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Re: [PrimeNumbers] New file uploaded to primenumbers
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 It's nice to see such clearly presented ideas, definitely worth having a look. I've noticed Marcel take a shot at one of the final theories, which implies that there were no obvious bloopers in the first part.
The early proofs are to an amateur like myself quite convincing  I particularly like the Twin prime theorem.
However, one thing that this has higlighted is my weakness in absolute rigor, and I am a little wary about a couple of things.
 firstly that just because something is true for every finite number does not (IIRC) mean that it is true as a limit.
[ The example I remember from University being that of the 'staircase diagonal'. For _every_ granularity of staircase, the length was 2, but the limiting length was 1.414... ]
 other things I can't quite formulate, just an unease about the 'rounding to integer' that takes place everywhere. I can't describe my worries yet, so I'll scratch a few things on paper first, and work out what I don't like. (However, I think that replacement of n by n# would make these worries go away.)
I really would like to see another critical eye cast over these. In particular, it might well be that my worry about limits are either unfounded or irrelevant (i.e. the proofs can be reformulated without the concept of a limit).
Phil
On Sun, 04 February 2001, primenumbers@yahoogroups.com wrote:
> Hello,
>
> This email message is a notification to let you know that
> a file has been uploaded to the Files area of the primenumbers
> group.
>
> File : /Josip Novakovic's Proof/primes.doc
> Uploaded by : novakovic_josip@...
Mathematics should not have to involve martyrdom;
Support Eric Weisstein, see http://mathworld.wolfram.com
Find the best deals on the web at AltaVista Shopping!
http://www.shopping.altavista.com  This theorem states that
PrimePi(x) = 1 + Sum[k = 2..x] foo(k)
Where foo(k) is a function which is 1 for primes or 4, and 0 otherwise.
In particular:
foo(k) = bar(k, baz(k) )
Where baz(k) is not in a closed form, and may require up to k steps to calculate.
I.e. this is /quadratic/ time method, or O(N^2)
Wheel factoring up to sqrt(N) is a O(sqrt(N)) method of testing primality. i.e. Prime counting can be done in O(N^1.5) by the most basic method known  by counting primes.
What on earth does introducing contrived (Caran D'Ache! :) ) functions gain us, in any way, in this regard?
Or is it just trying to do things in the most complicated way possible  like my Obfuscated C Code?
Phil
Mathematics should not have to involve martyrdom;
Support Eric Weisstein, see http://mathworld.wolfram.com
Find the best deals on the web at AltaVista Shopping!
http://www.shopping.altavista.com  Steven, Thank you for the upload of the Carol & Kynea Primes. Paul had
mentioned that we upload a file to the primenumber group and I suggested
that he look into it. Since you've done it then I guess that he will
maintain it. One suggestion though, I think that the numbers should be
ranked from highest to lowest and the rank be listed next to each number.
Also, it is understood that I found the ones for K>15000, but we need to
know the full names of the other two gentlemen as well. So I am suggesting
that their full names be included. Thirdly, i think that the number of
digits in the numbers should also be given. We can follow a format just
like Chris Caldwell's Largest known primes list. We can also
includeinformation like possible divisors of these numbers are == +/1 mod
8, just to name a few. We can even include a little biography of the
discoverers...
Let me know what you all think.
Cletus
>
_________________________________________________________________
>Hello,
>
>This email message is a notification to let you know that
>a file has been uploaded to the Files area of the primenumbers
>group.
>
> File : /Prime Tables/Carol&Kynea.txt
> Uploaded by : harvey563 <sharvey@...>
> Description : primes of form 4^n+2^n1
>
>You can access this file at the URL
>
>http://groups.yahoo.com/group/primenumbers/files/Prime%20Tables/Carol%26Kynea.txt
>
>To learn more about file sharing for your group, please visit
>
>http://help.yahoo.com/help/us/groups/files
>
>Regards,
>
>harvey563 <sharvey@...>
>
>
>
>
>
Send and receive Hotmail on your mobile device: http://mobile.msn.com  Mr Brennan said:
"It looks like the Pollard rho algorithm, but it appears to have
some minor changes from the "canonical" implementation.
I didn't try running it  I just had a quick glance at the code."
