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

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  • bobgillson
    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
    Message 1 of 6 , Feb 9, 2013
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      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 <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 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]
      >
      >


      [Non-text portions of this message have been removed]
    • Leonel Morales
      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
      Message 2 of 6 , Feb 10, 2013
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        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: "bobgillson@..." <bobgillson@...>
        To: James Merickel <moralforce120@...>
        Cc: "primenumbers@yahoogroups.com" <primenumbers@yahoogroups.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 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 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]
        >
        >

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




        [Non-text portions of this message have been removed]
      • Alan Powell
        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
        Message 3 of 6 , Feb 10, 2013
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          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]
          >
          >

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

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





          [Non-text portions of this message have been removed]
        • 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 4 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]
            >
            >

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

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

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




            [Non-text portions of this message have been removed]
          • 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 5 of 6 , Feb 11, 2013
            • 0 Attachment
              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]






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