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Number of prime numbers between the values of (m) (m1)

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  • Ситников Сергей
    Given the value (m) p_n^2
    Message 1 of 5 , Jul 6 11:51 PM
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      Given the value (m)
      p_n^2 < m < p_{n + 1}^2
      p_n - primes
      n – prime number
      E – error in calculating the number of primes in the interval
      (p_n,m)
      By formula
      m \cdot \prod\limits_{i = 1}^n {\left( {\frac{{p_i - 1}}{{p_i }}} \right)} - 1

      \frac{{m \cdot \prod\limits_{i = 1}^n {\left( {\frac{{p_i - 1}}{{p_i }}} \right)} }}{{\prod\limits_{i = 1}^{n + 1} {\left( {\frac{{p_i - 1}}{{p_i }}} \right)} }} = m_1


      m\frac{{p_{n + 1} }}{{p_{n + 1} - 1}} = m_1

      If between the values of (m) and (m1) is a prime number, the difference between the errors of E-E1 = 0 if there are no primes of E-E1 = 1 if the two primes of E-E1 = -1
      Prove that between the two values (m) and (m1) can only be either a prime or no prime or two primes
    • djbroadhurst
      ... Let n = 5. Then p_5 = 11 and p_6 = 13. I give the value m = 13^2 - 1, which lies between p_5^2 and p_6^2. Then m1 = m*13/(13 - 1) = 13^2 + 13, by the
      Message 2 of 5 , Jul 7 6:33 AM
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        --- In primenumbers@yahoogroups.com,
        <chitatel2000@...> wrote:

        > Given the value (m)
        > p_n^2 < m < p_{n + 1}^2
        > p_n - primes
        ....
        > m\frac{{p_{n + 1} }}{{p_{n + 1} - 1}} = m_1
        ....
        > Prove that between the two values (m) and (m1)
        > can only be either a prime or no prime or two primes

        Let n = 5. Then p_5 = 11 and p_6 = 13. I "give the value"
        m = 13^2 - 1, which lies between p_5^2 and p_6^2. Then
        m1 = m*13/(13 - 1) = 13^2 + 13, by the formula above.
        But there are precisely 3 primes between m and m1, namely
        13^2 + 4 = 173
        13^2 + 10 = 179
        13^2 + 12 = 181

        Conclusion: Either the proposition is absurd,
        of there is a undisclosed condition on m.

        David
      • djbroadhurst
        In primenumbers@yahoogroups.com, ... http://chitatel2000.blogspot.com suggests that this condition might be something like m*prod(k=1,pi(sqrt(m)),1 -
        Message 3 of 5 , Jul 7 10:41 AM
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          In primenumbers@yahoogroups.com,
          "djbroadhurst" <d.broadhurst@...> wrote:

          > undisclosed condition on m

          http://chitatel2000.blogspot.com
          suggests that this condition might be something like

          m*prod(k=1,pi(sqrt(m)),1 - 1/prime(k)) + n - 1 = pi(m) ... [1]

          where pi(m) is the number of primes not exceeding m.
          Clearly, every solution to [1] is rational.
          The largest such rational number that I have found is

          {m =
          47207235513753656979626880173334817417470530133964196593860667/
          1171662372654492721849024419009716384297071411200000000000;}

          with pi(m) = 4231, pi(sqrt(m)) = 46 and the following proof:

          {if(
          m > prime(46)^2 &&
          m < prime(47)^2 &&
          m > prime(4231) &&
          m < prime(4232) &&
          m*prod(k=1,46,1 - 1/prime(k)) + 46 - 1 == 4231,
          print(proven));}

          proven

          The approximate value of this rational number is

          default(realprecision,60);
          print(m*1.);

          40290.8180850785312968981599334268867233610311597619338580777

          which exceeds the value given by Sergey Sitnikov at the blog
          noted above.

          David
        • djbroadhurst
          ... It is rather easy to prove that the number of solutions to [1] is finite, in direct contradiction to a conjecture by Sergey. Consider the ratio R(m) =
          Message 4 of 5 , Jul 7 11:35 AM
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            --- In primenumbers@yahoogroups.com,
            "djbroadhurst" <d.broadhurst@...> wrote:

            > m*prod(k=1,pi(sqrt(m)),1 - 1/prime(k)) + n - 1 = pi(m) ... [1]
            >
            > where pi(m) is the number of primes not exceeding m.
            > Clearly, every solution to [1] is rational.
            > The largest such rational number that I have found is
            >
            > {m =
            > 47207235513753656979626880173334817417470530133964196593860667/
            > 1171662372654492721849024419009716384297071411200000000000;}

            It is rather easy to prove that the number of solutions to [1]
            is finite, in direct contradiction to a conjecture by Sergey.
            Consider the ratio

            R(m) = (m*prod(k=1,pi(sqrt(m)),1 - 1/prime(k)) + n - 1)/pi(m)

            of the left and right hand sides of [1]. Then as m tends to infinity,
            we know, from the prime number theorem and Mertens' formula,
            that R(m) tends to

            R(infinity) = 2*exp(-Euler) =
            1.1229189671337703396482864295817615735314207738503063363083...

            which is not equal to unity. Hence [1] has a finite number of solutions.

            David
          • djbroadhurst
            ... It is interesting to see how Sergey fooled himself into making his false conjecture. His mistake was to believe that we may rely on the sieve of
            Message 5 of 5 , Jul 7 4:10 PM
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              --- In primenumbers@yahoogroups.com,
              "djbroadhurst" <d.broadhurst@...> wrote:

              > m*prod(k=1,pi(sqrt(m)),1 - 1/prime(k)) + n - 1 = pi(m) ... [1]
              > It is rather easy to prove that the number of solutions to [1]
              > is finite, in direct contradiction to a conjecture by Sergey.

              It is interesting to see how Sergey fooled himself into
              making his false conjecture. His mistake was to believe
              that we may rely on the sieve of Eratosthenes to give
              the "right" constant in the prime number theorem. In fact,
              we should not: there is a mismatch by the celebrated factor
              2*exp(-Euler) > 1 that so perplexed Pafnuty Lvovich Chebyshev
              and Franz Carl Joseph Mertens.

              Elsewhere in this list, one may find an investigation of this issue.
              In particular, my remarks in
              http://tech.groups.yahoo.com/group/primenumbers/message/20936?var=0
              prompted Mike Oakes' impressive statistics in
              http://tech.groups.yahoo.com/group/primenumbers/message/20940?var=0

              David
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