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• ... Of course, I meant smooth monotonic functions. In fact, pi0(x) = R(x) - sum(R(x^r)) + arctan(pi/logx)/pi - 1/logx, where sum is over non-trivial zeta
Message 1 of 11 , Nov 2, 2002
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> Yes, I also believe they exist, but they must be close to R(x). I think the difference is about O[1/log x].

Of course, I meant smooth monotonic functions. In fact,

pi0(x) = R(x) - sum(R(x^r)) + arctan(pi/logx)/pi - 1/logx,

where sum is over non-trivial zeta zeros, and

pi0(x) = lim(eps->0,(pi(x-eps)+pi(x-eps))/2)

equals to pi(x) in non-prime points and less than pi(x) by 0.5 in prime points.

So we can take the sum over some first zeta zeros, e.g. with |r|<=T, and obtain good approximating function with "standard deviation" of about sqrt(2*sum(1/|r|^2)), where sum is over non-taken zeros (with |r|>T in our case). This "standard deviation" comes from the distribution of F(e^u)-pi(e^u) values over positive u's. For R(x) this deviation is about 0.21492188798... (Note that these thoughts are true assuming GSH).

Best,

Andrey

[Non-text portions of this message have been removed]
• ... Please read F(e^u)-pi0(e^u), where F(x) is our approximating function: R(x) + arctan(pi/logx)/pi - 1/logx - sum(R(x^r), some r) ... i.e. when F(x)=R(x)
Message 2 of 11 , Nov 2, 2002
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> This "standard deviation" comes from the distribution of F(e^u)-pi(e^u) values over positive u's.

R(x) + arctan(pi/logx)/pi - 1/logx - sum(R(x^r), some r)

> For R(x) this deviation is about 0.21492188798...

i.e. when F(x)=R(x) (the term "arctan(pi/logx)/pi - 1/logx" may be omitted since it's O[1/(log x)^3] and tends to zero when x->+inf and doesn't distort the distribution).

Sorry for so many messages. Good night... |-)

Andrey

[Non-text portions of this message have been removed]
• ... Please read (F(e^u)-pi0(e^u))*u*e^(-u/2) of course. Andrey [Non-text portions of this message have been removed]
Message 3 of 11 , Nov 3, 2002
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> > This "standard deviation" comes from the distribution of F(e^u)-pi(e^u) values over positive u's.
>

Andrey

[Non-text portions of this message have been removed]
• ... http://arxiv.org/abs/math.NT/0210312/ is more correct. Best wishes, Andrey [Non-text portions of this message have been removed]
Message 4 of 11 , Nov 3, 2002
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Sebastian Martín wrote:

> see you:
>
> http://arxiv.org/abs/math.NT/0210312

http://arxiv.org/abs/math.NT/0210312/ is more correct.

Best wishes,

Andrey

[Non-text portions of this message have been removed]
• Thanks see you: http://arxiv.org/abs/math.NT/0210312 Sincerely Sebastian Martín Ruiz ... _______________________________________________________________
Message 5 of 11 , Nov 3, 2002
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Thanks

see you:

http://arxiv.org/abs/math.NT/0210312

Sincerely

Sebastian Martín Ruiz

--- Phil Carmody <thefatphil@...> escribió: >
--- Sebastian Martin <sebi_sebi@...> wrote:
> > Hi all:
> > ¿which is the best bound for the prime counting
> > function
> >
> > Is to improve the program attached for obtain
> prime
> > numbers
>
>
> There are two fairly good approximations that are in
> common use.
> One is just 'li':
>
> li(x) =~ 1.045+ Integral{x = 2 .. +inf} [ x/ln(x) ]
> (the 1.045 is effecively the li(2) component, the
> integral from 0..2)
>
> The second is one of Riemann's, and is a
> modifications to the above
>
> R(x) = Sum {n = 1..+inf} [ mu(n)/n . li(x^(1/n)) ]
>
> mu(n) = the mobius function, which is 0 if n
> contains a repeated factor,
> -1 for n with an odd number of prime factors, and +1
> for n with an even
> number of prime factors. mu({1..7...}) =
> {1,-1,-1,0,-1,+1,-1...}
> so
> R(x) = li(x) - li(x^(1/2))/2 - li(x^(1/3))/3 -
> li(x^(1/5))/5 + li(x^(1/6))/6 - ...
>
> R(x)-pi(x) hovers around zero much more than
> li(x)-pi(x) does, and most of
> the time seems more accurate.
>
> I believe that there are more accurate functions,
> that are computationally
> harder to evaluate, perhaps Mathworld has more
> information.
>
>
> Phil
>
>
> =====
> First rule of Factor Club - you do not talk about
> Factor Club.
> Second rule of Factor Club - you DO NOT talk about
> Factor Club.
> Third rule of Factor Club - when the cofactor is
> prime, or you've trial-
> divided up to the square root of the number, the
> factoring is over.
>
> __________________________________________________
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>
>
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> The Prime Pages : http://www.primepages.org/
>
>
>
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>
>

