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

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• ... 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,
Message 1 of 16 , Jul 28, 2013
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
Message 2 of 16 , Jul 28, 2013
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
Message 3 of 16 , Jul 28, 2013

> 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))
Message 4 of 16 , Jul 28, 2013

> 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
Message 5 of 16 , Jul 30, 2013
>
> [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:
Message 6 of 16 , Jul 30, 2013
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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