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Re: [PrimeNumbers] Conjecture fails with new record at 80671

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  • Leonel Morales
    Thanks a lot Alan! I really appreciate you explanation. And then when 3 is involved d = 6k +- 2. I have calculated thousands of those and haven t noticed that.
    Message 1 of 6 , Feb 10, 2013
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      Thanks a lot Alan! I really appreciate you explanation.

      And then when 3 is involved d = 6k +- 2. I have calculated thousands of those and haven't noticed that.

      I have looked at the sequences of values of d: for a single prime p all the possible values of d, and for the set of primes, all the minimum values of d (http://oeis.org/A078611), all the maximum values, and several others that seem to be not listed in OEIS. All d for triplets involving 3 is http://oeis.org/A206037.


      Because d = p - pa = pb -p then pa*pb = p^2 - d^2 which leads to think of primes in a triplet as a right triangle:

      pa*pb + d^2 = p^2

      It can be demonstrated that no pair of such triangles is similar.

      I am also trying to look at primes as a network where two primes are connected if they are consecutive in a triplet. I am using Gephi to analyze such network and it consistently finds two communities in the network. I am still trying to figure what are the implications.

      Thanks a lot for your answer,

      Leonel


      ________________________________
      From: Alan Powell <AlanPowell@...>
      To: primenumbers@yahoogroups.com
      Sent: Sunday, February 10, 2013 4:49 PM
      Subject: Re: [PrimeNumbers] Conjecture fails with new record at 80671


       
      Leonel

      I think what you are seeing is a simple property
      of all prime numbers, excepting 3.

      If we ignore the special prime 3, all odd primes take
      the form 6n+1 or 6n-1 (since 6n+-0, 6n+-2, 6n+-3
      and 6n+-4 are obviously composite). Note that the
      form 6n+-5 can be written as 6n-+1):

      Case 1:
      -------
      If your central prime p has the form 6p+1, then
      2p has the form 12n+2 or 6m+2.
      The only combination of prime forms that add up
      to 6m+2 are 6a+1 plus 6b+1 for some {a,b}.
      Thus your triple is {6a+1, 6n+1, 6b+1}.
      Then your distance:
      d = pb-p = p-pa is
      = 6b+1-6n-1 = 6n+1-6a-1 or
      = 6(b-n) = 6(n-a) always a multiple of 6

      Case 2.
      -------
      If your central prime p has the form 6p-1,
      then 2p has the form 12n-2 or 6m-2.
      The only combination of prime forms that add
      up to 6m-2 are 6a-1 plus 6b-1 for some {a,b}.
      Then your triple is {6a-1, 6n-1, 6b-1}.
      Then your distance:
      d = pb-p = p-pa is
      = 6b-1-6n+1 = 6n-1-6a+1 or
      = 6(b-n) = 6(n-a) again always a multiple of 6

      Thus you are correct in that all d are multiples of 6,
      with the exception of when 3 is involved.

      Regards

      Alan Powell

      From: Leonel Morales
      Sent: Sunday, February 10, 2013 1:25 PM
      To: primenumbers@yahoogroups.com
      Subject: Re: [PrimeNumbers] Conjecture fails with new record at 80671

      Quite an interesting problem in fact, full of features and properties.

      I have been studying Goldbach partitions for even numbers of the form 2p where p is prime, that leads exactly to what Bob points. I have computed thousands of this partitions forming triplets of primes (pa, p, pb) where pa + pb = 2p, p at the center. For example, for p = 43 the triplets would be:

      (19, 43, 67), (13, 43, 73), (7, 43, 79), (3, 43, 83)

      The interesting part comes when the distances d = pb - p = p - pa are studied. For the primes above these are:

      24, 30, 36 and 40

      It is possible to demonstrate that all d are multiples of 6 except one at most, as is the case with 40, which then has to be the difference d between the prime p and 3, as is the case in the example for 43.

      I am preparing a manuscript with these findings that I plan to share in this group, but I am aware that many can be well known facts and not findings, so I am trying to verify each one.

      For example, I started with triplets, that can be considered arithmetic sequences of primes. There could be quartets, quintuplets, etc., in each case also arithmetic sequences. It can be demonstrated that arbitrarily long sequences of primes may exists but not infinite sequences. I thought this was a new result but found Green and Tao's article (http://arxiv.org/abs/math/0404188v6).

      Any help would be greatly appreciated.

      All the best,

      Leonel

      ________________________________
      From: "mailto:bobgillson%40yahoo.com" mailto:bobgillson%40yahoo.com>
      To: James Merickel mailto:moralforce120%40yahoo.com>
      Cc: "mailto:primenumbers%40yahoogroups.com" mailto:primenumbers%40yahoogroups.com>
      Sent: Saturday, February 9, 2013 11:50 AM
      Subject: Re: [PrimeNumbers] Conjecture fails with new record at 80671

      Dear James

      Why not apply your mind to some more basic aspects of prime number problems?

