Twin prime conjecture is precisely what I am trying to prove with this
conjecture. We can prove by Dirichlet Theorem that exist the same t?We know by 
Dirichlet Theorem that exists infinity many t1 y t2 that p+t1(p-q) and q+t2(p-q)
are both primes. But there are no two be equal?

--- El vie, 27/2/09, michael_b_porter <michael.porter@...>
escribió:

De: michael_b_porter <michael.porter@...>
Asunto: Re: primes in arithmetic sequences
Para: "Sebastian Martin Ruiz" <s_m_ruiz@...>
Fecha: viernes, 27 febrero, 2009 6:02

--- In primenumbers@yahoogroups.com, Sebastian Martin Ruiz
<s_m_ruiz@...> wrote:
> It is to say exists a positive integer t that p+t(p-q) and q+t(p-q)
are both primes?

Suppose that this conjecture is true.  Let (s,s+2) be a pair of twin
primes.  Then by the conjecture (with p=s+2, q=s), there is a positive
integer t such that s+2+2t and s+2t are both prime.  So for each pair
of twin primes, there is a greater pair of twin primes.

So the twin prime conjecture follows from your conjecture.

- Michael Porter










[Non-text portions of this message have been removed]

Dear Sebastian:
 
 
In a recent work J.M. Deshouillers and F. Luca [On the distribution of some
means concerning the densitiy, Funct. Approx. Comment. Math. Volume 39, Number 2
(2008), 335-344.] consider certain means of the values of the Euler function to
prove that they are dense modulo one. At the Czech-Slovak Number Theory
Conference in August 2007, F. Luca raised the question whether certain other
sequences of mean values of the Euler function are uniformly distributed modulo
one. Among these are the sequences of arithmetic and geometric means. Recently,
J.M. Deshouillers and H. Iwaniec gave a method leading to an affirmative answer
for Luca's question in the case of arithmetic mean, and a conditional answer for
the case of geometric mean. The aim is to be framiliar with this method. it
might be useful on your approach to prove twin prime number conjecture.
meanwhile you can keep up yourself in connection with Professor Iwaniec.
 
Sincerely Yours.
 
Saeed Ranjbar

















[Non-text portions of this message have been removed]

2. primes in arithmetic sequences
     Posted by: "Sebastian Martin Ruiz" s_m_ruiz@... s_m_ruiz
     Date: Thu Feb 26, 2009 11:15 am ((PST))

Let p and q odd prime numbers p>q by Dirichlet theorem exist t1 and t2
positives integers that  p+t1(p-q) and q+t2(p-q) are primes.Can someone
prove that they exist by the same t?It is to say exists a positive
integer t that p+t(p-q) and q+t(p-q) are both primes?
Sincerely
Sebastian Martin Ruiz



r1 = p+t1(p-q)
r2 = q+t2(p-q)

r1 - r2 = ( p+t1(p-q) ) - ( q+t2(p-q) )

r1 - r2 = (p - q) + t1 (p-q) - t2(p-q)

r1 - r2 = (1 + t1 - t2 ) (p - q)


If t1 = t2,

r1 - r2 = p - q


The existence of r1 and r2 is implied by the conjecture that
every even integer is the difference of two primes.


Kermit

In previous post, related to twin prime discussion by
Sebastian Martin Ruiz,

I showed sloppy thinking by saying,


The existence of r1 and r2 is implied by the conjecture that
every even integer is the difference of two primes.


I should not have said anything at all.

Kermit.

Yes, there are common values for t1 y t2, but no-one can prove it yet.
I don't expect that this more general statement could be proved prior
to proving the twin prime conjecture (which is the special case p-q=2
where the values of t1, t2 are most dense - in fact you have a t1 and
a t2 for each prime larger than p resp. q).
Maximilian

On Fri, Feb 27, 2009 at 3:58 AM, Sebastian Martin Ruiz
<s_m_ruiz@...> wrote:
>
> Twin prime conjecture is precisely what I am trying to prove with this
conjecture. We can prove by Dirichlet Theorem that exist the same t?We know by 
Dirichlet Theorem that exists infinity many t1 y t2 that p+t1(p-q) and q+t2(p-q)
are both primes. But there are no two be equal?
>
> --- El vie, 27/2/09, michael_b_porter <michael.porter@...>
escribió:
>
> De: michael_b_porter <michael.porter@...>
> Asunto: Re: primes in arithmetic sequences
> Para: "Sebastian Martin Ruiz" <s_m_ruiz@...>
> Fecha: viernes, 27 febrero, 2009 6:02
>
> --- In primenumbers@yahoogroups.com, Sebastian Martin Ruiz
> <s_m_ruiz@...> wrote:
>> It is to say exists a positive integer t that p+t(p-q) and q+t(p-q)
> are both primes?
>
> Suppose that this conjecture is true.  Let (s,s+2) be a pair of twin
> primes.  Then by the conjecture (with p=s+2, q=s), there is a positive
> integer t such that s+2+2t and s+2t are both prime.  So for each pair
> of twin primes, there is a greater pair of twin primes.
>
> So the twin prime conjecture follows from your conjecture.
>
> - Michael Porter
>
>
>
>
>
>
>
>
>
>
> [Non-text portions of this message have been removed]
>
>
>
> ------------------------------------
>
> Unsubscribe by an email to: primenumbers-unsubscribe@yahoogroups.com
> The Prime Pages : http://www.primepages.org/
>
> Yahoo! Groups Links
>
>
>
>

A positive odd prime integer which is equal to 1 mod 4,

is uniquely represented as the sum of two squares.

