Jarek,

I might be interested in helping to set up a distributed system.  Since each
segment is independent, this shouldn't take anything fancy.  I am a professional
web programmer and could manage this as a web project in a database.  That
way people could easily request segments and get them assigned.

Is your program just source at this point?  Or do you have a stand alone exe?
Someone else might be able to put together an windows version.

I have also looked at PrimeGrid and this is probably the best way to distribute
the project.  Does anyone here have experience on that system?

Paul


   ----- Original Message -----
   From: jarek372000
   To: primenumbers@yahoogroups.com
   Sent: Saturday, May 17, 2008 5:45 AM
   Subject: [PrimeNumbers] Re: AP25


   I am not sure what was the exact CPU power used, as Raanan was
   distributing the program among his computers, and also the number and
   kind of computers he had access to, varied. However, I understand you
   are asking what order of magnitude of CPU takes to find such a result.

   I think we weren't either particularly lucky or unlucky. Data gathered
   during the search indicates that it was about time to expect the first
   AP25.

   The search was in a natural way divided into many independent
   segments, each taking about 3 minutes on Athlon 64, it would take some
   10 times more on a 32-bit computer, otherwise it depends on the
   particular processor. Probably no RAM is used, processor cache should
   be enough. I think Raanan went through less than 10,000,000 such
   segments before finding the AP25 - this would be somewhat lower if we
   targeted AP25 from the start since we would have scheduled the search
   differently.

   To hunt AP26, the program could be speed up by a factor of 2, at the
   cost of missing about one third of AP25's on the way of current
   search, but with no loss of AP26's. I think that one would have to go
   through something like 500,000,000 segments without finding AP26, to
   be eligible to complain on bad luck.

   So we are talkig about at least 1000 CPU years of 64-bit computers to
   honestly expect an AP26.

   Now, I am not a professional programmer. I have written a C-code which
   can be run on a single computer, and in consequence it can be run on a
   local network with a proper script. I have no ability to make a
   distibuted project on the network. If someone is interested in running
   a wide distributed search, I can provide the code. Program segments
   are numbered, so in principle you can distribute the search ranges by
   sending out to everyone a New Year postcard once a year with the
   assigned range for the next year. Or with a few hundred million of
   computers you can find AP26 in a few minutes (provided you can
   distribute the numbers of segments).

   Jarek

   --- In primenumbers@yahoogroups.com, Phil Carmody <thefatphil@...> wrote:
   >
   > --- On Sat, 5/17/08, jarek372000 <Jaroslaw.Wroblewski@...> wrote:
   > > This morning the first known AP25 has been discovered:
   > >
   > > 6171054912832631 + 366384*23#*n, for n=0 to 24
   > > (Raanan Chermoni & Jaroslaw Wroblewski, May 17 2008)
   > >
   > > My contribution was the search program, while Raanan
   > > provided the computer power.
   >
   > Wow! A marvelous find. I think a lot of people here have been
   crossing their fingers for you over the last few months as the AP24s
   have been coming in, I certainly have.
   >
   > What kind of CPU power was behind the task? Does your program
   sensibly distribute? I'd certainly be willing to stick a CPU or two on
   a distributed project that was pushing for an AP26. However, my fear
   is that the bandwidth/comms issues might be cause Amdahl's law to
   frown at such a task.
   >
   > Phil
   >





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

Jarek wrote:
> This morning the first known AP25 has been discovered:
>
> 6171054912832631 + 366384*23#*n, for n=0 to 24
> (Raanan Chermoni & Jaroslaw Wroblewski, May 17 2008)
>
> My contribution was the search program, while Raanan provided the
> computer power.

Huge congratulations!
This is a very impressive and well deserved feat.
http://hjem.get2net.dk/jka/math/aprecords.htm is updated.
As the only known AP25 it gets 5 records: Longest known AP, largest
known AP25, and smallest known difference, start and end for an AP25.
Getting 5 records for one AP may sound like a lot but it's much more
than 5 times as hard as most of the other records.

--
Jens Kruse Andersen

Each Sierpinski (and Riesel) number and the dual of it always have the
same covering set as it is easy to see.

So we only need to find a prime for a candidate in any one of the
forms to eliminate it from the other side.