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• ... Sorry, but you re never going to persuade the person who made Bernstien s primegen twice as fast to move to a claimed O(n^(3/2)) time algorithm. You do
Message 6 of 11 , Nov 3, 2002
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--- Sebastian Martin <sebi_sebi@...> wrote:
> Thanks
>
> see you:
>
> http://arxiv.org/abs/math.NT/0210312

Sorry, but you're never going to persuade the person who made Bernstien's
primegen twice as fast to move to a claimed O(n^(3/2)) time algorithm.

You do know that O(n^(3/2)) is no faster than just testing each number in
turn using _nothing but trial division_, don't you? (Which is basically a
subset of what's taking place in the algorithm described therein.)

I'd be curious to see the proof that expresion (7) is O(n^(3/2)), because
it's basically
pi(x) = Sum { j=2..x } [ expression1 involving Sum { i=1..j } [ expresson2 ] ]
I can enumerate x(x-1)/2-1 evaluations of expression 2. And that's O(x^2).
Please enlighten us how you perform O(x^2) divisions and additions in O(x^(3/2))
time.

You've also completely neglected the asymptotic behaviour of the operations
that you're performing, so you'd need at least an extra epsilon in the
exponent.

Phil

=====
First rule of Factor Club - you do not talk about Factor Club.
Second rule of Factor Club - you DO NOT talk about Factor Club.
Third rule of Factor Club - when the cofactor is prime, or you've trial-
divided up to the square root of the number, the factoring is over.

__________________________________________________
Do you Yahoo!?
HotJobs - Search new jobs daily now
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• ... You can make the sum of d(i) only to Sqrt[i] and multiplied by 2 in F(n), the perfect square are not problem. On the other hand see you this relatively
Message 7 of 11 , Nov 4, 2002
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--- Phil Carmody <thefatphil@...> escribió: >
--- Sebastian Martin <sebi_sebi@...> wrote:
> > Thanks
> >
> > see you:
> >
> > http://arxiv.org/abs/math.NT/0210312
>
> Sorry, but you're never going to persuade the person
> primegen twice as fast to move to a claimed
> O(n^(3/2)) time algorithm.
>
> You do know that O(n^(3/2)) is no faster than just
> testing each number in
> turn using _nothing but trial division_, don't you?
> (Which is basically a
> subset of what's taking place in the algorithm
> described therein.)
>
> I'd be curious to see the proof that expresion (7)
> is O(n^(3/2)), because
> it's basically
> pi(x) = Sum { j=2..x } [ expression1 involving Sum
> { i=1..j } [ expresson2 ] ]

You can make the sum of d(i) only to Sqrt[i]
and multiplied by 2 in F(n), the perfect square
are not problem.

On the other hand see you this relatively fast
Mathematica program to calculate the prime numbers

based in the second formula of www.primepuzzles.net
Problem 38

(* Formula to Obtain the Next Prime to any Number p >
114 *)
DD[i_] := Sum[Quotient[i, j] - Quotient[i - 1, j], {j,
1, Sqrt[i]}]
G[i_] := -Quotient[2 - 2*DD[i], i]

(* Bound for Pi[p] prime counting function (Exist
other better?)*)
(* Rosser and Schoenfeld inequalities *)
(* Nested Product O(nlogn)^(3/2) *)
p = 114
114
While[p < 1000,
{n = Floor[N[1.045 + NIntegrate[1/Log[x], {x, 2,
p}], 15]];
A = Floor[(n + 1)*Log[n + 1] + (n + 1)*(Log[Log[n
+ 1]] - 0.9385)];
T = G[A]; m = A - 1; B = Timing[While[m > p, (T =
(T + 1)*G[m]; m--)]];
k = p; p = p + 1 + T;
Print["NXT(", k, ")=", p, " ", PrimeQ[p], " ",
B]}]

> I can enumerate x(x-1)/2-1 evaluations of expression
> 2. And that's O(x^2).
> Please enlighten us how you perform O(x^2) divisions
> time.
>
> You've also completely neglected the asymptotic
> behaviour of the operations
> that you're performing, so you'd need at least an
> extra epsilon in the
> exponent.
>
> Phil
>
>
>
> =====
> First rule of Factor Club - you do not talk about
> Factor Club.
> Second rule of Factor Club - you DO NOT talk about
> Factor Club.
> Third rule of Factor Club - when the cofactor is
> prime, or you've trial-
> divided up to the square root of the number, the
> factoring is over.
>
> __________________________________________________
> Do you Yahoo!?
> HotJobs - Search new jobs daily now
> http://hotjobs.yahoo.com/
>