Let p be the name of a prime equal to 1 mod 4.

Let  r and s be the names of the unique positive integers such that
p = r**2 + s**2.

Define F(p) = r + s.

Extend this mapping from the set of primes equal to 1 mod 4, to
all products of primes equal to 1 mod 4 by


F(p1**a1  p2**a2  . . . p_k**a_k) =  F(p1)**a1  F(p2)**a2   F(p3)**a3 .
. .  F(p_k)**a_k



F(5) = 2 + 1 = 3
F(13) = 2 + 3 = 5
F(17) = 4 + 1 = 5
F(25) = F(5*5) = F(5)*F(5) = 3 * 3 = 9
etc

Which odd positive integers are the sum of two squares?

We can't say much about this.

One of the things that we can say is that

If z is an odd positive integer which is the sum of two squares,

then one of the squares is odd, and one is even.

z   = ( n-m)**2 + (1 + n + m)**2 = n**2 - 2 n * m + m**2 + 1 + n**2 +
m**2 + 2 * n + 2 * m + 2 * n * m

z = 2* n**2 +2 *  m**2  + 2 * n + 2 * m + 1

z = 1 + 4 * sum of two distinct triangular numbers.

Where one of the triangular numbers is permitted to be equal to zero.

Does anyone know of   (or  able to discover)  any research previously
done on this function which maps sums of two opposite parity squares
onto the odd positive integers ?


Kermit Rose

Kermit Rose wrote:
>
> Which odd positive integers are the sum of two squares?
>
> We can't say much about this.
>

It is known that for an odd positive integer A, the
following two statements are equivalent:

- A is a sum of two squares.

- In the prime factorization of A, no prime of the form
    4x+3 appears an odd number of times.

Jack Brennen wrote:
>
> Kermit Rose wrote:
>>
>> Which odd positive integers are the sum of two squares?
>>
>> We can't say much about this.
>>
>
> It is known that for an odd positive integer A, the
> following two statements are equivalent:
>
> - A is a sum of two squares.
>
> - In the prime factorization of A, no prime of the form
>   4x+3 appears an odd number of times.
>
>
>

Hello Jack.

I overlooked that  multiplying a sum of squares times a square also
yields a sum of squares.

Thus I wish to amend my Function that maps the positive odd sums of two
squares onto the odd positive integers.

I had not yet defined to what the function would map squares of primes
equal to 3  mod  4.

I extend the function to cover this case by

F(9) = 3
F(49) = 7

F( q**2)  = q  if q is a prime equal to 3 mod 4.

Kermit Rose

Hi All

82751511*16229#+n*20333209*16229#+1 (n=0-4) describes an AP5 of primes

sieved n=10000000-15000000
17742343 tests
52888 prps
739001 ap3's
528 ap4's
0 ap5'
extended the 528 ap4's (up and down)gave 275 chances, no ap5's
Currently testing the extended Ap3's
Low  gave 247373 chances of an extendable ap4
High gave 245056 chances of an extendable ap4
at 23585 into the low chances 82751511 yielded this result.

cheers
Ken
p.s. Hopefully I'll get a few more


Primality testing 82751511*16229#+1 [N-1, Brillhart-Lehmer-Selfridge]
Running N-1 test using base 2
Calling Brillhart-Lehmer-Selfridge with factored part 33.34%
82751511*16229#+1 is prime! (45.1713s+0.0049s)
Primality testing 103084720*16229#+1 [N-1, Brillhart-Lehmer-Selfridge]
Running N-1 test using base 16249
Calling Brillhart-Lehmer-Selfridge with factored part 33.37%
103084720*16229#+1 is prime! (42.3828s+0.0042s)
Primality testing 123417929*16229#+1 [N-1, Brillhart-Lehmer-Selfridge]
Running N-1 test using base 3
Calling Brillhart-Lehmer-Selfridge with factored part 33.37%
123417929*16229#+1 is prime! (40.6010s+0.0041s)
Primality testing 143751138*16229#+1 [N-1, Brillhart-Lehmer-Selfridge]
Running N-1 test using base 2
Calling Brillhart-Lehmer-Selfridge with factored part 33.33%
143751138*16229#+1 is prime! (40.4721s+0.0040s)
Primality testing 164084347*16229#+1 [N-1, Brillhart-Lehmer-Selfridge]
Running N-1 test using base 7
Calling Brillhart-Lehmer-Selfridge with factored part 33.36%
164084347*16229#+1 is prime! (40.2943s+0.0038s)

Ken Davis wrote:
> 82751511*16229#+n*20333209*16229#+1 (n=0-4) describes an AP5 of primes

Congratulations!
http://users.cybercity.dk/~dsl522332/math/aprecords.htm#records is updated.
Ken now has the AP4 to AP7 records.

--
Jens Kruse Andersen

Will somebody please throw our silly website
censor a dictionary, and maybe a sandwich. Thanks.

The message of ENCRYPT3, hidden in a stream of 1009
integers posted here on January 21, can be deciphered as:

"The manuscript was a scroll sheathed in charred ceramic
and cloaked in ermines. There appeared to be a dead arctiid
moth inside. Though not of antiquity, it was made of papyrus
in the old style and was said to contain the lost secrets of
an exotic fifth force."

