Hi,

This is the result:

N Pi(x) R(x) Power

10 1E+10 455052511 455050683 420213980

11 1E+11 4118054813 4118052495 3951369042

12 1E+12 37607912018 37607910542 37155635094

13 1E+13 3.46066E+11 3.46066E+11 349383012475

14 1E+14 3.20494E+12 3.20494E+12 3285329105477

15 1E+15 2.98446E+13 2.98446E+13 30892707847616

16 1E+16 2.79238E+14 2.79238E+14 290491262067780

17 1E+17 2.62356E+15 2.62356E+15 2731556383920000

18 1E+18 2.474E+16 2.474E+16 25685455133563200

19 1E+19 2.34058E+17 2.34058E+17 241526262940073000

20 1E+20 2.22082E+18 2.22082E+18 2271127195778980000

21 1E+21 2.11273E+19 2.11273E+19 21355933208335000000

22 1E+22 2.01467E+20 2.01467E+20 200814769003914000000

23 1E+23 1.92532E+21 1.92532E+21 1888307621900430000000

24 1E+24 1.84356E+22 1.84356E+22 17756192398666400000000

25 1E+25 1.76846E+23 1.76846E+23 166965575334147000000000

alpha 0.07774984570

beta 0.9732770961

correlation R(x) 1.00000000000

correlation power 0.999997766

So, yes, Riemann function is better in this test.

I also attach the excel for those who can open it.

I based my testing on a book I was reading about unsolved problems that wrote that Riemann had used the formula x/ln(x) which was obviously an oversimplification.

I am sorry if I wasted your guys day.

Thanks & good night,

Chris

________________________________

From: djbroadhurst <

d.broadhurst@...>

To:

primenumbers@yahoogroups.com
Sent: Sunday, July 28, 2013 9:11 PM

Subject: [PrimeNumbers] Re: What if Riemann's prime-counting formula was not the best?

--- In

primenumbers@yahoogroups.com,

Chris De Corte <chrisdecorte@...> wrote:

> please provide me with 10 N, 10 pi(x) and Â 10 best Riemann's

> approximations and I will try to calculate a new alpha and beta.

Please use plain text in messages to this list, else

you may not be understood. Here I provide

[N, pi(10^N), R(10^N)]

[10, 455052511, 455050683]

[11, 4118054813, 4118052495]

[12, 37607912018, 37607910542]

[13, 346065536839, 346065531066]

[14, 3204941750802, 3204941731602]

[15, 29844570422669, 29844570495887]

[16, 279238341033925, 279238341360977]

[17, 2623557157654233, 2623557157055978]

[18, 24739954287740860, 24739954284239494]

[19, 234057667276344607, 234057667300228940]

[20, 2220819602560918840, 2220819602556027015]

[21, 21127269486018731928, 21127269485932299724]

[22, 201467286689315906290, 201467286689188773625]

[23, 1925320391606803968923, 1925320391607837268776]

[24, 18435599767349200867866, 18435599767347541878147]

[25, 176846309399143769411680, 176846309399141934626966]

where

R(x)=round(1+suminf(k=1,log(x)^k/(zeta(k+1)*k*k!)));

David

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