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polynomial complexity conjecture

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  • Sebastian Martin Ruiz
    Conjecture:   This polynomial expression is always zero for all triplets of consecutive prime numbers greater than 23.   Product[k
    Message 1 of 4 , Aug 1, 2012
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      Conjecture:
       
      This polynomial expression is always zero for all triplets of consecutive prime numbers greater than 23.
       
      Product[k Prime[n+1]+(m-k)*Prime[n-1]-m Prime[n],
      {m,2,Log[n]^(5-Pi)},{k,1,m-1}]=0
       
       
      It isinteresting to note that the order of complexity of this expression is less than
      log[n] ^ 4  and hence is a expression with polynomial complexity.
       
      Sincerely
       
      Sebastian Martin Ruiz

      [Non-text portions of this message have been removed]
    • bobgillson@yahoo.com
      I am pretty sure, but cannot be certain, that this was explained in Paul Underwood s recent reference to Richard s video, expounding and expanding Emil Artin s
      Message 2 of 4 , Aug 1, 2012
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        I am pretty sure, but cannot be certain, that this was explained in Paul Underwood's recent reference to Richard's video, expounding and expanding Emil Artin's theories regarding the counting of congruencies, in relation to recipricosity.

        If I am wrong, I apologise, but on the surface, it seems to be.

        Sent from my iPad

        On 1 Aug 2012, at 18:52, Sebastian Martin Ruiz <s_m_ruiz@...> wrote:

        > Conjecture:
        >
        > This polynomial expression is always zero for all triplets of consecutive prime numbers greater than 23.
        >
        > Product[k Prime[n+1]+(m-k)*Prime[n-1]-m Prime[n],
        > {m,2,Log[n]^(5-Pi)},{k,1,m-1}]=0
        >
        >
        > It isinteresting to note that the order of complexity of this expression is less than
        > log[n] ^ 4 and hence is a expression with polynomial complexity.
        >
        > Sincerely
        >
        > Sebastian Martin Ruiz
        >
        > [Non-text portions of this message have been removed]
        >
        >

        [Non-text portions of this message have been removed]
      • Sebastian Martin Ruiz
        I ve been watching the video and are different things. ________________________________ De: bobgillson@yahoo.com Para: Sebastian
        Message 3 of 4 , Aug 1, 2012
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          I've been watching the video and are different things.


          ________________________________
          De: "bobgillson@..." <bobgillson@...>
          Para: Sebastian Martin Ruiz <s_m_ruiz@...>
          CC: "primenumbers@yahoogroups.com" <primenumbers@yahoogroups.com>
          Enviado: Miércoles 1 de agosto de 2012 20:41
          Asunto: Re: [PrimeNumbers] polynomial complexity conjecture


           
          I am pretty sure, but cannot be certain, that this was explained in Paul Underwood's recent reference to Richard's video, expounding and expanding Emil Artin's theories regarding the counting of congruencies, in relation to recipricosity.

          If I am wrong, I apologise, but on the surface, it seems to be.

          Sent from my iPad

          On 1 Aug 2012, at 18:52, Sebastian Martin Ruiz <mailto:s_m_ruiz%40yahoo.es> wrote:

          > Conjecture:
          >
          > This polynomial expression is always zero for all triplets of consecutive prime numbers greater than 23.
          >
          > Product[k Prime[n+1]+(m-k)*Prime[n-1]-m Prime[n],
          > {m,2,Log[n]^(5-Pi)},{k,1,m-1}]=0
          >
          >
          > It isinteresting to note that the order of complexity of this expression is less than
          > log[n] ^ 4 and hence is a expression with polynomial complexity.
          >
          > Sincerely
          >
          > Sebastian Martin Ruiz
          >
          > [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]
        • Peter Kosinar
          ... Rewriting the conjecture into a bit more readable form, we get the following claim (where p[n] denotes n-th prime number): For n = 10, if we express
          Message 4 of 4 , Aug 1, 2012
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            > Conjecture:
            > �
            > This polynomial expression is always zero for all triplets of consecutive prime numbers greater than 23.
            > �
            > Product[k Prime[n+1]+(m-k)*Prime[n-1]-m Prime[n],
            > {m,2,Log[n]^(5-Pi)},{k,1,m-1}]=0

            Rewriting the conjecture into a bit more readable form, we get the
            following claim (where p[n] denotes n-th prime number):

            For n >= 10, if we express (p[n+1] - p[n-1])/(p[n] - p[n-1]) as
            reduced fraction, its numerator will not exceed Log(n)^(5-Pi).

            In term of prime gaps (where g[n] := p[n+1] - p[n]), the claim is
            equivalent to (g[n]+g[n-1])/gcd(g[n],g[n-1]) <= Log(n)^(5-Pi).

            Finally, taking logarithms, it can be rewritten as
            r(n) = Log[(g[n]+g[n-1])/gcd(g[n],g[n-1])] / Log[Log(n)]] <= 5-Pi.

            The conjecture itself looks both strange and weak at the same time; it
            asserts a quite strong property of two consecutive prime gaps, but since
            it deals with two gaps rather than a single one, there is no obvious
            relation to the other, well-known conjectures or theorems. For example,
            on one hand, Cramer's conjecture on gaps runs a bit short of yours -- its
            exponent is 2 and yours is smaller than that. On the other hand, it
            applies to all gaps uniformly, while yours only talks about two, implying
            the second one cannot be much bigger than the first.

            Numerically, 5-Pi is a bit more than 1.8584 and the closest challengers
            I've found so far are:

            n = 5949:
            p[n-1] = 58789
            p[n] = 58831
            p[n+1] = 58889
            g[n-1] = 42
            g[n] = 58
            r(n) = Log(50)/Log(Log(5949)) =
            1.8092074158688967510883160301868016764

            n = 8040878
            p[n-1] = 142414553
            p[n] = 142414669
            p[n+1] = 142414859
            g[n-1] = 116
            g[n] = 190
            r(n) = Log(153)/Log(Log(8040878)) =
            1.8184569954230012429937644772045398171

            Still seems to be in the "safe" region and my gut feeling is that the
            conjecture might be true. [yes, law of small numbers in practice ;-) ]

            Peter

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