No one won the $100  offered for this solution, and so the
prize will now transfer to the new message, ENxxxxx4,
shown below:


         2990 4517 8879 11220 7984 10726 7795 5244 4890 8120
	 10694 8436 6134 4305 5261 5384 6613 1653 5196 11208
	 11631 2641 2713 11715 6418 6970 1647 10766 11683 6229
	 8004 5325 12016 1535 10315 6300 6421 3395 4837 4913
	 6076 10940 11140 6134 4545 9500 3621 6645 7395 10524
	 3680 2341 7504 11288 6766 6163 8567 3028 8753 1344
	 8365 10509 613 5630 8895 5816 10434 7579 9696 7380
	 2448 6439 3800 10876 5623 7734 7154 9671 6464 8174
	 7130 10461 7217 1131 2419 4491 11792 894 5072 9974
	 9248 4179 1804 1710 6122 8680 10661 1865 11135 4323
	 10404 7645 6885 2817 6734 11009 7669 7820 6296 8070
	 6135 6579 6167 7398 3161 4840 6910 10589 11637 7153
	 10047 835 2641 1759 2275 967 2320 2264 3499 1593
	 8798 3492 10222 366 3704 10215 3124 3492 10222 2264
	 3499 10215 4346 3492 1936 8663 5792 11187 2858 9369
	 6145 7751 11604 963 7008 5044 2036 7347 8355 1416
	 4249 6406 8077 10594 2192 5639 1492 5003 4207 9860
	 7120 5584 1492 8575 8996 9860 7120 918 1492 3763
	 8996 1288 2782 2703 6910 7811 9276 6722 10002 1691
	 1864 4505 2886 5343 8087 2603 1467 1043 10596 370

	 4424 2872 10177 4799 6145 8192 8992 25 1621 6593
	 2557 6294 10929 1506 4806 6138 8199 8985 5547 1614
	 6600 9050 8201 4063 1658 2220 9579 495 10852 1786
	 8702 5289 371 10258 4681 1857 634 8435 1016 4908
	 3846 4230 11190 9459 6942 3614 4061 7689 6582 966
	 4386 9315 10560 2345 3175 1949 2915 11318 4696 11311
	 5159 5151 5692 4271 5238 2468 22 6741 2283 6293
	 5931 9468 9435 2162 1970 30 1808 2075 2527 2016
	 2872 1998 4435 3056 8002 5778 10263 10638 7437 5922
	 5749 8298 4342 6378 5979 6261 7419 7132 8071 6138
	 10516 7183 6381 1740 4333 5739 1102 2182 4325 7440
	 11095 11334 1928 1803 4546 4718 9526 9975 9155 6857
	 5991 5892 1836 2185 4627 758 11201 768 4362 5737
	 5226 10128 9007 9414 7117 1970 4054 10949 9103 4047
	 250 9096 9536 10949 9103 9543 10956 9096 4054 2090
	 4857 4047 797 9526 6780 9813 265 1241 3127 3967
	 7328 936 5376 10512 6921 4984 5591 8280 9110 166
	 743 260 3331 6086 7507 10609 7943 9227 497 7937
	 8890 8066 5745 8883 8073 7937 8890 8066 6044 8883
	 8073 597 940 10294 10537 9513 808 8538 8224 8450

	 4386 10400 10690 4379 10407 449 4386 10400 10690 1311
	 650 4303 10997 9171 10988 10387 2502 8770 6957 2680
	 3008 8769 2085 11060 1185 1791 1260 3907 2770 3987
	 6438 5403 5600 3644 7417 5738 1327 8373 7956 11004
	 10059 6804 9006 11133 2622 2148 5655 1067 2836 3606
	 1772 2359 9704 9878 6492 6906 7238 10933 649 10861
	 6659 3430 6196 208 1912 5342 8098 9972 6119 4026
	 9315 3843 5904 9775 4472 5968 6166 9614 9828 8748
	 6200 2851 4763 9931 2182 3990 9997 681 9154 1805
	 10405 8233 7747 5639 5992 1435 8806 7415 7207 7434
	 6198 10338 8271 3413 5147 2930 1275 3813 3817 10200
	 3259 7925 10224 7596 9969 4397 8719 6227 8485 3245
	 10235 7071 8138 4096 7375 8898 10663 3186 8039 8780
	 1423 1782 3746 5124 5703 8523 6115 9038 9720 10480
	 9411 1416 416 3739 10741 5106 7461 1863 6540 1671
	 9130 10892 735 9563 1351 3597 8652 6905 2536 8285
	 649 10430 10210 8787 5391 9258 4591 7711 4606 9258
	 8485 7711 5391 9258 8485 7711 5391 9258 9760 7711
	 5391 3003 2445 3984 9183 2922 5466 4043 2929 5459
	 4050 2922 5466 4043 2929 5459 4050 2922 5466 4043

	 2929 5460 2880 2425 10726 2415 4665 9859 8162 5230
	 10745 10538 9861 1863 10660 4129    1216 3119 8697    1333
	 8709 4965 99 2223 1974 4163 3465 10006 5776 5984
	 10026 9652 5180 3744 10003 9066 9074 7790 3902 5763
	 4521 6534 10497 7781 4552 10490   7788 4545 10497   7781
	 4552 10490 7788 4545 10497 7781 3331 2001 1138 4976
	 10677 2365 9847 5010 4847 10135   4489 3735 4800 5764
	 8552 10221 711 1636 10129 3782 4408 4352 2590 2729
	 6376 4947 9968 206 1632 1625 2695 1625 1632 1625
	 1632 1625 1632 2001 1632 1625 1789 2555 8928 9434
	 4664 8516 1815 8778 3299 10342   9202 6763 5962 8558
	 1378 8531 3465 951 1624 9287 8368 2855 7857 9741
	 2224 4616 3697 10283 10426 8663 3397 6464 5544 8136
	 2064 2702 9273 7701 92 4858 1355 1277 7833  3186
	 5717 9231 9426 4696 1915 3997  7680  10200
         1318    4163
	 8094 6021 7833 7457 6871 2930  6871 2930 6871 2930
	 6871 2930 6871 5724 6871 2930   6030 2930 6871 2930
	 6871 2930 975 1099 8368 4706   4727 8572 3831 6755
         2180    4227    2417    8572    1627    6755  2180    4227
         2417    8572
	 1627 4502 9385 2466 10311 9401  8905   10283 6093    6141

