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Re: [PrimeNumbers] polynomial complexity conjecture

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  • Sebastian Martin Ruiz
    I ve been watching the video and are different things. ________________________________ De: bobgillson@yahoo.com Para: Sebastian
    Message 1 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 2 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

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