Looking at the remaining candidates in SoB and in the dual Sierpinski
search of Payam Samidoost we see that there is no number belonging to
both lists, so no one of the 6 remaining candidates at SoB can be a
solution to the problem, and the same to Payam's as well.

In fact, the only two common candidates when Payam launched his
project were 19249 and 28433.

Payam found a prime for the dual of 19249 on August 17, 2002 so when
an anonymous member of TeamPrimeRib submitted a prime for 28433 on
December 30, 2004 to SoB, the search could have been called quits.

Lélio

Jens Kruse Andersen wrote:
> Huge congratulations!
> This is a very impressive and well deserved feat.
> http://hjem.get2net.dk/jka/math/aprecords.htm is updated.

I agree!  Decided to change my 2006 banner notes about the last
Mersenne to this AP on my "main" page primes.utm.edu/
and primes.utm.edu/largest.html

> Getting 5 records for one AP may sound like a lot but it's
> much more than 5 times as hard as most of the other records.

We know they exist with arbitrary length, but 25 seems a long
way from infinity doesn't it?

CC

Chris Caldwell wrote:

> I agree!  Decided to change my 2006 banner notes about the last
> Mersenne to this AP on my "main" page primes.utm.edu/
> and primes.utm.edu/largest.html

Nice. I see it's already in Prime Curios!:
http://primes.utm.edu/curios/page.php/6171054912832631.html
This can also be updated:
http://primes.utm.edu/glossary/page.php?sort=ArithmeticSequence

> We know they exist with arbitrary length, but 25 seems a long
> way from infinity doesn't it?

Yes it does. I made a similar "but" when Jarek's earlier record was
mentioned on Wikipedia's main page in 2007 with the text:
"Did you know...
...that existence of arbitrarily many primes in arithmetic progression was
proven in 2004, but it took 75 computers to find an example with 24 primes?"

It's archived at
http://en.wikipedia.org/w/index.php?title=Template:Did_you_know&oldid=138995374
(People are not supposed to know the "Did you know" facts)

I have updated http://en.wikipedia.org/wiki/Primes_in_arithmetic_progression

--
Jens Kruse Andersen

From: Lélio Ribeiro de Paula
> Each Sierpinski (and Riesel) number and the dual of it
> always have the
> same covering set as it is easy to see.
>
> So we only need to find a prime for a candidate in any one
> of the
> forms to eliminate it from the other side.
>
> Looking at the remaining candidates in SoB and in the dual
> Sierpinski
> search of Payam Samidoost we see that there is no number
> belonging to
> both lists, so no one of the 6 remaining candidates at SoB
> can be a
> solution to the problem, and the same to Payam's as
> well.
>
> In fact, the only two common candidates when Payam launched
> his
> project were 19249 and 28433.
>
> Payam found a prime for the dual of 19249 on August 17,
> 2002 so when
> an anonymous member of TeamPrimeRib submitted a prime for
> 28433 on
> December 30, 2004 to SoB, the search could have been called
> quits.

By Sierpinski's definitions, finding a prime in the dual set does not remove the
k value as a candidate to be what was later termed a Sierpinski number.

Phil

Phil Carmody wrote:
>
> By Sierpinski's definitions, finding a prime in the dual set does not remove
the k value as a candidate to be what was later termed a Sierpinski number.
>


Indeed, the prime of the form k+2^n might be in the covering set for the numbers
of the form k*2^n+1.

Your idea is very interesting.

My readings about near-repdigit numbers has made me fascinated by
Sierpinski and Riesel numbers and especially their covering sets.

If you look at the search of Payam Samidoost, can we prove any of the
  sixty-three or so possible candidates below the smallest provable
Riesel number 509203 are not Riesel numbers. I would be especially
interested in 2293 and 9221 because they are so small.

--- In primenumbers@yahoogroups.com, Lélio Ribeiro de Paula
<lelio73@...> wrote:
>
> Each Sierpinski (and Riesel) number and the dual of it always have the
> same covering set as it is easy to see.
>
> So we only need to find a prime for a candidate in any one of the
> forms to eliminate it from the other side.
>
> Looking at the remaining candidates in SoB and in the dual Sierpinski
> search of Payam Samidoost we see that there is no number belonging to
> both lists, so no one of the 6 remaining candidates at SoB can be a
> solution to the problem, and the same to Payam's as well.
>
> In fact, the only two common candidates when Payam launched his
> project were 19249 and 28433.
>
> Payam found a prime for the dual of 19249 on August 17, 2002 so when
> an anonymous member of TeamPrimeRib submitted a prime for 28433 on
> December 30, 2004 to SoB, the search could have been called quits.
>
> Lélio
>

Jack Brennen wrote:
>
> Indeed, the prime of the form k+2^n might be in the covering set for
> the numbers of the form k*2^n+1.
>

No, it cannot.