	 6700    8174 3252    8858    7575 3426 7515 2523 5466 3405
	 9109    1715 7759 8861    7789 4535 6037 1619 5122 7151
	 2544    7442 6064 9131    9034 1883 8176 8423 8916 5674
	 3778    4561 10203   7597    9306 2721 5396 927 8110 3043
	 4798    9897 5534    8724    1256 4340 9904 5527 2969 569
	 5030    6490 3298 3910    8856 883 4813 6787 3627 659
	 2701    421 6276 5965   6122 9269 1656 5955 7525 9864
	 2023    3614 8467 7988   7495 3330 4069 7988 7495 3614
	 8467    7988 315 3614   8467 7988 2765 10014 6745 3997
	 5117    5828 2695 2970   9116 5255 9449 5379 4140 9837
	 6544    9873 9378 4155  5806 10012 2725 6587 6289 2521
	 1915    7079 6261 8152  3622 4086 9894 7454 824 8645
	 998 8087 2278    2667  6414 2649 3853 5975 5371 7250
	 6111    3233 7246    8330  8473 3756 874 4364 6732 4653
	 10026   2327 5395 6939  6679 8801 7103 5759 1499 7798
	 2942    9215 6429    5938   4815 4966 8455 9438 5958 868
	 8468    9243 3229 5249   4410 7984 8485 8295 8750 7794
	 2262    9298 8762   4673    5501 6970 2841 2715 8381 2173
	 5770    1394 7900   6948    8331 6948 8331 6948 8331 6948
	 8331    6948 8331   6948    8331 6948 8331 6948 8331 6948
         5275   6095 3537 4087    5793 7426 1885  6014    3393

Does anyone know why my neat columns of integers invariably get
mangled by this text-editor?

Aldrich Stevens

Further to my work on Pat(f(n),n)

http://www.rankyouragent.com/primes/primes_simple.htm
http://www.rankyouragent.com/primes/primes.htm

which is a recursive self complicating algorithm which lays down the prime
numbers, I've come up with some interesting formulae which appear on the title
page of:

http://mwiner.files.wordpress.com/2009/02/primeconstellations2.pdf

I've tested these formulae for the first 400 million primes, spanning 10 billion
numbers.

An interesting characteristic of these formulae is that they are forward
looking.  That is, with knowledge of the first 9003 primes, I am able to make
predictions of the prime distributions over 10 billion numbers ( P(9003)^2 ).

I hope they are of interest.

Best regards...MCW

Hi everyone,

Apologies if the subject has been raised before.

(1) What should we call numbers that stay prime for all permutations of their
digits (if the name does not exist already)?

(2) Is the number of such primes infinite? I am into the 6-digit primes now but
I have not yet found anymore than the list given below.

2, 3, 5, 7, 11, 13, 17, 31, 37, 71, 73, 79, 97, 113, 131, 199, 311, 337, 919,
991

(3) What happens when we change the rules slightly, like include the even
numbers? For example include numbers such as 149, 491 and 941. In this case the
definition changes slightly because the digit 4 should not become a unit.

I am hoping I can find more especially if they have some more interesting
properties.

Thank you.
Peter.

[Non-text portions of this message have been removed]

> (1) What should we call numbers that stay prime for all
> permutations of their digits (if the name does not exist already)?

See http://primes.utm.edu/glossary/page.php?sort=PermutablePrime .
Names include permutable prime, and absolute prime.

CC

Let P(z) be the set of all the primes whose squares do not exceed an integer z.

Can anyone point me towards (or provide me with) a proof that pi(z) converges to
phi(prod(P(z)))*z/prod(P(z)) as z --> oo, where phi is the Euler totient?
Presumably it'll entail reference to the Prime Number Theorem; perhaps, in a
scholarly paper, one need simply put 'it follows by the Prime Number Theorem
that pi(z) converges to phi(prod(P(z)))*z/prod(P(z)) as z --> oo'...?

With thanks,

Tom

Hi -- I am ust wondering why the last eight posts of mine to this group, over a
period of 2 years, have produced precisely zero replies. I know there's no
obligation on anyone to reply, and I know I have been naive here and there
but... has someone flicked a switch?

Tom

--- In primenumbers@yahoogroups.com, "gulland68" <tmgulland@...> wrote:
>
> Let P(z) be the set of all the primes whose squares do not exceed an integer
z.
>
> Can anyone point me towards (or provide me with) a proof that pi(z) converges
to phi(prod(P(z)))*z/prod(P(z)) as z --> oo, where phi is the Euler totient?
Presumably it'll entail reference to the Prime Number Theorem; perhaps, in a
scholarly paper, one need simply put 'it follows by the Prime Number Theorem
that pi(z) converges to phi(prod(P(z)))*z/prod(P(z)) as z --> oo'...?
>
> With thanks,
>
> Tom
>

Your notations are a bit confusing.
"z" usually stands for a complex variable.
Please clarify also what you mean with "converges to".
This term has a precise meaning which cannot be applied here.
Both expressions tend to +oo, but I assume you did not mean "have the same
limit".