      For example, if you inspect the natural numbers line, be they odd or even, N inevitably appears at the centre of two prime numbers.

      For example (N) 7 is at the centre of 3 and 11, (N) 19 is at the centre of the primes 7 and 31, (N) 8 is at the centre of the primes 3 and 13.

      If you could tell us why this is so, not only would you explain the gaps between the primes, you would also prove Goldbach's Conjecture.

      Kind regards

      Bob

      On 9 Feb 2013, at 16:50, James Merickel mailto:moralforce120%40yahoo.com> wrote:

      > Dear groupmembers,
      >
      > The conjecture, which did not make a whole lot of sense anyway and was already showing itself unsupported empirically before the following result, turns out to be false. I had thought that perhaps there would be an argument modulo 19 or 19#.
      >
      > The next record of 62 primes in succession occurs with 80671 (arrived c. 8AM today):
      >
      > With the 19 numbers
      > a=2*103*157, b=3*173*233, c=5*41*193*239*271, d=7*113*167*257,
      > e=11*53*107*199*241*263, f=13*29*101, g=17*43*109*181*229, h=19*97*=
      > 281, i=23*59, j=31, k=37*151, l=47*179*277, m=61*73*137, n=67*83*191*211*283, o=71*79*127*131*163*197*293, p=89, q=139, r=223*269 and s=227*251,
      >
      > 293#*(1/a+1/b+1/c+1/d+1/e+...+1/q+1/r+1/s+80652)=
      >
      > 164835030772135218479491263344471877765987572654819835433403742
      > 893710632923823067093079399895258247669010672441740020208086391
      >
      > is the largest of a sequence of 62 primes starting with 80671 addends of 1 =
      > and produced by multiplying each addend save one by the primes in sequence =
      > from 2 through 293.
      >
      > A more reasonable conjecture is that requiring the largest prime in such a =
      > sequence to have no or at most one term in the sum as the primorial of the =
      > largest prime multiplied has a solution for 19 and for no larger prime. This I have not checked (and may not, since it isn't that interesting).
      >
      > JGM
      >
      > [Non-text portions of this message have been removed]
      >
      >

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    • James Merickel
      Sorry, I had initially posted as a forward on my own recent post and then erased more than I had meant to on a bounce caused by using yahoo.com rather than
      Message 2 of 6 , Feb 11, 2013
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        Sorry, I had initially posted as a forward on my own recent post and then erased more than I had meant to on a bounce caused by using 'yahoo.com' rather than 'yahoogroups.com'.  This refers to my post preceding.  Not sure if 'out of context' is a really sound way to consider it, though, being so soon after the relevant post.  Apologies in any case, because it was slightly off.   

        Below, edited to add the accidentally omitted prime 149 and change '62' to '63'.
         
        Note: I have had to increase the stack and start over to find the next record, if I might.
         
        JGM
         
        --- On Sun, 2/10/13, Maximilian Hasler <maximilian.hasler@...> wrote:


        From: Maximilian Hasler <maximilian.hasler@...>
        Subject: Re: [PrimeNumbers] Conjecture fails with new record at 80671
        To: "James Merickel" <moralforce120@...>
        Date: Sunday, February 10, 2013, 12:04 PM


        I refrain from asking "which conjecture",
        but your post is out of context.


        Maximilian



        On Sat, Feb 9, 2013 at 12:50 PM, James Merickel <moralforce120@...> wrote:


         



        Dear groupmembers,

        The conjecture, which did not make a whole lot of sense anyway and was already showing itself unsupported empirically before the following result, turns out to be false.  I had thought that perhaps there would be an argument modulo 19 or 19#.

        The next record of 63 primes in succession occurs with 80671 (arrived c. 8AM today):

        With the 19 numbers
        a=2*103*157, b=3*173*233, c=5*41*193*239*271, d=7*113*167*257,
        e=11*53*107*199*241*263, f=13*29*101, g=17*43*109*149*181*229, h=19*97*281, i=23*59, j=31, k=37*151, l=47*179*277, m=61*73*137, n=67*83*191*211*283, o=71*79*127*131*163*197*293, p=89, q=139, r=223*269 and s=227*251,

        293#*(1/a+1/b+1/c+1/d+1/e+...+1/q+1/r+1/s+80652)=

        164835030772135218479491263344471877765987572654819835433403742
        893710632923823067093079399895258247669010672441740020208086391

        is the largest of a sequence of 63 primes starting with 80671 addends of 1 =
        and produced by multiplying each addend save one by the primes in sequence =
        from 2 through 293.

        A more reasonable conjecture is that requiring the largest prime in such a =
        sequence to have no or at most one term in the sum as the primorial of the =
        largest prime multiplied has a solution for 19 and for no larger prime.  This I have not checked (and may not, since it isn't that interesting).

        JGM

        [Non-text portions of this message have been removed]






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