The covering sets of all known Riesel and Sierpinski numbers are
exactly the same as that of their duals, as can be easily seen.

So if a prime for a dual of a Sierpinski number were to be in the
covering set of that particular Sierpinski number, it should also be
in the dual covering set as well, which contradicts the definition of
covering sets.

Lélio

--- On Mon, 5/19/08, Lélio Ribeiro de Paula <lelio73@...> wrote:
> Jack Brennen wrote:
> > Indeed, the prime of the form k+2^n might be in the
> covering set for
> > the numbers of the form k*2^n+1.
>
> No, it cannot.
>
> The covering sets of all known Riesel and Sierpinski
> numbers are
> exactly the same as that of their duals, as can be easily
> seen.
>
> So if a prime for a dual of a Sierpinski number were to be
> in the
> covering set of that particular Sierpinski number, it
> should also be
> in the dual covering set as well, which contradicts the
> definition of
> covering sets.

Now reread what Jack wrote (which was also going be in my original post too, but
thinking that it was a bit obvious I removed it for brevity), and think a bit
more.

Phil

Phil Carmody wrote:
>
> Now reread what Jack wrote (which was also going be in my original post too,
but thinking that it was a bit obvious I removed it for brevity), and think a
bit more.
>


I tried (in private email) giving the example of the dual sequences:

    10^n-7
    7*10^n-1

They both have the same covering set (it can be found in under a
minute with just a little thought).  One of the sequences has a
very easy to find prime; the other one is easily proven to have no
primes.

I have prepared the first version of the description of the algorithm
I have used for the AP24 search (and which was used on Raanan's
computers for the AP25 search). This can be also applied for an AP26
search.

If you are interested in trying to learn more details, please go to

http://www.math.uni.wroc.pl/~jwr/AP26/

I must confess, that my (lack of) programming skills makes it
impossible to further utilize the algorithm. Therefore I would like a
group of relatively good programmers interested in the subject to
learn what are the tricks I have applied, and to take over all the
programming stuff.

Jarek

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

Formulas for producing chains of primes may be derived
from known examples. If you have a quadratic equation
that already has a notable chain , such as  10x^2 - 20x + 29,
then chances are good that it will be a fairly simple matter
to develop several more equations that will crank out chains
approximately the same length.

The process is straightfoward.

The first two terms of 10x^2 - 20x + 29 are 19 and 29. Choose
a small square and multiply each term by it, then find the
difference.
29 * 2^2 - 19 * 2^2 = 40. This distance between the terms must
be allocated in such a way as to allow another term to be inserted
that is consistant with the constant increment characteristic of the
original equation.We will have now three terms instead of two:
19 * 2^2 ,19 * 2^2 + 10, 19 * 2^2 +30... which defines a new related
quadratic that could in this instance also be expressed as
10x^2 - 20x + 86. Some terms of this new equation will have
factors from the small square that was used to derive it, and these
are easily removed, leaving a new chain of primes.

To summarize the calculations used in deriving the first five new
equations from 10x^2 - 20x + 29 :

29 * 2^2 - 19 * 2^2 = 40. = 10 + 30 (for a one term insertion).
The new sequence is 76, 86, 116....
29 * 3^2 - 19 * 3^2 = 90. = 10 + 30 + 50 (for a two term insertion).
The new sequence is 171, 181, 211....
29 * 4^2 - 19 * 4^2 = 160. = 10 + 30 + 50 + 70 (for a three term
insertion).
The new sequence is 304, 314, 344....
29 * 5^2 - 19 * 5^2 = 250. = 10 + 30 + 50 + 70 + 90
(for a four term insertion). The new sequence is 475, 485, 515....
29 * 6^2 - 19 * 6^2 = 360. = 10 + 30 + 50 + 70 + 90 + 110
(for a five term insertion). The new sequence is 684, 694,724....