I can guess what you mean but if you are wondering why you don't get an answer
to a question, you might first try to rephrase it more carefully. This shows
that you are really interested in an answer.
(And might actually lead you to it.)

Hints : use the prime number theorem, replace z by sqrt(n), look for an
asymptotic expression of primorial and what can be said about phi (wikipedia
and/or mathworld should do).

Maximilian


--- In primenumbers@yahoogroups.com, "gulland68" <tmgulland@...> wrote:
>
> Hi -- I am ust wondering why the last eight posts of mine to this group, over
a period of 2 years, have produced precisely zero replies. I know there's no
obligation on anyone to reply, and I know I have been naive here and there
but... has someone flicked a switch?
>
> Tom
>
> --- In primenumbers@yahoogroups.com, "gulland68" <tmgulland@> wrote:
> >
> > Let P(z) be the set of all the primes whose squares do not exceed an integer
z.
> >
> > Can anyone point me towards (or provide me with) a proof that pi(z)
converges to phi(prod(P(z)))*z/prod(P(z)) as z --> oo, where phi is the Euler
totient? Presumably it'll entail reference to the Prime Number Theorem; perhaps,
in a scholarly paper, one need simply put 'it follows by the Prime Number
Theorem that pi(z) converges to phi(prod(P(z)))*z/prod(P(z)) as z --> oo'...?
> >
> > With thanks,
> >
> > Tom
> >
>

--- In primenumbers@yahoogroups.com, "Maximilian Hasler" <maximilian.hasler@...>
wrote:
>
> Your notations are a bit confusing.
> "z" usually stands for a complex variable.

It was suggested to me by a topologist that I use z to denote an integer; I had
aleady used a great many others to denote other things. I'll give it some more
thought.

> Please clarify also what you mean with "converges to".

I thought this was the right term. I had previously used 'tends to' but an
algebraist told me it was not a mathematical term. By converges to, I mean that,
if a(z)/b(z) converges to 0 as z --> oo, b(z) increases in relation to a(z),
within an error the value of which decreases as a proportion of a(z)/b(z) as z
--> oo. You're saying that's not right?

> This term has a precise meaning which cannot be applied here.

How about (pi(sqrt(n))*sqrt(n))/(phi(prod(P(sqrt(n))))*prod(P(sqrt(n))) --> 1 as
z --> oo?

> Both expressions tend to +oo, but I assume you did not mean "have the same
limit".
>
> I can guess what you mean but if you are wondering why you don't get an answer
to a question, you might first try to rephrase it more carefully. This shows
that you are really interested in an answer.
> (And might actually lead you to it.)
>
> Hints : use the prime number theorem, replace z by sqrt(n), look for an
asymptotic expression of primorial

I wouldn't know how to go about that; but I am fairly familiar with phi.
Many thanks for your help - any further such would be very much appreciated.

Cheers,
Tom

  and what can be said about phi (wikipedia and/or mathworld should do).
>
> Maximilian
>
>
> --- In primenumbers@yahoogroups.com, "gulland68" <tmgulland@> wrote:
> >
> > Hi -- I am ust wondering why the last eight posts of mine to this group,
over a period of 2 years, have produced precisely zero replies. I know there's
no obligation on anyone to reply, and I know I have been naive here and there
but... has someone flicked a switch?
> >
> > Tom
> >
> > --- In primenumbers@yahoogroups.com, "gulland68" <tmgulland@> wrote:
> > >
> > > Let P(z) be the set of all the primes whose squares do not exceed an
integer z.
> > >
> > > Can anyone point me towards (or provide me with) a proof that pi(z)
converges to phi(prod(P(z)))*z/prod(P(z)) as z --> oo, where phi is the Euler
totient? Presumably it'll entail reference to the Prime Number Theorem; perhaps,
in a scholarly paper, one need simply put 'it follows by the Prime Number
Theorem that pi(z) converges to phi(prod(P(z)))*z/prod(P(z)) as z --> oo'...?
> > >
> > > With thanks,
> > >
> > > Tom
> > >
> >
>

Hi,

Until no, I have received no help on this topic from this group as well as the
openpfgw group.
I there really no one using openpfgw on NX enabled Linux kernels ?

I'm investigating the primality of number of the form 138^p-137^p. I have
checked all prime exponents p <= 179317 with openpfgw on a quad core Windows
server and found only composite numbers!

Now I want to go further (at least check all p <= 500000) but computation time
for each new exponent p is becoming quite long.

I have several dual and quad core Linux machines I would like to use to
parallelize and speed up my search but I can't use them until I find a version
of openpfgw that don't crash because of the NX flag :-(

In case there is really no operational version of openpfgw, what software could
I use on Linux to do the same job openpfgw does with approximately the same
performance ?

Best regards,

j_chrtn

--- On Tue, 3/10/09, j_chrtn <j_chrtn@...> wrote:
> Hi,
>
> Until no, I have received no help on this topic from this
> group as well as the openpfgw group.

That's odd, I remember at least 2 people giving you some assistance.

> I there really no one using openpfgw on NX enabled Linux kernels ?
...
> In case there is really no operational version of openpfgw,
> what software could I use on Linux to do the same job
> openpfgw does with approximately the same performance ?

I don't remember you reporting back on whether the versions from 2003 worked for
you. Do they?