The code was an attempt at a modifcation of the Number Field Sieve method (find w and v such that w^2 mod operand = v^2, then find the gcd of w  v and operand). It looks for a solution in the area where w is less than the integer square root of the operand plus the remainder. So, if we start with 21 as our operand, it'll begin at:
4 + (21  16)
4 + 5 = 9
Now, 9^2 mod 21 = 81 mod 21 = (81  63) = 18
the integer square root of 18 is 4, thus
w  v = 9  4 = 5, and gcd of 5 and 21 is 1.
The program, when it sees a gcd of 1 or 2 (2 I use to filter out any even number that may show up as a factor), will then loop through the same operation, only with a different w. The new
value of w is determined by an algorithm that takes the old w and divides it by 10, then subtracts the result from w. Thus we
have 9 / 10 = .9 In this instance, the program will go into a loop that iterates through all the values between 9 and 1. To give a flavor of what happens with a realworld number, here's the output of the program (with the value of w being reported each time through the loop) for a semirandom number:
Now factoring 1234567894891...
0
1351681
1216513
1081345
1216513
1202997
1189481
1175965
1162449
1148933
1135417
1121901
1108385
1121901
1120550
1119199
1117848
1116497
1115146
1113795
1112444
1111093
1112444
1112309
1112174
1112039
1111904
1111769
1111634
1111499
1111364
1111229
1111094
1111229
1111216
1111203
1111190
1111177
1111164
1111151
1111138
1111125
1111112
1111099
1111112
1111111
1111111
1111110
1111109
1111108
1111107
1111106
1111105
1111104
1111103
1111102
No answer found.
If you look at the tenth entry from the end, you'll notice that it approaches the integer square root of the input number (the square root of 1234567894891 is 1111111.10825651). Anyway, for certain numbers (I'd really like to know if they have some common properties...) the program will find a factor of the input number operand in its descent towards the square root. I don't know exactly why that occurs, but I plan on investigating it further when I have more time (I'm suffering from Very Large Work Project syndrome that's cutting into my recreational programming and research time). Thanks for the info, btw, about Pollard Rho  I imagine that many of the answers I'm looking for are in the theory behind that algorithm.
Andrew Plewe  Andrew: The {m,n} numbers were presumably chosen using
an "isprime" test.
If one keeps {m,n} as close as possible,
to make h=m^2m*n+n^2 as small as possible,
for fixed d=m+n, then one gets:
{forprime(d=3,10^4,t=0;ns=(d1)/2;
forstep(n=ns,1,1,
if(t==0,m=dn;h=d^23*m*n;
if(isprime(h),t=1;print([m,n,h,d]))));
if(t==0,print("fail" d)))}
[2, 1, 3, 3]
[3, 2, 7, 5]
[4, 3, 13, 7]
[6, 5, 31, 11]
[7, 6, 43, 13]
[9, 8, 73, 17]
[11, 8, 97, 19]
[13, 10, 139, 23] [?]
[15, 14, 211, 29]
[16, 15, 241, 31]
[20, 17, 349, 37]
[21, 20, 421, 41]
....
[no failures to up to 10^4]
[?] Like you I am mystified why the author selected
[16, 7, 193, 23] with {16,7}> presi in modo "distinto"
instead of the closest pair {13,10}
I guess the conjecture is that it can always
be done one way or another.
The heuristics are resonable:
you get (d1)/2 attempts at a prime h
with d^2 > h > d^2/2.
PNT says that this is "very easy on average",
but that is quite different from proving
what I infer to be:
una congettura?
David  Edgar:
> Conjecture is not one.
To be clear, let me make a conjecture
> They are convinced that its development is a theorem.
> Non =E8 una congettura. Sono convinto che il suo
> sviluppo sia un teorema.
which I believe to be highly probable,
but cannot prove.
Conjecture: For every integer x with gcd(x,6)=1
there exists an integer y such that
h=(x^2+3*y^2)/4
is a prime less than x^2.
Comment: Set x=m+n, y=mn to get your quadratic form
h=(m^3+n^3)/(m+n)=m^2m*n+n^2
which I prefer to diagonalize.
Note that I conjecture that there is at least one
prime of your form when we fix m+n to be any
odd number not divisible by 3.
I cannot prove my conjecture.
