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Re: The number of primes in the interval (0, m)

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  • djbroadhurst
    ... Dorogoi chitatel / Dear reader It s worse than that: you have no result for m 40291. For real m and positive integer n let Q(m,n) = m*prod(k=1,n,1 -
    Message 1 of 2 , Jul 12, 2010
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      --- In primenumbers@yahoogroups.com,
      <chitatel2000@...> wrote:

      > David. You are skeptical about my result.

      Dorogoi chitatel' / Dear reader

      It's worse than that: you have no "result" for m > 40291.

      For real m and positive integer n let
      Q(m,n) = m*prod(k=1,n,1 - 1/prime(k)) + n - 1,
      pi(m) = number of primes not exceeding m.

      In http://tinyurl.com/2w73pef you conjectured that
      Q(m,pi(sqrt(m))) = pi(m) ................... [1]
      has an infinite number of solutions:
      > The Number Theory "Number of primes in intervals"
      > "A million dollar problem"
      > Friday, 2 April 2010
      > Infinite number of figures (m_Q)

      I claim that this is false, since
      Q(m,pi(sqrt(m))) > pi(m), for m > 40291 .... [2]

      You keep posting messages about "error in calculation".
      But yours was an "error in understanding": the left
      and right hand sides of [1] are clearly incommensurable,
      at large m, for the very good reason that 2 > exp(Euler).

      Little is served by comparing things that *always* differ,
      for m > 40291.

      Let's take an example, using http://primes.utm.edu/nthprime/

      m = 29996224275833
      pi(m) = 1000000000000
      n = pi(sqrt(m)) = 379312
      Q(m,n) = 1085398179034.020151103297662104429...
      Q(m,n) - pi(m) > 85398179034

      So here your "error in calculation" exceeds 85 billion.

      Konets istorii? / End of story?

      David
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