Phil

--- In primenumbers@yahoogroups.com, Phil Carmody <thefatphil@...> wrote:
>
> That's odd, I remember at least 2 people giving you some assistance.
>
> I don't remember you reporting back on whether the versions from 2003 worked
for you. Do they?
>
> Phil
>

Hi Phil,

Your reply surprises me : what assistance are you talking about?
If some people here sent me answers, it seems that I never received them or else
I missed them I'm afraid :-(

I've tested the following versions:
- "OpenPFGW v1.2 - Official Release Feb 11, 2005 Linux binary Static linked
binary" downloaded from http://tech.groups.yahoo.com/group/primeform

- 20041011_Linux_Dev_Beta_pfgw from
http://tech.groups.yahoo.com/group/openpfgw/files

- 20041023_Linux_Dev_Beta_pfgw also from
http://tech.groups.yahoo.com/group/openpfgw/files

but all three versions produce a SIGSEGV if I try to do a PRP test or even if I
just do a pfgw -i

For a 2003 version, I fact, I don't know where I can get one. If you have it,
could you please send it to me? Thanks.


I had the same problem on Windows 2003 server SP2 and XP SP3 but here I found a
solution : I added an exception for pfgw in the "Data execution prevention" tab
of the "Performance/Advanced" tab of "My computer" properties panel.

On Linux with NX (or XD) flag, I could not find the same kind of escape. I tried
to play with the exec-shield kernel parameter and with the execstack command but
with no success.

j_chrtn

I just got another prime 18-tuplet:

51342365971531191697537333 + d,
d = 0, 4, 6, 10, 16, 18, 24, 28, 30, 34, 40, 46, 48, 54, 58, 60, 66, 70
(26 digits, Mar 11, 2009, Jaroslaw Wroblewski)

This is the largest known 18-tuplet and the largest known case of 18
Simultaneous Primes.

Jarek

Hi Phil,
 
Thank you for your binary.
 
Unfortunately, your version gives the same result as the ones I tested; that is
"segmentation fault" (SIGSEGV) on system with NX enabled.
On systems without the NX flag, all versions (including yours) work just fine.
It's a bit ennoying.
 
Could you please tell me on which Linux distribution and version you run your
pfgw and also send me the result of the command : cat /proc/cpuinfo
 
I really can't imagine that no one uses pfgw on recent NX enabled Linux systems.
So I believe there is a solution to the problem. Maybe the lastest version of
pfgw fixes the issue but unfortunately, I don't know which version it is and
even less where I can download it.
 
j_chrtn


--- On Tue, 3/10/09, Phil Carmody <thefatphil@...> wrote:

From: Phil Carmody <thefatphil@...>
Subject: Re: [PrimeNumbers] Re: OpenPFGW on Linux with NX CPU flag enabled
To: "j_chrtn" <j_chrtn@...>
Date: Tuesday, March 10, 2009, 4:26 PM

> Phil Carmody <thefatphil@...> wrote:
> >
> > That's odd, I remember at least 2 people giving you
> some assistance.
> >
> > I don't remember you reporting back on whether the
> versions from 2003 worked for you. Do they?
>
> Your reply surprises me : what assistance are you talking
> about?

In that case, someone else asked exactly the same question!

> If some people here sent me answers, it seems that I never
> received them or else I missed them I'm afraid :-(
>
> I've tested the following versions:
> - "OpenPFGW v1.2 - Official Release Feb 11, 2005 Linux
> binary Static linked binary" downloaded from
http://tech.groups.yahoo.com/group/primeform
>
> - 20041011_Linux_Dev_Beta_pfgw from
http://tech.groups.yahoo.com/group/openpfgw/files
>
> - 20041023_Linux_Dev_Beta_pfgw also from
http://tech.groups.yahoo.com/group/openpfgw/files
>
> but all three versions produce a SIGSEGV if I try to do a
> PRP test or even if I just do a pfgw -i
>
> For a 2003 version, I fact, I don't know where I can get
> one. If you have it, could you please send it to me?

Good luck finding the right system libraries, but this is the version I always
used to use, and trusted.

Phil








[Non-text portions of this message have been removed]

 
Hi Steven
 
Yes, I've got the same crash with statically linked binaries. I recompiled v1.2
source code with -g to have debugging information and now I know where the crash
occurs : it's in _ecpuidsupport function in ./packages/woltobj/cpuid.o
 
Here is the call stack :
 
(gdb) run -i
Starting program: /tmp/pfgw -i
Program received signal SIGSEGV, Segmentation fault.
0x082e7cd9 in _ecpuidsupport ()
(gdb) where
#0  0x082e7cd9 in _ecpuidsupport ()
#1  0x080d328a in guessCpuType () at cpuid.cxx:154
#2  0x080d4956 in _getCpuInfo () at gwcpuinit.inl:286
#3  0x0805245d in pfgw_main_init () at pfgw_main.cpp:1602
#4  0x0804bcc0 in pfgw_main (argc=2, argv=0xbffffaf4) at pfgw_main.cpp:575
#5  0x0804a8ba in main (argc=2, argv=0xbffffaf4) at newmain.cpp:94
 