  Listed here is the original prime list and its first five
derivations:

original : 19 29  59  109  179  269  379  509  659  829  1019  1229
1459  1709  1979  2269  2579  2909 3529
derived:
2^2 : 19 43 29  83  59  163  109  283  179  443  269  643
379  883  509  1163  659  1483  829
.
3^2 : 19 181 211 29  331  421  59  661  811  109  1171  1381
179  1861  2131  269  2731  3061  379

4^2 : 19 157 43  197  29  277  83  397  59  557  163  757
109  997  283  1277  179  1597  443

5^2 : 19  97 103  113  127  29  167  193  223  257  59  337
383  433  487  109  607  673  743

6^2 : 19 347  181  43  211  467  29  587  331  83  421  947
59  1187  661  163  811  1787 109

Longer lists of primes are generated by the well-known
equations x^2 - x + 41, and x^2 - 79x + 1601. The
same basic procedure is used on these equations
as in the first example to develop new prime chains,
but they are a little more complex to figure out. The
derived sequence  may not actually start on the
first term, but on a prior or later one, and the constant
increment of the originals that we are working with is
now 2, not 20.

With x^2 - x + 41  the basic calculations are:
43 * 2^2 - 41* 2^2 = 8. = 3 + 5 (for a one term insertion).
The new sequence is 164, 167, 172..... However, a prior
term 163 exists and really should be on the list so the
sequence adjusts to 163,164, 167.. :

2^2 : 163   41  167  43  179  47 199  53  227  61
263 71  307  83   359  97 419 113 487 131
563 151 647 173 739 197 839 223 947 251
1063 281 1187 313 1319 347 1459 383 1607 421

With x^2 - 79x + 1601  the basic calculations are:
1623 * 2^2 - 1523* 2^2 =  -312. = -157 - 155 (for a one term
insertion).
The new sequence is 6404, 6247, 6092..... However, the chain
does not really start until the 41st  term so the sequence adjusts
to 1684, 1607, 1532.. :

2^2 : 421 1607 383 1459 347 1319 313 1187 281 1063
251 947  223  839  197  739  173  647 151  563
131 487 113  419   97  35 9 83  307  71  263
61  227  53   199  47  179  43   167  41  163
41 167  43  179  47  179  53   227   61   263
71  307 83  359  97  419  113   487 131 563
151 647 173   739 197 839  223  947 251 1063
281 1187  313  1319 347 1459 383  1607 421

Aldrich Stevens

I have found the following AP19
2027663243 + 1002606*19#*n, n=0..18
which, I believe, improves 24 years old "minimal known start AP19" record.

The total CPU time I have used for the search was about 3-4 hours of a
64-bit Athlon 3000.

The details of the search as well as homework exercises for those who
would like to learn more about the algoritm used, will be uploaded
shortly into the directory
http://www.math.uni.wroc.pl/~jwr/AP26/
as AP26v3.pdf (AP19 search is described in the last section).

Jarek

Jarek wrote:
> I have found the following AP19
> 2027663243 + 1002606*19#*n, n=0..18
> which, I believe, improves 24 years old "minimal known start AP19" record.

Congrats. The record table at
http://hjem.get2net.dk/jka/math/aprecords.htm#minimalstart is updated.
The old record was
8297644387 + 431*19#*n, n=0..18, 1984, Paul Pritchard
The ending prime had 11 digits there and 15 in the new record.

--
Jens Kruse Andersen

In http://www.upforthecount.com/math/mandrill.html, Walter Nissen has
defined a primitive friendly pair (pfp):
1. Abundancy is defined as the ratio of the multiplicative sum-of-
divisors function to the integer itself.
2. Integers m and n are friendly iff they have the same abundancy.
3. Friends m and n are primitive friendly iff they have no common
prime factor of the same multiplicity.

In http://ftp.cwi.nl/CWIreports/MAS/MAS-R0307.pdf, some pfps with a
deficient abundancy are listed (dpfp).
They allow to derive new amicable pairs from existing ones.

Other pfps can be found in
http://wwwhomes.uni-bielefeld.de/achim/mpn.html also.
They allow to derive new multiperfect numbers from existing ones.

The lowest value for a pfp/dpfp known to us is:
384/361 19^2*127 19^4*151*911

Is it possible to find a pfp/dpfp with a lower abundancy?

Michel

(1) [ Definition ]

A positive integer p is prime
if and only if

a > 0  divides p implies a = 1  or a = p.


