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RE: [PrimeNumbers] reciprocal consecutive primes

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  • Milton Brown
    Actually, it doesn t seem that hard: 1/p =p+2 r =q+2 =p+4 It must be shown that 1/p
    Message 1 of 5 , Nov 2, 2005
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      Actually, it doesn't seem that hard:

      1/p < 1/q + 1/r

      q>=p+2
      r>=q+2>=p+4

      It must be shown that 1/p < 1/(p+2) + 1/(p+4)

      equivalently (p+2)(p+4)/p < 2p +6 or

      p+6+8/p < 2p+6 or

      8/p < p or

      p^2 > 8 which happens when p>= 3

      Notice it also works for 2, 3, and 5 since
      1/2 < 1/3 + 1/5 ( 7.5 < 8 )

      Milton L. Brown

      > [Original Message]
      > From: theo2357 <Theo.3.1415@...>
      > To: <primenumbers@yahoogroups.com>
      > Date: 11/2/2005 9:46:17 AM
      > Subject: [PrimeNumbers] reciprocal consecutive primes
      >
      > Let p,q,r be three consecutive primes. How can be proven that
      > 1/p < 1/q + 1/r? It's NOT trivial!
      >
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    • Wojciech.Florek@amu.edu.pl
      ... This is equivalent to a statment that for a given prime q the preceding prime p =2q/3. Supose it is true. Moreover, it is evident (and there is threom
      Message 2 of 5 , Nov 2, 2005
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        On Wed, 2 Nov 2005, theo2357 wrote:

        > Let p,q,r be three consecutive primes. How can be proven that
        > 1/p < 1/q + 1/r? It's NOT trivial!
        >
        >

        This is equivalent to a statment that for a given prime q the preceding
        prime p>=2q/3.
        Supose it is true. Moreover, it is evident (and there is threom regarding
        it, I think) r<2q. These inequalities lead:
        2 3
        a) p>=-q => 1/p <=--
        3 2q

        1 2 1 3
        b) r<2q => 1/r > -- => 1/q+1/r > -- + -- = -- >= 1/p
        2q 2q 2q 2q

        If p>=2q/3 is true, then r <= 3q/2. So 1/q+1/r >= 5/3q = 1.666...*(1/q)
        whereas 1/p <= 1.5*(1/q).

        So, my formulation is:
        Show that for any prime p the next prime q is not larger than 1.5*p
        (equality is valid for p=2, only).
        Maybe it has been proven?

        Wojtek

        ===============================================
        Wojciech Florek (WsF)
        Adam Mickiewicz University, Faculty of Physics
        ul. Umultowska 85, 61-614 Poznan, Poland

        Phone: (++48-61) 8295033 fax: (++48-61) 8257758
        email: Wojciech.Florek@...
      • theo2357
        I don t know if it s proven. It s my own observation. 1/p qr/(q+r) which means p is greater than the half of the harmonic
        Message 3 of 5 , Nov 3, 2005
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          I don't know if it's proven. It's my own observation.
          1/p < 1/q + 1/r is equivalent to:
          p > qr/(q+r) which means p is greater than the half of the harmonic
          mean of q and r.
          Let
          H the harmonic mean
          G the geometric mean
          A the arithmetic mean

          Then Cauchy says that H < G < A
          Can you use that?
        • jbrennen
          ... I think this is implied by the Iwaniec & Pintz finding, namely that there is always a prime between x and x-x^(23/42), for any real number x 11. The
          Message 4 of 5 , Nov 3, 2005
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            --- theo2357 wrote:
            >
            > Let p,q,r be three consecutive primes. How can be proven that
            > 1/p < 1/q + 1/r? It's NOT trivial!
            >

            I think this is implied by the Iwaniec & Pintz finding, namely
            that there is always a prime between x and x-x^(23/42), for
            any real number x > 11.

            The Iwaniec & Pintz theorem can be used to show the weaker result,
            that for prime p >= 5, nextprime(p)/p < Phi, where Phi is the
            golden ratio, (sqrt(5)+1)/2 = 1.618...

            And if q/p < Phi, and r/q < Phi, then the original inequality
            1/p < 1/q + 1/r follows quite easily.
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