You can prove this thing?
Non posso dimostrare la mia congettura.
Potete dimostrare questa cosa?
David  Thanks Jim: All well now. I also checked that a signed
cert goes OK: the trailing
[Signature]
1$=0DFF68B645EEB5780475FFD833CD8E984E73D39D
2$=58B83826080F6CF6CB62CCC14288EC15D4EB7FF1
was wisely ignored.
David  Zaher:
> Description : symbols i want thier description
http://groups.yahoo.com/group/primenumbers/files/Symbols.jpg
1) integral round closed contour in the complex plane
2) principal value of integral with a pole in the integrand
3) [looks like an integral needing viagra, but is probably]
an integral avoiding a pole on the real axis
by a small semicircular detour to the upper half plane,
in which case its real part is as in (2) and
there is an imaginary part i*Pi*residue_at_pole
[assuming that f(z)*=f(z*)]
But none of these Cauchy symbols
has much to do with primes.
David  In a message dated 12/01/03 03:54:29 GMT Standard Time,
primenumbers@yahoogroups.com writes:
> File : /Articles/zetazerosdistrib.pdf
Andrey: there would appear to be an error in your second formula for N_0(T)
> Uploaded by : andrey_601 <Andrey_601@...>
> Description : On the distribution of zeta zeros
>
> You can access this file at the URL
>
>
>
>
> http://groups.yahoo.com/group/primenumbers/files/Articles/zetazerosdistrib.pdf
>
>
["another exact form"], as n is a free variable and undefined. Is there a
"sigma over n" missing, or what?
Mike Oakes
[Nontext portions of this message have been removed] > > http://groups.yahoo.com/group/primenumbers/files/Articles/zetazerosdistrib.pdf
The sigma is over prime powers p^n < x. Of course, n is positive integer, as defined in the first formula and can be seen from context.
> >
> >
>
> Andrey: there would appear to be an error in your second formula for N_0(T)
> ["another exact form"], as n is a free variable and undefined. Is there a
> "sigma over n" missing, or what?
Best wishes,
Andrey
[Nontext portions of this message have been removed] In a message dated 20/01/03 00:41:59 GMT Standard Time, ajw01@...
writes:
> Anyway, at the top of the list are a trio of zeros with very low
Andrew: These are very interesting results!
> imaginary part, beating the 2.817...*I zero found a while back by
> David Broadhurst. These were initially found with 9 sig. figure
> precision, and then computed to greater precision with more terms of
> the series, so are almost certainly real.
>
However, the phrase "are almost certainly real" implies that you don't
believe the imag part is 0.002... but is rather 0.0, so to quote results to 9
figures is a bit misleading... To put it another way: what precision _are_
you claiming?
Could you maybe expand on what "more terms of the series" means: how many
terms did you take, and of what series? This would help others trying to
duplicate/refine your figures.
Mike Oakes
[Nontext portions of this message have been removed]   In primenumbers@yahoogroups.com, mikeoakes2@a... wrote:
> In a message dated 20/01/03 00:41:59 GMT Standard Time, ajw01@u...
of
> writes:
>
>
> > Anyway, at the top of the list are a trio of zeros with very low
> > imaginary part, beating the 2.817...*I zero found a while back by
> > David Broadhurst. These were initially found with 9 sig. figure
> > precision, and then computed to greater precision with more terms
> > the series, so are almost certainly real.
don't
> >
>
> Andrew: These are very interesting results!
>
> However, the phrase "are almost certainly real" implies that you
> believe the imag part is 0.002... but is rather 0.0, so to quote
results to 9
> figures is a bit misleading... To put it another way: what precision
_are_
> you claiming?
The bit in quotes means I believe they are almost certainly "actual"
zeros, in the earlier messages there was one case in which a zero
I found wasn't actual/real as pari's suminf doesn't like this series.
So although Newton's method converged, the value was bogus! Now I use
the sum function.
They don't have any nice Re(z)=0.5 properties like the normal zeta
function!> Could you maybe expand on what "more terms of the series" means: how
many
> terms did you take, and of what series? This would help others
trying to
> duplicate/refine your figures.
The series I used is formula (4) in
>
http://mathworld.wolfram.com/PrimeZetaFunction.html
the normal (1) doesn't converge over a wide range.