(2)

A positive integer p is prime
if and only if

p divides  the product of integers b c
implies  p divides b  or  p divides c.


(3) A positive integer p is prime
if and only if

there is exactly way in which p is the difference of two square integers.

(4)  A positive integer p is prime
if and only if

b**n = 1 mod p  implies  n divides (p-1)


(5)  A positive integer p is prime
if and only if
p = b + c implies b and c are relatively prime.



Kermit  Rose


kermit@...

I actually didn't find "bad primes" defined in any number theory way...
they were associated with a Calabi-Yau based L-function in
a paper with no real good definition.
So I did a web search and an OEIS search and some experimentation.
A little thought of the geometric average basis of the defintion gave
me an higher order "bad" I call "worse".
An a good I call "Angel".


%I A000001
%S A000001 19, 43, 47, 61, 73, 79, 83, 89, 109, 113, 137, 139, 157,
167, 181, 197, 199,
211, 229, 233, 239, 241, 271, 281, 283, 293, 313, 317, 353, 359, 383, 389,
401, 439, 443, 449, 463, 467, 503, 509, 521, 523
%N A000001 Worse primes<-bad primes (A130903):
If[Prime[n]^2 - Prime[n + 2]*Prime[n - 2] < 0,
            Prime[n]]
%F A000001 a(n) = If[Prime[n]^2 - Prime[n + 2]*Prime[n - 2] < 0,
            Prime[n]]
%t A000001 Flatten[Table[If[Prime[n]^2 - Prime[n + 2]*Prime[n - 2]
< 0,Prime[n], {}], {n, 3, 100}]]
%Y A000001 Cf. A130903
%O A000001 1
%K A000001 ,nonn,
%A A000001 Roger Bagula and Gary W. Adamsom (rlbagulatftn@...),
May 23 2008
RH
RA 192.20.225.32
RU
RI


%I A000001
%S A000001 5, 7, 11, 13, 17, 23, 29, 31, 37, 41, 53, 59, 67, 71, 97,
101, 103, 107,
127, 131, 149, 151, 163, 173, 179, 191, 193, 223, 227, 251, 257, 263, 269,
277, 307, 311, 331, 337, 347, 349, 367, 373, 379, 397, 409, 419, 421, 431,
433, 457, 461, 479, 487, 491, 499, 541
%N A000001 Angel primes<-good primes (A028388):
If[Prime[n]^2 - Prime[n + 2]*Prime[n - 2] > 0,
            Prime[n]]
%C A000001 This sequence set is a next level higher of good and bad
primes.
The bad primes come from modular forms and L-Series theory.
http://en.wikipedia.org/wiki/Arithmetic_of_elliptic_curves:
Quote:
"Reduction mod p
Reduction of an abelian variety A modulo a prime ideal of (the
integers of) K - say, a prime number p - to get an abelian variety Ap
over a finite field, is possible for almost all p. The 'bad' primes,
for which the reduction degenerates by acquiring singular points, are
known to reveal very interesting information. As often happens in
number theory, the 'bad' primes play a rather active role in the theory.
Here a refined theory of (in effect) a right adjoint to reduction mod
p - the Néron model - cannot always be avoided. In the case of an
elliptic curve there is an algorithm of John Tate describing it.
For abelian varieties such as Ap, there is a definition of local
zeta-function available. To get an L-function for A itself, one takes
a suitable Euler product of such local functions; to understand the
finite number of factors for the 'bad' primes one has to refer to the
Tate module of A, which is (dual to) the étale cohomology group
H1(A), and the Galois group action on it. In this way one gets a
respectable definition of Hasse-Weil L-function for A. In general its
properties, such as functional equation, are still conjectural - the
Taniyama-Shimura conjecture was just a special case, so that's hardly
surprising.
It is in terms of this L-function that the conjecture of Birch and
Swinnerton-Dyer is posed. It is just one particularly interesting
aspect of the general theory about values of L-functions L(s) at
integer values of s, and there is much empirical evidence supporting it."
%D A000001 http://en.wikipedia.org/wiki/Arithmetic_of_elliptic_curves
%F A000001 a(n) = If[Prime[n]^2 - Prime[n + 2]*Prime[n - 2] <>0,
            Prime[n]]
%t A000001 Flatten[Table[If[Prime[n]^2 - Prime[n + 2]*Prime[n - 2]
> 0,Prime[n], {}], {n, 3, 100}]]
%Y A000001 Cf. A130903,A028388
%O A000001 1
%K A000001 ,nonn,
%A A000001 Roger Bagula and Gary W. Adamsom (rlbagulatftn@...),
May 23 2008
RH
RA 192.20.225.32
RU
RI


--
Respectfully, Roger L. Bagula
  11759Waterhill Road, Lakeside,Ca 92040-2905,tel: 619-5610814
:http://www.geocities.com/rlbagulatftn/Index.html
alternative email: rlbagula@...

