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Re: found errors in predicting xth prime#

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  • 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 1 of 3 , Feb 27 3:35 PM
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      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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