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A prime number in the interval (m, m_1)

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  • Ситников Сергей
    P_n^2
    Message 1 of 1 , Jul 9 11:17 PM
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      P_n^2 <m <P _ (n+1) ^2
      For the left restriction (P_n^2) value (m) can pass. It is a question of an error of calculation. For the right restriction it is impossible to pass. There the formula accepts compound numbers for simple numbers. From here:
      m*P _ (n+1)/P _ (n+1)-1=m_1
      m=t*P_n
      t=2.3.4... P _ (n-1)
      On an interval (m, m_1) m=t*P_n
      t=n
      P_n ^ 2 <m <P_ (n +1) ^ 2
      For the left limit (P_n ^ 2) value (m) can move. This is a question of error calculation. For the right limit switch is impossible. There, the formula takes the number of components for prime numbers. Hence:
      m * P_ (n +1) / P_ (n +1) -1 = m_1
      m = t * P_n
      t = 2.3.4 ... P_ (n-1)
      The interval (m, m_1) m = t * P_n
      t = n
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