The dpfp with the highest abundancy that I have found is

5^3*7^3*11*13^2*17*43*61^2*67*71*79*97*461^3*1279*106261
5^4*7^4*13*19*23*31*41*43^2*61*71^2*79^2*149*151*229*461^4*467*631*1279
^2*2383*2557*2801*5113*23971

that has abundancy 60444556520325120/30222279399705313
(approx 1.999999925)

Can we find others with a higher abundancy ?

A single dpfp is known with the same primes for both members
3^5*11*13^2 3^4*11^3*13
abundancy 6832/3861

Is it possible to find other deficient pairs with the same property ?
Is it possible to find abundant pairs with the same property ?

Some dpfps have the property that both members start with the same
prime.
Here are same instances:

2^3*19^2*127
2*7^2*13*19^3*181
abundancy 720/361

3^3*5
3^2*7*13
abundancy 16/9

7^5*11*19*43
7^8*13*19^3*37*181*1063
abundancy 23040/16807

11^3*13*43*61^2*79*97
11^2*19*31*43^2*61*79^2*631
abundancy 31610880/25073257

13^2*31^3*37*61*83
13*23*31^2*83^2*331*367
abundancy 1225728/1036919

17*19*37*43^2*79*307*631
17^2*23*37^2*43^3*67*307^2*367*733
abundancy 425779200/357047447

19^2*127
19^4*151*911
abundancy 384/361

23^2*37*79*137
23*37^3*73*137^2
abundancy 127680/116587

and last, only one found for 31
31*43*53*73^2*83*103*137^2*151^2*307*331*367*461^4*547*617*1093^2*1801
*23971*398581
31^2*43^2*67*73*79*137*151*163*307^2*331^2*367^2*433*461^3*547^2*613*6
31*733*1093*2617*3463*5233*106261
abundancy
5052878450522398695315178782720/4397372153897210688772404368287

Is it possible to find dpfps where both members start with a higher
prime like 37, 41, ... ?

Well, got the official word from D.B.:
linear transforms are of no value and
egregious factors are annoying. But
then again he says that about
almost everything. I really thought
I was on to something with

X^2 - 80X + 1763  --- 79 consecutive primes
X^2 - 237X + 14409  --- 76 consecutive primes

but apparently, no.

I still want to try to beat the limit though,
and use transforms
to find a way to make longer prime chains.

Can anyone factor

     X^4 - 97X^3 + 3294X^2 -45458X + 213589
     (49 consecutive primes)

into quadratics?

Now that is an equation that really looks like
fun to me, and a factorization would be a good
place to start.

Aldrich Stevens

--- In primenumbers@yahoogroups.com, "aldrich617" <aldrich617@...> wrote:
>
> Well, got the official word from D.B.:
> linear transforms are of no value and
> egregious factors are annoying. But
> then again he says that about
> almost everything. I really thought
> I was on to something with
>
> X^2 - 80X + 1763  --- 79 consecutive primes
> X^2 - 237X + 14409  --- 76 consecutive primes
>

When "X" is odd the expressions are even. They are not equal to 2, so
they are then composite.

> but apparently, no.
>
> I still want to try to beat the limit though,
> and use transforms
> to find a way to make longer prime chains.
>
> Can anyone factor
>
>     X^4 - 97X^3 + 3294X^2 -45458X + 213589
>     (49 consecutive primes)
>
> into quadratics?

