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Factorization of 6 consecutive integers > 10^1100
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 Let
m=67440294559676054016000
y=(m*(10^71+145589)+22)^2
N=(y11^4)*(y35^2)*(y47^2)*(y94^2)*(y146^2)*(y148^2)/m5
then N+k is factorized for k in [0,5].
Proof:
http://physics.open.ac.uk/~dbroadhu/cert/ifac6.zip
Software used:
GMPECM
Msieve
OpenPFGW
PariGP
Primo
Neighbouring composites:
c1094=(N1)/(5*1948359547)
c1090=(N+6)/(2*11969*4471247659)
subjected to 300 ECM curves at B1=100000
David Broadhurst
14 August 2007  David Broadhurst wrote:
> m=67440294559676054016000
Congratulations!
> y=(m*(10^71+145589)+22)^2
> N=(y11^4)*(y35^2)*(y47^2)*(y94^2)*(y146^2)*(y148^2)/m5
>
> then N+k is factorized for k in [0,5].
Very neat. I was wondering whether you would go for 6 record
numbers with a construction inspired by Jaroslaw's P(X).
Thanks for showing such an interest in my record page.
http://hjem.get2net.dk/jka/math/consecutive_factorizations.htm is updated.
I assume you don't object to 94digit factorizations being
"relatively easy".

Jens Kruse Andersen   In primeform@yahoogroups.com, "Jens Kruse Andersen" <jens.k.a@...>
wrote:
> I assume you don't object to 94digit factorizations being
No objection at all: I took a lazy route for k=6,
> "relatively easy".
staying well within the comfort zone of Msieve.
Update on the wings:
N = 6 mod 460041067299095322777708682147
with p35 still running.
David  Hello !
I have a collection about 210 6k twins.
for n+2 and n2 I have scan all primfactors up to
1.6E9,
without success.
Perhaps find one of you the 4th candidate. Any ideas
or interests ?
regards
Norman
Wissenswertes für Bastler und Hobby Handwerker. BE A BETTER HEIMWERKER! www.yahoo.de/clever   In primeform@yahoogroups.com, "N.L." <nluhn@...> wrote:
> I have a collection about 210 6k twins.
If you run ECM up to p20 level you have a good chance:
> for n+2 and n2 I have scan all primfactors up to
> 1.6E9, without success.
mu = 2*210*exp(Euler)*20/6200 = 2.4
and then you may well end up with a long haul,
usin ECPP at 6.2k digits, for a /genuine/ record.
[Note that, uncharacteristically, Jens allowed himself
to cheat, by not waiting for Primo at 4178 digits.]
David  David Broadhurst wrote:
>  In primeform@yahoogroups.com, "N.L." <nluhn@...> wrote:
Around half would find no factor from 1.6E9 to p20 level ECM,
>
>> I have a collection about 210 6k twins.
>> for n+2 and n2 I have scan all primfactors up to
>> 1.6E9, without success.
>
> If you run ECM up to p20 level you have a good chance:
>
> mu = 2*210*exp(Euler)*20/6200 = 2.4
so that half doesn't contribute to the chance.
And p20 ECM on 420 6k numbers is not easy:
420*74 curves taking 5 GHz minutes on one of my cpu cores
gives 108 GHz days.
With one (dual core) computer and lots of projects, I would
only give it p15 ECM if I got the twins.
> and then you may well end up with a long haul,
I don't know if anybody would do 6k ECPP for my novel record page.
> usin ECPP at 6.2k digits, for a /genuine/ record.
> [Note that, uncharacteristically, Jens allowed himself
> to cheat, by not waiting for Primo at 4178 digits.]
I wouldn't do 4178 digits, so it's up for grabs for a record share:
(240819405*2^13879+3)/(3*13*43*358877)
I may have cheated a little by allowing a prp factor but I also list the
proven 4number record which I also have.
By the way, 6k with Primo would make the Primo top20:
http://www.ellipsa.net/public/primo/top20.html
There are no certifications since 2005.
The Prime Pages ECPP top20 at http://primes.utm.edu/top20/page.php?id=27
has a Primo submission from July 12 2007:
http://primes.utm.edu/primes/page.php?id=81647
It's 17443#/22^17443 with 7508 digits by Markus Hiltbrunner
who has no other primes. It would be third on the Primo top20.
Is there a published certificate? It appears Marcel requires it but not
Chris.

