- 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...

--- In primenumbers@yahoogroups.com, "leavemsg1" <leavemsg1@...>

wrote:>

sometimes the 'Hu'man element gets in the way of the experience; and

> 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...

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.

>

using...

> one, rather small, example...

>

> let x = 10,000;

> but before computing Ii(x), input for Ii(x) should be trimmed

> summation from k = 0 to floor(sqrt(ln(x))),

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

> x; Ij(x) = Ii(x) - {Eta [(k+2)*(sqrt(x/(10^(k+1))))]}.

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?

>

pull-

> so, compute Ii(9889.12) to predict the 10,000th prime number, by

> back:

i

> 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

> 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

--- In primenumbers@yahoogroups.com, "leavemsg1" <leavemsg1@...>

wrote:>

and

> 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;

> that's good, sometimes amazing! I got so 'post' happy that I typed

1]

> 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 *

> =

enough!

> > 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

> >

pullback

> > 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

> > method similar to those used in advanced calculus.

the

> >

> > i have a similar method for computing an accurate count for all

> > prime numbers before a certain 'x' but w/pullback of the output.

and

> >

> > give me a number, less than 1,000,000,000, with unique digits,

> i

of

> > will show you its nearly exact position, and give an explanation

> > how I calc'd it... you'll wonder how I found such a neat equation.

> >

> > regards,

> > Bill Bouris, Aurora, IL USA

> >

>