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New formula for approximating pi(n)

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  • Rusi Kolev
    Hey guys, a few weeks ago I made a post in which I showed a proof for the formula: pi(n) ~ n( .5 -sqrt(.25 - 1/ln(n))) and also showed that it was a better
    Message 1 of 1 , May 28, 2003
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      Hey guys,
      a few weeks ago I made a post in which I showed a proof for the
      formula: pi(n) ~ n( .5 -sqrt(.25 - 1/ln(n))) and also showed that it
      was a better approximation to pi(n) over n/ln(n) and n/(ln(n)-1).
      In my paper (posted at http://rusi.greatnow.com/Math/ApproximatingPi
      (n).pdf) I proved a few other formulae. Last week I realized that I
      can actually use them to improve (or at least try to) over Li(n) and
      R(n) - the Riemann Function.
      So far I have the formula
      K(n) = Li(n) / ( 1 - sum( mobius(k)/n^((k-1)/k), k=2..infinity) )
      and of course K(n) ~ Pi(n).
      I can prove that it is better than Li(n) and I have tested and it is
      better than R(n) for at least half the cases I tried. The problem is
      that I do not have the computational power to test this formula for
      large values of k. The largest I can afford is k(max)=10^5.
      Well I have made a table that you can view at:
      http://rusi.greatnow.com/Math/Primes.mht
      Feel free to e-mail me at rusi_kolev@... to tell me what you
      think.

      Bye

      P.S. oh and if for some reason you can not open, you can access
      http://rusi.greatnow.com/Math/K(n).gif
      to view the formula at least ....
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