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24344Re: [PrimeNumbers] polynomial complexity conjecture

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  • Peter Kosinar
    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)) =

      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)) =

      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 ;-) ]


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