## predicting xth prime #, accurately

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• 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
• 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...
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

>
> 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
>
• 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

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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