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predicting xth prime #, accurately

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  • leavemsg1
    Hello, Group. a new function Ii(x) for... predicting the xth prime number, accurately, given the number x: without proof, just an exercise in calculation. it
    Message 1 of 3 , Feb 25, 2007
      Hello, Group.

      a new function Ii(x) for...
      predicting the xth prime number, accurately, given the number x:
      without proof, just an exercise in calculation.

      it took me over two years to perfect it, but it's solid!

      summation from k = 0 to floor(sqrt(ln(x))),
      x; Ii(x) = Eta [{(exp(1))*(x/(10^k))*log(x/(10^k))} / {(log(1+
      ((10^k)/x)))^y}] where y = {(x*(e^k))/(10^(k+1))}.
      complicated, I know... but read on...

      one, rather small, example...

      let x = 10,000;
      but before computing Ii(x), input for Ii(x) should be trimmed using...
      summation from k = 0 to floor(sqrt(ln(x))),
      x; Ij(x) = Ii(x) - {Eta [(k+2)*(sqrt(x/(10^(k+1))))]}.

      k = 0 to 3;
      x = 10000; Ij(10000) = [2 * 31.62] + [3* 10] + [4 * 3.16] + [5 * 1]=
      110.89;

      so, compute Ii(9889.12) to predict the 10,000th prime number, by pull-
      back:
      k = 0 to 3;
      x = 9889.12; Ii(9889.12) = 98265.50 + 6204.68 + 259.99 + 3.99 =
      104,734.17;

      the 10,000th prime number is 104,729, but my answer is close enough!

      the error,... i used 10,000 instead of 9889.12 to calculate the
      denominators over the summation, because I did it by hand using a
      cheap calculator.
      interesting?

      i wanna say that it's accurate for any value x-- I used a pullback
      method similar to those used in advanced calculus.

      i have a similar method for computing an accurate count for all the
      prime numbers before a certain 'x' but w/pullback of the output.

      give me a number, less than 1,000,000,000, with unique digits, and i
      will show you its nearly exact position, and give an explanation of
      how I calc'd it... you'll wonder how I found such a neat equation.

      regards,
      Bill Bouris, Aurora, IL USA
    • leavemsg1
      2 corrections below... ... sometimes the Hu man element gets in the way of the experience; and that s good, sometimes amazing! I got so post happy that I
      Message 2 of 3 , Feb 27, 2007
        2 corrections below...
        --- In primenumbers@yahoogroups.com, "leavemsg1" <leavemsg1@...>
        wrote:
        >
        > Hello, Group.
        >
        > a new function Ii(x) for...
        > predicting the xth prime number, accurately, given the number x:
        > without proof, just an exercise in calculation.
        >
        > it took me over two years to perfect it, but it's solid!
        >
        > summation from k = 0 to floor(sqrt(ln(x))),
        > x; Ii(x) = Eta [{(exp(1))*(x/(10^k))*log(x/(10^k))} / {(log(1+
        > ((10^k)/x)))^y}] where y = {(x*(e^k))/(10^(k+1))}.
        > complicated, I know... but read on...

        sometimes the 'Hu'man element gets in the way of the experience; and
        that's good, sometimes amazing! I got so 'post' happy that I typed
        in another 'log' in the denominator that shouldn't be there...

        Ii(x) should = Eta [{(exp(1))*(x/(10^k))*log(x/(10^k))} / {(1+
        ((10^k)/x))^y}] where y = {(x*(e^k))/(10^(k+1))}.
        otherwise it's verifiable now... just look below for the other cor-
        rection and someone can comment on the calculation.

        >
        > one, rather small, example...
        >
        > let x = 10,000;
        > but before computing Ii(x), input for Ii(x) should be trimmed
        using...
        > summation from k = 0 to floor(sqrt(ln(x))),
        > x; Ij(x) = Ii(x) - {Eta [(k+2)*(sqrt(x/(10^(k+1))))]}.

        Ij(x) should = {Eta [(k+1)*(sqrt(ln(x/(10^(k+1)))))]}... to be the
        proper pullback!

        >
        > k = 0 to 3;
        > x = 10000; Ij(10000) = [2 * 31.62] + [3* 10] + [4 * 3.16] + [5 * 1]
        =
        > 110.89;

        no! Ij(10000) = 11.47
        finally if x = 10000; then compute Ii(10000-11.47) = Ii(9988.5) =
        104,728.9 ~= 104,729
        now, we have an appreciable result for the 10,000th prime!

        Phil Carmody?? or Chris Caldwell?? Paul Underwood?? or David
        Broadhurst? anyone? ... cool huh?

