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Re: [PrimeNumbers] Re: Brocard's Conjecture, and other notes

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  • Jeremy Wood
    After work I ll revisit everything again. In the meantime I kept the file online, but put red notes around the incorrect section. But the paper still points
    Message 1 of 5 , Oct 7, 2005
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      After work I'll revisit everything again. In the
      meantime I kept the file online, but put red notes
      around the incorrect section.

      But the paper still points out -- although it doesn't
      formally prove -- that:
      pi(p(i+1)^2)>=r(i)*(p(i+1)^2)+i-1
      where r(i) =
      (p(1)-1)/p(1)*(p(2)-1)/p(2)*...(p(i)-1)/p(i)

      If anyone has any thoughts as to how one could apply
      this towards Brocard's Conjecture, please let me know.
      Or if this is also flawed, please let me know.

      Cheers
      - Jeremy Wood

      --- Patrick Capelle <patrick.capelle@...> wrote:

      > --- In primenumbers@yahoogroups.com, Jeremy
      > <mickleness@y...> wrote:
      > >
      > > Hi everyone... I just joined the list.
      > >
      > > I wrote a little paper on primes recently,
      > offering an informal proof
      > > of Brocard's Conjecture. a few notes on twin
      > primes. and other
      > > observations.
      > >
      > > I was wondering if people on this list could look
      > it over and let me
      > > know... well... if it has any merit. I'm
      > competent at math, but
      > > proofs and high level math are a little foreign to
      > me...
      > >
      > > http://homepage.mac.com/bricolage1/essays/
      > >
      >
      >
      > Hello Jeremy,
      >
      > At the beginning of your proof of Brocard's
      > conjecture,you wrote :
      > "Well if d-b >= k, and a >= b and c >= d, then
      > surely c-a >= k ".
      > Surely not.There are cases where c-a < k.
      > Take for instance a = 5, b = 2, c = 7, d = 6 and k =
      > 3.
      >
      > Regards,
      > Patrick Capelle.
      >
      >
      >
      >
    • Jeremy Wood
      Sigh. Nevermind... taking the link down. The other expression is also flawed... it fails badly when i exceeds 35 (that is, (p(i))^2=22201). Back to the
      Message 2 of 5 , Oct 7, 2005
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        Sigh. Nevermind... taking the link down.

        The other expression is also flawed... it fails badly
        when i exceeds 35 (that is, (p(i))^2=22201).

        Back to the drawing board...

        --- Jeremy Wood <mickleness@...> wrote:

        > After work I'll revisit everything again. In the
        > meantime I kept the file online, but put red notes
        > around the incorrect section.
        >
        > But the paper still points out -- although it
        > doesn't
        > formally prove -- that:
        > pi(p(i+1)^2)>=r(i)*(p(i+1)^2)+i-1
        > where r(i) =
        > (p(1)-1)/p(1)*(p(2)-1)/p(2)*...(p(i)-1)/p(i)
        >
        > If anyone has any thoughts as to how one could apply
        > this towards Brocard's Conjecture, please let me
        > know.
        > Or if this is also flawed, please let me know.
        >
        > Cheers
        > - Jeremy Wood
        >
        > --- Patrick Capelle <patrick.capelle@...>
        > wrote:
        >
        > > --- In primenumbers@yahoogroups.com, Jeremy
        > > <mickleness@y...> wrote:
        > > >
        > > > Hi everyone... I just joined the list.
        > > >
        > > > I wrote a little paper on primes recently,
        > > offering an informal proof
        > > > of Brocard's Conjecture. a few notes on twin
        > > primes. and other
        > > > observations.
        > > >
        > > > I was wondering if people on this list could
        > look
        > > it over and let me
        > > > know... well... if it has any merit. I'm
        > > competent at math, but
        > > > proofs and high level math are a little foreign
        > to
        > > me...
        > > >
        > > > http://homepage.mac.com/bricolage1/essays/
        > > >
        > >
        > >
        > > Hello Jeremy,
        > >
        > > At the beginning of your proof of Brocard's
        > > conjecture,you wrote :
        > > "Well if d-b >= k, and a >= b and c >= d, then
        > > surely c-a >= k ".
        > > Surely not.There are cases where c-a < k.
        > > Take for instance a = 5, b = 2, c = 7, d = 6 and k
        > =
        > > 3.
        > >
        > > Regards,
        > > Patrick Capelle.
        > >
        > >
        > >
        > >
        >
        >
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