 
And the CPU information from a cat /proc/cpuinfo :
 
processor       : 0
vendor_id       : GenuineIntel
cpu family      : 15
model           : 4
model name      : Intel(R) Pentium(R) D CPU 2.80GHz
stepping        : 8
cpu MHz         : 2833.920
cache size      : 1024 KB
fdiv_bug        : no
hlt_bug         : no
f00f_bug        : no
coma_bug        : no
fpu             : yes
fpu_exception   : yes
cpuid level     : 5
wp              : yes
flags           : fpu vme de pse tsc msr pae mce cx8 apic sep mtrr pge mca cmov
pat pse36 clflush dts acpi mmx fxsr sse sse2 ss nx pni ds_cpl
bogomips        : 5423.10
processor       : 1
vendor_id       : GenuineIntel
cpu family      : 15
model           : 4
model name      : Intel(R) Pentium(R) D CPU 2.80GHz
stepping        : 8
cpu MHz         : 2833.920
cache size      : 1024 KB
fdiv_bug        : no
hlt_bug         : no
f00f_bug        : no
coma_bug        : no
fpu             : yes
fpu_exception   : yes
cpuid level     : 5
wp              : yes
flags           : fpu vme de pse tsc msr pae mce cx8 apic sep mtrr pge mca cmov
pat pse36 clflush dts acpi mmx fxsr sse sse2 ss nx pni ds_cpl
bogomips        : 5570.56
 
 
Any idea ?
 
j_chrtn
 


--- On Wed, 3/11/09, Steven Harvey <harvey563@...> wrote:

From: Steven Harvey <harvey563@...>
Subject: Re: [PrimeNumbers] Re: OpenPFGW on Linux with NX CPU flag enabled
To: "j_chrtn" <j_chrtn@...>
Date: Wednesday, March 11, 2009, 12:38 PM







Have you tried a statically linked version?
 
Steven Harvey

harvey563@...

--- On Tue, 3/10/09, j_chrtn <j_chrtn@...> wrote:


From: j_chrtn <j_chrtn@...>
Subject: [PrimeNumbers] Re: OpenPFGW on Linux with NX CPU flag enabled
To: primenumbers@yahoogroups.com
Date: Tuesday, March 10, 2009, 10:43 AM






Hi,

Until no, I have received no help on this topic from this group as well as the
openpfgw group.
I there really no one using openpfgw on NX enabled Linux kernels ?

I'm investigating the primality of number of the form 138^p-137^p. I have
checked all prime exponents p <= 179317 with openpfgw on a quad core Windows
server and found only composite numbers!

Now I want to go further (at least check all p <= 500000) but computation time
for each new exponent p is becoming quite long.

I have several dual and quad core Linux machines I would like to use to
parallelize and speed up my search but I can't use them until I find a version
of openpfgw that don't crash because of the NX flag :-(

In case there is really no operational version of openpfgw, what software could
I use on Linux to do the same job openpfgw does with approximately the same
performance ?

Best regards,

j_chrtn




















[Non-text portions of this message have been removed]

Companion Primes.
If p is an odd positive prime, and q is a prime >p,

define q to be a companion to p

if and only if

there exist a divisor d1 of (q+p)
and a divisor d2 of (q-p)  such that

d1 + d2 = p.


Of the 753378  odd prime pairs among
the odd positive primes < 10000,

24921 are pairs of companions,
and
728457 are pairs of non-companions.

Kermit Rose

Hi Peter,
 
You know what ? I'm happy ! ;-)
 
I've tried to modify george woltman's elf objects with _TEXT32=code,alloc,load
as you propose and then the miracle happens : pfgw now works perfectly well; no
more SIGSEGV !
 
A great great thanks to you for your advice.
 
Best regards,
 
j_chrtn


--- On Wed, 3/11/09, Peter Kosinar <goober@...> wrote:

From: Peter Kosinar <goober@...>
Subject: Re: [PrimeNumbers] Re: OpenPFGW on Linux with NX CPU flag enabled
To: "j. charton" <j_chrtn@...>
Date: Wednesday, March 11, 2009, 10:09 PM

Hello,

Although I have failed to reproduce the problem you're describing on my
machine (which apparently does support the NX flag, but the kernel I'm using
is not enforcing it), I have an idea that might help.

Could you try the following command(s) on your machine?

cd packages/woltobj
for f in *.o; do objcopy --set-section-flags _TEXT32=code,alloc,load $f; done

(the for-command is in one line, but the mailer might have split it).
Naturally, try it on a copy of the source directory tree, if things go weird.
Afterwards, try to compile the program just as you did before and see if the
change helped.

If no, could you send me the compiled binary?

Peter





[Non-text portions of this message have been removed]

--- In primenumbers@yahoogroups.com,
Kermit Rose <kermit@...> wrote:

> If p is an odd positive prime, and q is a prime >p,
> define q to be a companion to p
> if and only if
> there exist a divisor d1 of (q+p)
> and a divisor d2 of (q-p) such that
> d1 + d2 = p.
> Of the 753378 odd prime pairs among
> the odd positive primes < 10000,
> 24921 are pairs of companions

Extending the upper limit, N, I obtained

[ N, [your pairs, all pairs], fraction]

[ 10000, [ 24921,   753378], 0.0330790]
[ 20000, [ 52164,  2554930], 0.0204170]
[ 30000, [ 80351,  5260146], 0.0152754]
[ 40000, [108829,  8826301], 0.0123301]
[ 50000, [137833, 13166146], 0.0104687]
[ 60000, [167470, 18334540], 0.0091341]
[ 70000, [196287, 24036711], 0.0081661]
[ 80000, [226348, 30697530], 0.0073734]
[ 90000, [255888, 37945116], 0.0067436]
[100000, [286267, 45988845], 0.0062247]

with a fraction falling off faster than 1/sqrt(N).

David

I discovered an iterative algorithm for finding the factors of a
positive integer.

Unfortunately, it has two serious flaws.

