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15700RE: [PrimeNumbers] Re: another way to calculate primes

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  • Mazzarello Gianni
    Dec 6, 2004
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      It's likely that they are asymptotically equal.
      In this moment I'm running a routine (I'm over 8.5 x 10^6) that counts
      separately the number of primes ending with the digit 1, 3, 7, 9.
      The fraction of the total number of primes found (that is: p1/PN, p3/PN,
      p7/PN, p9/PN, where pk is a prime with k as last digit) is < 0.2500 for
      the primes ending with 1 and 9, is > 0.2500 for the primes ending with 3
      and 7.
      They are all approaching 0.25: always from below p1 and p9 (0.2499),
      always from above p3 and p7 (0.2501).

      Gianni


      >-----Original Message-----
      >From: jbrennen [mailto:jack@...]
      >Sent: Monday, December 06, 2004 3:19 PM
      >To: primenumbers@yahoogroups.com
      >Subject: [PrimeNumbers] Re: another way to calculate primes
      >
      >
      >
      >--- Mike Oakes wrote:
      >>
      >> In a message dated 06/12/2004 11:57:15 GMT Standard Time,
      >> g.mazzarello@t... writes:
      >>
      >> >I was wondering this: for a fixed M, are there more primes
      >< M ending
      >> >with 3 and 7 than primes < M ending with 1 and 9?
      >>
      >> Everyone believes that asymptotically (as M -> oo) the numbers are
      >> equal, but no-one can prove this.
      >
      >I thought that they were proven to be asymptotically equal by
      >an extension of Dirichlet's Theorem.
      >
      > http://www.utm.edu/research/primes/notes/Dirichlet.html
      >
      >See the last equation on that page, which gives bounds for an
      >error term.
      >
      >
      >
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