I found for the initial zero finding, using /p9 in pari, 10000 terms
was enough. Looking at the sum of the series for terms 10001 to 15000
confirms this. Generally more terms are required for smaller Re(z),
and perhaps the number of terms needed also slowly increases with
Im(z), but it has been ok in the range I looked at. For Re(z)<0.005
the number of terms would be even greater.
You'll also see a number of the zeros have been computed to greater
precision ( /p28 or the default in pari). So far I've found 20000
terms has been enough for this. Comparing the more accurate results
with the /p9 value, I believe the zeros should be all accurate except
in the last digit of the real and imaginary parts. This should also be
the case for the /p28 values.
Andrew  In a message dated 21/01/03 01:01:02 GMT Standard Time, ajw01@...
writes:
> The series I used is formula (4) in
Andrew: your table of zeros is fascinating.
> http://mathworld.wolfram.com/PrimeZetaFunction.html
> the normal (1) doesn't converge over a wide range.
>
>
But I continue to be worried: all the references I've found that give the
formula
prime_zeta(n) = {sigma k=1 to oo}(mu(k)/k)*ln(zeta(k*n))
quote the range of validity one would expect, namely: Re(n) > 1.
You are assuming this formula is valid for 0 < Re(n) < 1 also.
For Re(n) < 1, some of the zeta()'s in this sum have Re(argument) < 1, and
are therefore analytic continuations of the "original" zeta() (defined in
terms of the Euler product), and it is only the latter definition which
allows the Mobius inversion formula to be used.
For n = 1/m (m >= 2), one of the zeta()'s has argment 1.0, and so is infinite.
So, although you are computing a formal sum of zeta functions which converges
to prime_zeta(n) for Re(n) > 1, it isn't obvious that, for Re(n) < 1, said
sum is in fact prime_zeta(n).
I also have worries about the computation of ln(zeta(k*n)) when n is not
real: remembering that ln(z), for complex z, is multivalued, with period
2*pi*i, are you sure the correct value is always being taken by the software
you use?
Mike
[Nontext portions of this message have been removed]  Mike Oakes said to Andrew Walker:
> But I continue to be worried: all the references
Yes, but I quickly found and Andrew massively confirmed
> I've found that give the formula
> prime_zeta(n) = {sigma k=1 to oo}(mu(k)/k)*ln(zeta(k*n))
> quote the range of validity one would expect,
> namely: Re(n) > 1.
that if you are prepared to live with conditional convergence
and can smartly exploit PariGp, then you handle smaller Re(n),
with a price that increases as you get closer to zero.
What goes for you is the random process im moebius :)
David   In primenumbers@yahoogroups.com, mikeoakes2@a... wrote:
> In a message dated 21/01/03 01:01:02 GMT Standard Time, ajw01@u...
give the
> writes:
>
>
> > The series I used is formula (4) in
> > http://mathworld.wolfram.com/PrimeZetaFunction.html
> > the normal (1) doesn't converge over a wide range.
> >
> >
>
> Andrew: your table of zeros is fascinating.
> But I continue to be worried: all the references I've found that
> formula
< 1, and
> prime_zeta(n) = {sigma k=1 to oo}(mu(k)/k)*ln(zeta(k*n))
> quote the range of validity one would expect, namely: Re(n) > 1.
>
> You are assuming this formula is valid for 0 < Re(n) < 1 also.
>
> For Re(n) < 1, some of the zeta()'s in this sum have Re(argument)
> are therefore analytic continuations of the "original" zeta()
(defined in
> terms of the Euler product), and it is only the latter definition
which
> allows the Mobius inversion formula to be used.
is infinite.
>
> For n = 1/m (m >= 2), one of the zeta()'s has argment 1.0, and so
>
which converges
> So, although you are computing a formal sum of zeta functions
> to prime_zeta(n) for Re(n) > 1, it isn't obvious that, for Re(n) <
1, said
> sum is in fact prime_zeta(n).
is not
>
> I also have worries about the computation of ln(zeta(k*n)) when n
> real: remembering that ln(z), for complex z, is multivalued, with
period
> 2*pi*i, are you sure the correct value is always being taken by
the software
> you use?