(1) It is usually less efficient than trial division by all positive
integers < sqrt of number to be factored.

(2) It sometimes finds the trivial factors, 1 and the number itself.

When this happens,  if you retry with a different set of initial values,
it might find the non trivial factors.

I wonder if there is anyway to improve the iteration process.

Here is my algorithm.

Suppose z is the positive integer to be factored.

The goal is to find A,B,C, such that

(A + C) * (A + B) = z.

I initially randomly choose 3 values between 1 and sqrt(z) for A,B,C.

Then

I set deltaA as large as I can and still have

( (A+deltaA) + C) * ( (A + delta A) + B)  not exceed z.

If it exactly equals z, then I have found my factors.

I redefine A = A + deltaA.

Then
I set deltaB as large as I can and still have

(A + C) * (A + (B + deltaB) ) not exceed z.

If it equals z, then I have found my factors.

I redefine B = B + deltaB.

Then I set deltaC as large as I can and still have

(A + (C + deltaC))*(A + B) not exceed z.

If it equals z, then I have found my factors.

I redefine C = C + deltaC.

I repeat until I find the factors.

Example for trivial case:

To factor 21 by iteration.

Suppose we
Randomly choose
A = 4, B = 1, C  = 4.

(A + deltaA + C) * (A + deltaA + B ) - 21 = 0

(4 + deltaA + 4) * (4 + deltaA + 1) - 21 = 0

(deltaA + 8) * (deltaA + 5) - 21 = 0

deltaA**2 + 13 deltaA + 40 - 21 = 0

deltaA**2 + 13 deltaA + 19 = 0

deltaA = -1

A -->  A + deltaA
A = 4 - 1 = 3
B and C are still the same;
B = 1, C = 4

(A + C)* (A+B+deltaB) = 21

(3+4)*(3+1+deltaB) = 21

7 * (deltaB + 4) = 21

deltaB + 4 = 3
deltaB = -1

B---> B + deltaB

B = 1 + -1 = 0

A and C stay the same,

A = 3, B = 0, C = 4

(A + C + deltaC)*(A+B) - 21 = 0

(3 + 4 + deltaC)*(3 + 0) - 21 = 0

(7 + deltaC)*3 - 21 = 0

7*3 + deltaC * 3 - 21 = 0

3 * deltaC = 0

Factors are (A+C, A+B) = (3+4,3+0) = (7,3)




Kermit Rose

--- In primenumbers@yahoogroups.com, Kermit Rose <kermit@...> wrote:
> I discovered an iterative algorithm for finding the factors of a
> positive integer. Unfortunately, it has two serious flaws.

nice start...

> (1) It is usually less efficient than trial division by
> all positive integers < sqrt of number to be factored.

Then I suggest to use rather that, at least it terminates.

> (2) It sometimes finds the trivial factors, 1 and the number itself.

and most often gets stuck after the first iteration.

> When this happens,  if you retry with a different set of initial
values,
> it might find the non trivial factors.

with almost 0 probability

> I wonder if there is anyway to improve the iteration process.

redesign from scratch

> I set deltaA as large as I can and still have
> ( (A+deltaA) + C) * ( (A + delta A) + B)  not exceed z.
> I redefine A = A + deltaA.

why not say: choose A as large as possible to have (A+B)(A+C) <= z ?

> Then
> I set deltaB as large as I can and still have
> (A + C) * (A + (B + deltaB) ) not exceed z.
> (...) B <- B+deltaB

i.e. B = [z/(A+C)]-A !

> Then I set deltaC as large as I can and still have
> (A + (C + deltaC))*(A + B) not exceed z.

i.e. C = [z/(A+B)]-A

> I repeat until I find the factors.

WHAT DOES THIS MEAN ???
ALL ARE AS LARGE AS POSSIBLE !!

> Example for trivial case:
> Suppose we
> Randomly choose
> A = 4, B = 1, C  = 4.
> (A + deltaA + C) * (A + deltaA + B ) - 21 = 0

you should write <= 0 since you said "not exceed"

> (4 + deltaA + 4) * (4 + deltaA + 1) - 21 = 0
> (deltaA + 8) * (deltaA + 5) - 21 = 0
> deltaA = -1

aha so 7*4 does not exceed 21.

Then of course everything is possible.

Hello Max.


2b. Re: Factor Iteration
     Posted by: "Maximilian Hasler" maximilian.hasler@... maximilian_hasler
     Date: Thu Mar 12, 2009 10:21 pm ((PDT))

--- In primenumbers@yahoogroups.com, Kermit Rose <kermit@...> wrote:

Max:  and most often gets stuck after the first iteration.


Yes.  I forgot to mention this flaw because I had programmed around it.

This getting stuck after the first iteration is the most serious flaw.



> > I wonder if there is anyway to improve the iteration process.
>

Max:   redesign from scratch


Ya.   I will have to.


> > I set deltaA as large as I can and still have
> > ( (A+deltaA) + C) * ( (A + delta A) + B)  not exceed z.
> > I redefine A = A + deltaA.
>

Max:   why not say: choose A as large as possible to have (A+B)(A+C) <= z ?


Yes.   This would have been much simpler.  I see now that it is equivalent.


> > Example for trivial ca
Max:   you should write <= 0 since you said "not exceed"


I tried to avoid using <= because it looks like an arrow to me.




Max:   aha so 7*4 does not exceed 21.

Then of course everything is possible.



Oops.

My original concept was to have the delta values always be positive, but in
implementation,
they sometimes came out negative.


Kermit