Jens Kruse Andersen  Hello !
That is the problem !
A ECM test for a single number ( I have 420!) takes to
many time. Better is a Pollard Test for 6k numbers but
I don't have a programm that handle this type.
I can test factors up to 10^11..if I have luck,okay ,I
don't have luck >pitch!

Norman
 Jens Kruse Andersen <jens.k.a@...> schrieb:
> David Broadhurst wrote:
__________________________________
> >  In primeform@yahoogroups.com, "N.L."
> <nluhn@...> wrote:
> >
> >> I have a collection about 210 6k twins.
> >> for n+2 and n2 I have scan all primfactors up to
> >> 1.6E9, without success.
> >
> > If you run ECM up to p20 level you have a good
> chance:
> >
> > mu = 2*210*exp(Euler)*20/6200 = 2.4
>
> Around half would find no factor from 1.6E9 to p20
> level ECM,
> so that half doesn't contribute to the chance.
> And p20 ECM on 420 6k numbers is not easy:
> 420*74 curves taking 5 GHz minutes on one of my cpu
> cores
> gives 108 GHz days.
> With one (dual core) computer and lots of projects,
> I would
> only give it p15 ECM if I got the twins.
>
> > and then you may well end up with a long haul,
> > usin ECPP at 6.2k digits, for a /genuine/ record.
> > [Note that, uncharacteristically, Jens allowed
> himself
> > to cheat, by not waiting for Primo at 4178
> digits.]
>
> I don't know if anybody would do 6k ECPP for my
> novel record page.
> I wouldn't do 4178 digits, so it's up for grabs for
> a record share:
> (240819405*2^13879+3)/(3*13*43*358877)
> I may have cheated a little by allowing a prp factor
> but I also list the
> proven 4number record which I also have.
>
> By the way, 6k with Primo would make the Primo
> top20:
> http://www.ellipsa.net/public/primo/top20.html
> There are no certifications since 2005.
> The Prime Pages ECPP top20 at
> http://primes.utm.edu/top20/page.php?id=27
> has a Primo submission from July 12 2007:
> http://primes.utm.edu/primes/page.php?id=81647
> It's 17443#/22^17443 with 7508 digits by Markus
> Hiltbrunner
> who has no other primes. It would be third on the
> Primo top20.
> Is there a published certificate? It appears Marcel
> requires it but not
> Chris.
>
> 
> Jens Kruse Andersen
>
>
Yahoo! Clever  Der einfachste Weg, Fragen zu stellen und Wissenswertes mit Anderen zu teilen. www.yahoo.de/clever  Norman wrote:
> A ECM test for a single number ( I have 420!) takes to
I don't know why GMPECM stopped giving recommended p15 parameters.
> many time. Better is a Pollard Test for 6k numbers but
> I don't have a programm that handle this type.
> I can test factors up to 10^11..if I have luck,okay ,I
> don't have luck >pitch!
Here is some of the documentation for a former version:

This is the README file of gmpecm5.0, a new version of gmpecm,
replacing version 4c.
The ECM method is a probabilistic method, and can be viewed in some sense
as a generalization of the P1 and P+1 method, where we only require that
P+t is smooth, with t random of order P^(1/2). The optimal B1 and B2 bounds
have to be chosen according to the (usually unknown) size of P. The
following
table gives a set of neartooptimal B1 and B2 pairs, with the corresponding
expected number of curves to find a factor of given size (this table does
not
take into account the "extra factors" found by BrentSuyama's extension, see
below).
digits D optimal B1 B2 expected curves N(B1,B2,D)
15 2e3 1.2e5 30
20 11e3 1.4e6 90
25 5e4 1.2e7 240
30 25e4 1.1e8 500
35 1e6 8.4e8 1100
40 3e6 4.0e9 2900
45 11e6 2.6e10 5500
50 43e6 1.8e11 9000
55 11e7 6.8e11 22000
60 26e7 2.3e12 52000
65 85e7 1.3e13 83000
70 29e8 7.2e13 120000
Table 1: optimal B1 and expected number of curves to find a
factor of D digits.
Important note: the expected number of curves is significantly smaller
than the "classical" one we get with B2=100*B1. This is due to the
fact that this new version of gmpecm uses a default B2 which is much
larger than 100*B1 (for large B1), thanks to the improvements in step 2.
After performing the expected number of curves from Table 1, the
probability that a number of D digits was missed is exp(1), i.e.
about 37%. After twice the expected number of curves, it is exp(2),
i.e. about 14%, and so on.
Example: after performing 9000 curves with B1=43e6 and B2=1.8e11,
the probability to miss a 50digit factor is about 37%.
In summary, we advise the following method:
0  choose a target factor size of D digits
1  choose "optimal" B1 and B2 values to find factors of D digits
2  run once P1 with those B1 and B2
3  run 3 times P+1 with those B1 and B2
4  run N(B1,B2,D) times ECM with those B1 and B2, where N(B1,B2,D) is the
expected number of ECM curves with step 1 bound B1, step 2 bound B2,
to find a factor of D digits (cf above table)
5  if no factor is found, either increase D and go to 0, or use another
factorization method (MPQS, GNFS)