Can you give some values? I would hope that Pari does this
>
properly...
I think it would be valuable if someone with better maths knowledge
in this area than me could check through the paper
38 #4421 10.41
Fröberg, CarlErik
On the prime zeta function.
Nordisk Tidskr. Informationsbehandling (BIT) 8 1968 187202.
The function in the title is defined by the Dirichlet series
$P(s)=\sum p\sp {s}$, $s=\sigma+it$,
$\sigma>1$, where the sum runs over all primes $p$.
The author tabulates $P(s)$ for various values of
$s$. Table I lists four roots of the equation $P(s)=0$
which as far as I know is the only reference in print on this topic.
Hopefully it will put my mind at ease as well!
Using the mobius formula in Pari with Newton's method did find the
zeros given in the paper, I won't be able to check until Tuesday
what values they had for Re(z). All I did last year was check that
the two prime zeta series did converge to the same value for complex
values with R(z)>1. While I do seem to be finding zeros for the
mobius series, it would be nice to know they are also zeros for
the prime series. Perhaps David can elaborate some more on this?
I think also one of the problems is that unlike the normal zeta
function, the prime zeta function has only been minimally studied,
so it would be nice to have moe info to check this is all above
board!
Andrew  Andrew Walker wrote:
>
Yes, I can illustrate the problem.
> > I also have worries about the computation of ln(zeta(k*n)) when n
> is not
> > real: remembering that ln(z), for complex z, is multivalued, with
> period
> > 2*pi*i, are you sure the correct value is always being taken by
> the software
> > you use?
> >
>
> Can you give some values? I would hope that Pari does this
> properly...
>
If the complex number z, in polar coordinates, is z = r*exp(i*theta), the
usual definition of ln(z) is:
that function which is real for arguments on the real axis and continuous in
the zplane cut along the negative real axis, i.e. ln(z) = ln(r) + i*theta,
with 2*pi <= theta <= 2*pi
However, when z is complex, this definition does not respect one of the
algebraic identities involved in going from the Euler product to the
sumofseries form of the zeta function, since it is not true in general that
ln(z^s) = s*ln(z), even for real (and a fortiori for complex) exponents s.
Example:
z = i, s = 5
ln(z^s) = ln(i^5) = ln(i) = pi/2
whereas
s*ln(z) = 5*ln(z) = 5*(pi/2)
The argument of ln(zeta(k*n)) certainly doesn't "respect" this cut along the
negative real axis, taking as it does not only zero but even (as I pointed
out before) even infinite values. So I claim that the ln() is a priori{Lat.}
uncertain modulo{Lat.} (2*pi*i). If this uncertainty is not pinned down, your
formal series may well still converge, but that fact alone does NOT prove
that it converges to the "right" (in the sense of analytic continuation)
values.
For settling these formal questions of uniqueness and analytic continuation,
appeals (pace{Lat.} David B.) to the convergence of Pari computations are
irrelevant.
Mike
[Nontext portions of this message have been removed]  Hi, John:
a) What is this program exactly about?
b) Could you put a stop button on it? (It didn't want to be closed, then it tried to crash my operating system :P).
Jose Brox Original Message 
From: primenumbers@yahoogroups.com
To: primenumbers@yahoogroups.com
Sent: Monday, February 03, 2003 11:08 AM
Subject: [PrimeNumbers] New file uploaded to primenumbers
Hello,
This email message is a notification to let you know that
a file has been uploaded to the Files area of the primenumbers
group.
File : /Jon Perry's Stuff/randomprimewalkproject.zip
Uploaded by : jon_perryuk <perry@...>
Description : Random Prime Walks
You can access this file at the URL
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To learn more about file sharing for your group, please visit
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[Nontext portions of this message have been removed]  a)It's part of my prime squares research. I wondered (after staring at
Ulam's Prime Spiral for about an hour), if any patterns might show
themselves. For example, with primes=2, at n=5000, the patterns look like
smoke from a cigarette.
b)I probably could, although I find Delphi threads quite difficult. Use
Ctrl+Alt+Delete to safely close a program.
Jon Perry
perry@...
http://www.users.globalnet.co.uk/~perry/maths/
http://www.users.globalnet.co.uk/~perry/DIVMenu/
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