 *11^1 *23^1 +/- 2^5 *3^2 *5^1 *13^1 *17^1 *19^1 ) then we would get 599 and 479 as solutions if they are prime (those are just example exponents - I'm not sure which exponents we would need, but I could find them if you'd like). Actually, we may not get them with that exact setup - we might have to move the 11 to the right side of the +/-, or the 2 to the left side. It's even possible the 3 would be on one side by itself with all of the rest on the other side.

    To see how this works better, take a simple case:
    abs( 3*5 +/- 2^x ). Here I am keeping the exponents on 3 and 5 set at 1 and only changing the exponent on 2. Notice this will find all primes less than the square of the next prime, 7^2=49.
    15...
    +2=17, +4=19, +8=23, +16=31, +32=47, +64=79 - and 79>49 so we stop.
    15...
    -2=13, -4=11, -8=7, -16=-1, -32=-17, -64=-49 and we stop.
    notice the absolute value converts the -17 to a prime.
    We got 1 as a solution - if 1 isn't considered prime, then we must state that the solutions we get are simply the primes + 1, or we could call them the non-composites, or we could simply call 1 a "special prime".
    Notice we didn't get all primes < 49. To get the rest, we could start with abs( 3^2 * 5 +/- 2^x ).
    45...
    +2=47, +4=49 stop.
    45...
    -2=43, -4=41, -8=37, -16=29, -32=13, -64=-19, -128=-83 too big, stop.
    And the two equations, 15+/-2^x and 45+/-2^x got all primes less than 49. If they didn't, we could continue playing with exponents, or we could change the grouping like this:
    abs( 2*5 +/- 3^x ).
    10...
    +3=13, +9=19, +27=37, +81=91 too big, stop.
    10...
    -3=7, -9=1, -27=-17, -81=-71 too big, stop.
    and play with exponents with that setup.
    You try it: abs( 3*5*7 +/- 2^x ) is abs( 105 +/- 2^x ). We're looking for solutions between 11 and 121.
    this gets 107,109,113,103,101,97,89,73,41, and -23
    quite an impressive number of primes for one line of the algorithm.
    you could try abs (3^x * 7 +/- 2^y * 5^z ) and so on to get more.
    Well, this is a big enough entry. I'll be glad to answer any other questions, or if I didn't answer your q's, please ask again. Thanks again for your interest!
    Aaron
     
  6. Oct 15, 2003 #5
    Your conjecture is indeed quite intriguing, however one aspect of it plagues me with doubt as to its usefulness. The need to "play around" with different exponents that are picked at random leaves quite a lot of room for error. I believe there must be some way to relate how far one must take the exponents with the value of the prime that determines the set of primes you use.

    What I am trying to say is that if we define the exponent of all primes in the set to be y then there must be some relationship between y and (x-1) primes or simply x.

    If such a relationship could be discovered then it would greatly improve upon your current method of arbitrarily picking exponents and would thus go a long way in proving that your theorem can indeed solve all the primes between x and x^2.


    So basically:

    x = 13

    set of all primes < x^2
    abs( 7^y * 11^y +/- 2^y * 3^y * 5^y)

    where

    y = some relationship to x


    I hope you can see the usefulness in this contribution. If such a relationship between y and x could be discovered, your theorem would be greatly improved.

    Hope this helps,
    Reddhawk
     
  7. Oct 15, 2003 #6
    Also, I forgot to ask: how do you know where to place the +/- within in your problem. It seems to me that the location of this has a great deal to do with the solution set you get.

    Thanks,
    Reddhawk
     
  8. Oct 16, 2003 #7
    I'll try to give a more complete answer later as I only have a minute now. In my experience solving these exponential equations, I have found that the exponents don't need to be very large at all, maybe about 2*log(base 10) of the largest prime involved. Also, the place you put your +/- doesn't matter. Indeed, my conjecture is that (1.) one position of the +/- will get all primes if you allow arbitrarily large exponents and (2.) small exponents are sufficient if you allow the +/- to shift around and the primes to change locations from one side of the +/- to the other side.
    By the way, I think I may have a proof now - maybe - I just need to formalize it and see if there are any holes in it.
    Thanks for the questions - keep them coming
    Aaron
    p.s. don't forget, all primes in the equation can (usually do) have different exponents - I didn't know if you meant that when you were using y above or not.
    p.p.s. Notice that these are exponential equations, and if you plot them with the exponents as the axes, then integer exponents always produce an output of primes on the output axis as long as the output is below the next-prime-squared "cap". These equations thus give a piecewise function that gives the distribution of the primes in integer exponents. Also, things above the "cap" form a set of integers "enriched" with primes (kind of like doing a 10X10 sieve of Aristothenes but only eliminating the first couple of primes).
    see you later
     
    Last edited: Oct 17, 2003
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