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• ... So you agree that the big-Oh expression in the document doesn t apply to the formula in the document? That was my claim. The formula you now allude to is
Message 8 of 11 , Nov 4, 2002
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--- Sebastian Martin <sebi_sebi@...> wrote:
> --- Phil Carmody <thefatphil@...> escribi�: >
> --- Sebastian Martin <sebi_sebi@...> wrote:
> > > Thanks
> > >
> > > see you:
> > >
> > > http://arxiv.org/abs/math.NT/0210312
> >
> > Sorry, but you're never going to persuade the person
> > primegen twice as fast to move to a claimed
> > O(n^(3/2)) time algorithm.
> >
> > You do know that O(n^(3/2)) is no faster than just
> > testing each number in
> > turn using _nothing but trial division_, don't you?
> > (Which is basically a
> > subset of what's taking place in the algorithm
> > described therein.)
> >
> > I'd be curious to see the proof that expresion (7)
> > is O(n^(3/2)), because
> > it's basically
> > pi(x) = Sum { j=2..x } [ expression1 involving Sum
> > { i=1..j } [ expresson2 ] ]
>
> You can make the sum of d(i) only to Sqrt[i]
> and multiplied by 2 in F(n), the perfect square
> are not problem.

So you agree that the big-Oh expression in the document doesn't
apply to the formula in the document? That was my claim.

The formula you now allude to is now effectively trial dividing every
number, including all composites, to its square root in order to see
if it's prime, with _no_ early abort if you find a factor.

This is still crazier than the trial-division method with an instant abort,
which is still slower than any other method sensibly proposed in the last
200 years.

> On the other hand see you this relatively fast
> Mathematica program to calculate the prime numbers

No I don't. I see an _incredibly slow_ Mathematica program.

<<<
In[9]:= p=100000

Out[9]= 100000

In[10]:= While[p < 100100,
{n = Floor[N[1.045 + NIntegrate[1/Log[x], {x, 2, p}], 15]];
A = Floor[(n + 1)*Log[n + 1] + (n + 1)*(Log[Log[n+ 1]] - 0.9385)];
T = G[A]; m = A - 1; B = Timing[While[m > p, (T =(T + 1)*G[m]; m--)]];
k = p; p = p + 1 + T;
Print["NXT(", k, ")=", p, " ", PrimeQ[p], " ",B]}]
NXT(100000)=100003 True {1.42 Second, Null}
NXT(100003)=100019 True {1.33 Second, Null}
NXT(100019)=100043 True {1.32 Second, Null}
NXT(100043)=100049 True {1.32 Second, Null}

Interrupt> a

Out[10]= \$Aborted

In[11]:= p=1000000

Out[11]= 1000000

In[12]:= While[p < 1000100,
{n = Floor[N[1.045 + NIntegrate[1/Log[x], {x, 2, p}], 15]];
A = Floor[(n + 1)*Log[n + 1] + (n + 1)*(Log[Log[n+ 1]] - 0.9385)];
T = G[A]; m = A - 1; B = Timing[While[m > p, (T =(T + 1)*G[m]; m--)]];
k = p; p = p + 1 + T;
Print["NXT(", k, ")=", p, " ", PrimeQ[p], " ",B]}]
NXT(1000000)=1000003 True {19.01 Second, Null}
NXT(1000003)=1000033 True {18.96 Second, Null}
NXT(1000003)=1000033 True {18.96 Second, Null}
NXT(1000033)=1000037 True {18.94 Second, Null}
>>>

Notice the time of day in the following Pari/GP prompts, and the size of the
arguments I'm giving it.
<<<
(15:35) gp > isprime(nextprime(10^200))
%17 = 1
(15:35) gp > isprime(nextprime(10^210))
%18 = 1
(15:35) gp > isprime(nextprime(10^220))
%19 = 1
(15:35) gp > isprime(nextprime(10^230))
%20 = 1
(15:35) gp > isprime(nextprime(10^240))
%21 = 1
(15:36) gp > isprime(nextprime(10^250))
%22 = 1
(15:36) gp > isprime(nextprime(10^260))
%23 = 1
(15:36) gp > isprime(nextprime(10^270))
%24 = 1
(15:36) gp > isprime(nextprime(10^280))
%25 = 1
(15:36) gp > isprime(nextprime(10^290))
%26 = 1
>>>

To suggest that the Mathematica program is a fast program is the same
as suggesting that everyone should be using Turing Machines for speed.

Phil

=====
First rule of Factor Club - you do not talk about Factor Club.
Second rule of Factor Club - you DO NOT talk about Factor Club.
Third rule of Factor Club - when the cofactor is prime, or you've trial-
divided up to the square root of the number, the factoring is over.

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