        >
        > so, compute Ii(9889.12) to predict the 10,000th prime number, by
        pull-
        > back:
        > k = 0 to 3;
        > x = 9889.12; Ii(9889.12) = 98265.50 + 6204.68 + 259.99 + 3.99 =
        > 104,734.17;
        >
        > the 10,000th prime number is 104,729, but my answer is close enough!
        >
        > the error,... i used 10,000 instead of 9889.12 to calculate the
        > denominators over the summation, because I did it by hand using a
        > cheap calculator.
        > interesting?
        >
        > i wanna say that it's accurate for any value x-- I used a pullback
        > method similar to those used in advanced calculus.
        >
        > i have a similar method for computing an accurate count for all the
        > prime numbers before a certain 'x' but w/pullback of the output.
        >
        > give me a number, less than 1,000,000,000, with unique digits, and
        i
        > will show you its nearly exact position, and give an explanation of
        > how I calc'd it... you'll wonder how I found such a neat equation.
        >
        > regards,
        > Bill Bouris, Aurora, IL USA
        >
      • leavemsg1
        go to the yahoo group primecalculations via yahoo group search to see the other formulas that coincide with this prediction function! thx, Bill ... and ...
        Message 3 of 3 , Feb 27, 2007
          go to the yahoo group 'primecalculations' via yahoo group search to
          see the other formulas that coincide with this prediction function!
          thx, Bill

          --- In primenumbers@yahoogroups.com, "leavemsg1" <leavemsg1@...>
          wrote:
          >
          > 2 corrections below...
          > --- In primenumbers@yahoogroups.com, "leavemsg1" <leavemsg1@>
          > wrote:
          > >
          > > Hello, Group.
          > >
          > > a new function Ii(x) for...
          > > predicting the xth prime number, accurately, given the number x:
          > > without proof, just an exercise in calculation.
          > >
          > > it took me over two years to perfect it, but it's solid!
          > >
          > > summation from k = 0 to floor(sqrt(ln(x))),
          > > x; Ii(x) = Eta [{(exp(1))*(x/(10^k))*log(x/(10^k))} / {(log(1+
          > > ((10^k)/x)))^y}] where y = {(x*(e^k))/(10^(k+1))}.
          > > complicated, I know... but read on...
          >
          > sometimes the 'Hu'man element gets in the way of the experience;
          and
          > that's good, sometimes amazing! I got so 'post' happy that I typed
          > in another 'log' in the denominator that shouldn't be there...
          >
          > Ii(x) should = Eta [{(exp(1))*(x/(10^k))*log(x/(10^k))} / {(1+
          > ((10^k)/x))^y}] where y = {(x*(e^k))/(10^(k+1))}.
          > otherwise it's verifiable now... just look below for the other cor-
          > rection and someone can comment on the calculation.
          >
          > >
          > > one, rather small, example...
          > >
          > > let x = 10,000;
          > > but before computing Ii(x), input for Ii(x) should be trimmed
          > using...
          > > summation from k = 0 to floor(sqrt(ln(x))),
          > > x; Ij(x) = Ii(x) - {Eta [(k+2)*(sqrt(x/(10^(k+1))))]}.
          >
          > Ij(x) should = {Eta [(k+1)*(sqrt(ln(x/(10^(k+1)))))]}... to be the
          > proper pullback!
          >
          > >
          > > k = 0 to 3;
          > > x = 10000; Ij(10000) = [2 * 31.62] + [3* 10] + [4 * 3.16] + [5 *
          1]
          > =
          > > 110.89;
          >
          > no! Ij(10000) = 11.47
          > finally if x = 10000; then compute Ii(10000-11.47) = Ii(9988.5) =
          > 104,728.9 ~= 104,729
          > now, we have an appreciable result for the 10,000th prime!
          >
          > Phil Carmody?? or Chris Caldwell?? Paul Underwood?? or David
          > Broadhurst? anyone? ... cool huh?
          >
          > >
          > > so, compute Ii(9889.12) to predict the 10,000th prime number, by
          > pull-
          > > back:
          > > k = 0 to 3;
          > > x = 9889.12; Ii(9889.12) = 98265.50 + 6204.68 + 259.99 + 3.99 =
          > > 104,734.17;
          > >
          > > the 10,000th prime number is 104,729, but my answer is close
          enough!
          > >
          > > the error,... i used 10,000 instead of 9889.12 to calculate the
          > > denominators over the summation, because I did it by hand using a
          > > cheap calculator.
          > > interesting?
          > >
          > > i wanna say that it's accurate for any value x-- I used a
          pullback
          > > method similar to those used in advanced calculus.
          > >
          > > i have a similar method for computing an accurate count for all
          the
          > > prime numbers before a certain 'x' but w/pullback of the output.
          > >
          > > give me a number, less than 1,000,000,000, with unique digits,
          and
          > i
          > > will show you its nearly exact position, and give an explanation
          of
          > > how I calc'd it... you'll wonder how I found such a neat equation.
          > >
          > > regards,
          > > Bill Bouris, Aurora, IL USA
          > >
          >
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