Some things have changed in later versions. I'm not sure how
they affect p15 work.
This is a guess at reasonable p15 parameters with GMPECM 6.1.2:
P1 with B1=20000 and default B2: ecm pm1 20000
3 times P+1 with B1=10000 and default B2: ecm c 3 pp1 10000
26 curves with B1=2000 and default B2: ecm c 26 2000
This may take around 2 GHz weeks for 420 6k numbers.
Suppose you have saved far more primes n1 where n is trivially
factored and n+1 was composite.
Then n1, n, 2n2, 2n, 3n3,3n are factored. You could then
try trial factoring and prp (no ecm) on
n3, n2, 2n3, 2n1, 2n+1, 3n2, 3n1.
If one of these gives a prp cofactor then you have 3 of 4 and
could do p15 ecm on the last (the last two if 2n1 split).
If successful, you would probably have two 6k prp's that
require ECPP, but I accept them as prp records.

Jens Kruse Andersen   In primeform@yahoogroups.com, "Jens Kruse Andersen" <jens.k.a@...>
wrote:
> http://www.ellipsa.net/public/primo/top20.html
Marcel missed this one, from July 2007:
> There are no certifications since 2005.
http://primes.utm.edu/primes/page.php?id=81647
David   In primeform@yahoogroups.com, "Jens Kruse Andersen" <jens.k.a@...>
wrote:
> 420*74 curves taking 5 GHz minutes on one of my cpu cores
That sounds like less than a week's work, for Sean Irvine.
> gives 108 GHz days.
Best advice to Norman: share your finds with Sean!
David   In primeform@yahoogroups.com, "N.L." <nluhn@...> wrote:
> A ECM test for a single number ( I have 420!) takes to
I see that p15 was enough:
> many time.
http://hjem.get2net.dk/jka/math/consecutive_factorizations.htm> n+3 = 2^3*943127*5020192965913*prp6203
Congrats to Norman and Christophe.
David  David Broadhurst wrote:
> I see that p15 was enough:
Yes, congratulations.
>
> http://hjem.get2net.dk/jka/math/consecutive_factorizations.htm
>> n+3 = 2^3*943127*5020192965913*prp6203
>
> Congrats to Norman and Christophe.
They found two cases with p12 and p13 as penultimate factors.
There is no plan to prove prp6203.
After 3 weeks with the page, my 4 records are already beaten, and 3 further
lengths have been found. I'm surprised by so many great results so quickly.
Thanks to everybody. I expect things to cool down now but what do I know?

Jens Kruse Andersen   In primeform@yahoogroups.com, "Jens Kruse Andersen" <jens.k.a@...>
wrote:
> I'm surprised by so many great results so quickly.
That's easy to explain: your idea for a record page was
so clearly and cleanly executed that anyone who had
anything up their sleeve was almost bound to respond.
> I expect things to cool down now but what do I know?
Well, at least the k=6 and k=8 records are soft targets
for Phil's GNFS ==> SNFS improvement. (k=7 and k=10
may be harder to improve since there I had beginner's
luck, with the PFGW + GMPECM parts of the problem.)
David   In primeform@yahoogroups.com, "Jens Kruse Andersen"
<jens.k.a@...> wrote:
> I expect things to cool down now but what do I know?
Well, it may be that k=3 is also under threat:
http://www.primegrid.com/all_news.php#11> Your computer has made a significant discovery,
But maybe that message is about a Woodall/Cullen find?
> and we want to know your details before we publicize it.
David   In primeform@yahoogroups.com,
"Jens Kruse Andersen" <jens.k.a@...> wrote:
> The Prime Pages has a Primo submission from July 12 2007:
...
> http://primes.utm.edu/primes/page.php?id=81647
> Is there a published certificate?
This apparently unvalidated claim is still on list.
> It appears Marcel requires it but not Chris.
Perhaps Chris might ask the submitter for evidence?
David
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