? factor(x^4-97*x^3+3294*x^2-45458*x+213589)
[x^4 - 97*x^3 + 3294*x^2 - 45458*x + 213589 1]

Of course this is factored into linear factors over the complex
numbers ;-)

>
> Now that is an equation that really looks like
> fun to me, and a factorization would be a good
> place to start.
>

Please see:
http://www.maa.org/editorial/mathgames/mathgames_07_17_06.html

As far as I can tell, if only monic polynomials[1] with unique
positive values were permissible, then Euler would have won the
quadratic race.[2]

Paul

[1] http://mathworld.wolfram.com/MonicPolynomial.html
[2] http://mathworld.wolfram.com/Prime-GeneratingPolynomial.html

> Aldrich Stevens
>

Here are a few things I have done about 2nd kind Cunningham Chains.

1. I have determined that the known solution
3203000719597029781
is the only start of a 2nd kind CC16 below
67280843307747189120=6.278*10^19.
The search required about 70 hours CPU (3 hours real time using many
computers).

2. By a selective search I have found the following new solutions:

CC14, 2nd kind: 20299226100350079536373721 (26 digits)

CC15, 2nd kind: 13029362719535767707961 (23 digits)
CC15, 2nd kind: 11270578849664703769201 (23 digits)
CC15, 2nd kind: 7339063830988412866201 (22 digits)

CC16, 2nd kind: 15745040405007366603151 (23 digits)
CC16, 2nd kind: 2447621166480001151641 (22 digits)

Jarek

1a. Quadratic Prime Chains
     Posted by: "aldrich617" aldrich617@... aldrich617
     Date: Sun May 25, 2008 6:28 am ((PDT))



Can anyone factor

     X4 - 97X3 + 3294X2 -45458X + 213589
     (49 consecutive primes)

into quadratics?




A basic theorem of algebra implies that
a quartic polynomial that takes on more than eight prime values is
irreducible.


The basic reason is that if
  X4 - 97X3 + 3294X2 -45458X + 213589


could be factored into quadratics then whenever the product is prime,

one of the factors must be plus or minus that prime,
and the other factor must be plus or minus one.

Jarek wrote:
> 3203000719597029781
> is the only start of a 2nd kind CC16 below
> 67280843307747189120=6.278*10^19.

> CC14, 2nd kind: 20299226100350079536373721 (26 digits)
> CC15, 2nd kind: 13029362719535767707961 (23 digits)
> CC16, 2nd kind: 15745040405007366603151 (23 digits)

Congratulations! If you use just some of the cpu power from
AP hunting on Cunningham chains then I guess you could
dominate the record tables.

http://hjem.get2net.dk/jka/math/Cunningham_Chain_records.htm
will be updated when Dirk Augustin has updated the source.

--
Jens Kruse Andersen

--- In primenumbers@yahoogroups.com, "Jens Kruse Andersen"
<jens.k.a@...> wrote:

> Congratulations! If you use just some of the cpu power from
> AP hunting on Cunningham chains then I guess you could
> dominate the record tables.

Thanks. After about one day of running my program on 27 computers at
Mathematical Institute of Wroclaw University, I think I know what it
is capable of.

CC13 record of 39 digits is safe, as far as my methods are concerned.
My sieving is fairly efficient, but very shallow (it ends at 3-digit
primes and there is no way to extend it further). It would be insane
to try for a record well over 30 digits.

Likewise AP17 and AP18 records, you can take CC14 back, Jens :-) If
you add a few digits to it, I am unlikely to think of trying to
recapture it :-)

I feel pretty comfortable with longer chains. My feeling is that CC17
is just a matter of time. The real challenge is CC18, but I think that
with CPU power comparable to used in AP25 search it is worth trying.

Jarek

I have just found a new high:

CC15, 2nd kind: 105998574608115401372161 (24 digits)

Jarek

I got it!

CC17, 1st kind: 2759832934171386593519 (22 digits)

This is the only one CC16+ 1st kind I have found so far.
I have found 4 new CC16 2nd kind and rediscovered the old one.

Current records divided by 10^20 are as follows:

CC15 1st: 119.93 (Alm/Andersen 2004)
CC15 2nd: 1059.98 (JW 2008)

CC16 1st: 55.19 (part of CC17)
CC16 2nd: 157.45 (JW 2008)

CC17 1st: 27.59 (JW 2008)

16 Simultaneous Primes Record is
Tuplet (Andersen 2004) and it would
require CC16: 800.96 to beat.

17 Simultaneous Primes Record is
Tuplet (Waldvogel/Leikauf 2000) and it would
require CC17: 111.15 to beat.

Jarek