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• Re: What if Riemann's prime-counting formula was not the best?

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• Alan, I said in my document that I only made a calculation up to 49978001 because I don t have other data available. If you all agree on a new trial then
Jul 28 1 of 16
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Alan,

I said in my document that I only made a calculation up to 49978001 because I don't have other data available.

If you all agree on a new trial then please provide me with 10 N, 10 pi(x) and  10 best Riemann's approximations and I will try to calculate a new alpha and beta.

Best,
Chris

________________________________
From: Alan Powell <AlanPowell@...>
Cc: Chris De Corte <chrisdecorte@...>
Sent: Sunday, July 28, 2013 1:42 PM
Subject: Re: What if Riemann's prime-counting formula was not the best?

Chris

In the Wikipedia article titled “Prime-counting function”

http://en.wikipedia.org/wiki/Prime-counting_function

the value of Pi(10^24) is exactly 18,435,599,767,349,200,867,866.

Pi(x) = alpha*x^beta  with alpha=0.2083666 and beta=0.9294465
gives

only 4,222,251,563,919,881,535,488  which is out by a
factor of 4+ !!

You may want to study some elementary books on Number Theory to
understand why Riemann’s function gives the best asymptotic value
as the number of primes tends towards infinity.

Regards

Alan

[Non-text portions of this message have been removed]
• ... 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,
Jul 28 1 of 16
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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
• 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
Jul 28 1 of 16
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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

________________________________
Sent: Sunday, July 28, 2013 9:11 PM
Subject: [PrimeNumbers] Re: What if Riemann's prime-counting formula was not the best?

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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[Non-text portions of this message have been removed]
• ... That was, of course, a log-lover s spoof on log-haters. However, this log-hater: http://arxiv.org/pdf/1307.4444.pdf attempts to fix up the obvious lunacy
Jul 28 1 of 16
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> Poly=polinterpolate(vector(25,k,10^k),data);

That was, of course, a log-lover's spoof on log-haters.

However, this log-hater:
http://arxiv.org/pdf/1307.4444.pdf
attempts to fix up the obvious lunacy of polynomial
approximation of pi(x).

Andrey Kulsha may offer us tighter conjectural
bounds on pi(10^26) ?

David
• ... On the assumption that Delta(x) defined by Andrey in http://www.primefan.ru/stuff/primes/table.html#theory continues to satisfy abs(Delta(x))
Jul 28 1 of 16
View Source

> Andrey Kulsha may offer us tighter conjectural
> bounds on pi(10^26) ?

On the assumption that Delta(x) defined by Andrey in
http://www.primefan.ru/stuff/primes/table.html#theory
continues to satisfy abs(Delta(x)) < 1, I estimate that

pi(10^26) = 1699246750872419991992147 +/- 167036339194

David (subject to error, as ever)
• djbroadhurst wrote: ... For large values ​​of x, this algorithm is inconvenient, eg for x = 10^250 requires over 1868 terms, Much faster can be calculated
Jul 30 1 of 16
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>
> [N, pi(10^N), R(10^N)]
........
> [25, 176846309399143769411680, 176846309399141934626966]
>
> where
>
> R(x)=round(1+suminf(k=1,log(x)^k/(zeta(k+1)*k*k!)));
>
For large values ​​of x, this algorithm is inconvenient, eg for x = 10^250 requires over 1868 terms,
Much faster can be calculated as

pi(x) ~= pli(x) = round(Li(x) - 1/2 Li(sqrt(x)))

where Li(x) is the Logarithm integral

pli(10^25) = 176846309399141938590795
(pli(10^25)/R(10^25)) - 1 = 2 10^-17
(pli(10^25)/pi(10^25)) -1 = -1 10^-14

pli{10^250)= 1740206254656916846774941665048386410178028975968929264655269395003484\
7365084787720410883002915274182213664956284195372937010842285191263145\
7678993892420170619475710388189158537825404886895382231933346054713467\
85875358018952542776800464839768387582

--
marian otremba
• ... No. The Gram formula is still very convenient at this size. Pari-GP, gives the exact value of R(10^250) in 0.1 seconds:
Jul 30 1 of 16
View Source
Chroma <chromatella@...> wrote:

>> R(x)=round(1+suminf(k=1,log(x)^k/(zeta(k+1)*k*k!)));
> For large values of x, this algorithm is inconvenient,
> eg for x = 10^250 requires over 1868 terms

No. The Gram formula is still very convenient at this size.
Pari-GP, gives the exact value of R(10^250) in 0.1 seconds:

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

{default(realprecision,260);print(R(10^250));
print(" took "gettime" milliseconds");

17402062546569168467749416650483864101780289759689292646552693950034847365084787720410883002915274182213664956284195372937010842285191263145767899389242017061947571038442681072462756632213511422607548574658029047365218974809766827365028215685475746
took 98 milliseconds

Perhaps you are paying for inferior software?
If so, the general rule is: the less you pay,
the better the deal.

Pari-GP is totally free and hence rather hard to beat :-)

David
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