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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
Message 1 of 18 , Feb 5, 2001
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

Mathematics should not have to involve martyrdom;
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• 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) =
Message 2 of 18 , Aug 21, 2001
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

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• 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
Message 3 of 18 , Mar 11, 2002
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^n-1
>
>You can access this file at the URL
>
>
>
>http://help.yahoo.com/help/us/groups/files
>
>Regards,
>
>harvey563 <sharvey@...>
>
>
>
>
>

_________________________________________________________________
• 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
Message 4 of 18 , Sep 18 9:55 AM
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 real-world number, here's the output of the program (with the value of w being reported each time through the loop) for a semi-random 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

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^2-m*n+n^2 as small as possible,
Message 5 of 18 , Oct 16, 2002
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^2-m*n+n^2 as small as possible,
for fixed d=m+n, then one gets:

{forprime(d=3,10^4,t=0;ns=(d-1)/2;
forstep(n=ns,1,-1,
if(t==0,m=d-n;h=d^2-3*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 (d-1)/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
• ... To be clear, let me make a conjecture which I believe to be highly probable, but cannot prove. Conjecture: For every integer x with gcd(x,6)=1 there exists
Message 6 of 18 , Oct 16, 2002
Edgar:

> Conjecture is not one.
> They are convinced that its development is a theorem.

> Non =E8 una congettura. Sono convinto che il suo
> sviluppo sia un teorema.

To be clear, let me make a conjecture
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.

h=(m^3+n^3)/(m+n)=m^2-m*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$=0DFF68B645EEB5780475FFD833CD8E984E73D39D Message 7 of 18 , Nov 14, 2002 Thanks Jim: All well now. I also checked that a signed cert goes OK: the trailing [Signature] 1$=0DFF68B645EEB5780475FFD833CD8E984E73D39D
2$=58B83826080F6CF6CB62CCC14288EC15D4EB7FF1 was wisely ignored. David • ... 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 Message 8 of 18 , Dec 14, 2002 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, ... Andrey: there would appear to be an error in your second formula for N_0(T) [ another exact form ], Message 9 of 18 , Jan 12, 2003 In a message dated 12/01/03 03:54:29 GMT Standard Time, primenumbers@yahoogroups.com writes: > File : /Articles/zetazerosdistrib.pdf > 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 > > 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? Mike Oakes [Non-text portions of this message have been removed] • ... The sigma is over prime powers p^n Message 10 of 18 , Jan 12, 2003 > > http://groups.yahoo.com/group/primenumbers/files/Articles/zetazerosdistrib.pdf > > > > > > 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? 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. Best wishes, Andrey [Non-text portions of this message have been removed] • In a message dated 20/01/03 00:41:59 GMT Standard Time, ajw01@uow.edu.au ... Andrew: These are very interesting results! However, the phrase are almost Message 11 of 18 , Jan 20, 2003 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 > 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. > Andrew: These are very interesting results! 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 [Non-text portions of this message have been removed] • ... of ... don t ... results to 9 ... _are_ ... The bit in quotes means I believe they are almost certainly actual zeros, in the earlier messages there was Message 12 of 18 , Jan 20, 2003 --- In primenumbers@yahoogroups.com, mikeoakes2@a... wrote: > In a message dated 20/01/03 00:41:59 GMT Standard Time, ajw01@u... > 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 of > > the series, so are almost certainly real. > > > > Andrew: These are very interesting results! > > 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? 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@uow.edu.au ... Andrew: your table of zeros is fascinating. But I continue to be worried: all the Message 13 of 18 , Jan 25, 2003 In a message dated 21/01/03 01:01:02 GMT Standard Time, ajw01@... 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 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 multi-valued, with period 2*pi*i, are you sure the correct value is always being taken by the software you use? Mike [Non-text portions of this message have been removed] • ... Yes, but I quickly found and Andrew massively confirmed that if you are prepared to live with conditional convergence and can smartly exploit Pari-Gp, then Message 14 of 18 , Jan 25, 2003 Mike Oakes said to Andrew Walker: > 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. Yes, but I quickly found and Andrew massively confirmed that if you are prepared to live with conditional convergence and can smartly exploit Pari-Gp, 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 • ... give the ... Message 15 of 18 , Jan 25, 2003 --- In primenumbers@yahoogroups.com, mikeoakes2@a... wrote: > In a message dated 21/01/03 01:01:02 GMT Standard Time, ajw01@u... > 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 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 multi-valued, 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, Carl-Erik On the prime zeta function. Nordisk Tidskr. Informationsbehandling (BIT) 8 1968 187--202. 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
• ... Yes, I can illustrate the problem. If the complex number z, in polar coordinates, is z = r*exp(i*theta), the usual definition of ln(z) is:- that function
Message 16 of 18 , Jan 26, 2003
Andrew Walker wrote:

>
> > 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 multi-valued, 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...
>

Yes, I can illustrate the problem.

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 z-plane 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
sum-of-series 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

[Non-text 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
Message 17 of 18 , Feb 3, 2003
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 -----
Sent: Monday, February 03, 2003 11:08 AM

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
Description : Random Prime Walks

You can access this file at the URL

http://help.yahoo.com/help/us/groups/files

Regards,

jon_perryuk <perry@...>

Unsubscribe by an email to: primenumbers-unsubscribe@yahoogroups.com
The Prime Pages : http://www.primepages.org/

[Non-text 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
Message 18 of 18 , Feb 3, 2003
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/