... be ... the ... are ... stands ... Proth/NewPGen ... I ... 10000 ... Paul, I created a simple file in the new ABC2 format (usable by PFGW developmentMessage 1 of 5 , Feb 1, 2001View Source--- In primenumbers@y..., paulmillscv@y... wrote:
> Hi to all,be
> By popular request, a new Yahoo Groups number. This one could
> It is 3*2^n 1 n is prime. Note a certain similarity to
> one and only 2^n 1 (Mersenne). Note also the 3,2,1. So hereare
> the first 3Mn numbers of note.stands
> (To start the first one for any n is 3*2^18 +- 1 A twin pair!)
> Then the first 3Mn is 3*2^43 1 then n=103 and the record
> at n=827Proth/NewPGen
> Yes, 3*2^827 1 is prime.
> Let's give a big hand to Proth and NewPGen courtesy of Gallot and
> Jobling. (I knew it had to be easier to find titans) If you can't
> beat them join them. So here are my first 2 titans
> 3*2^4204 1 (1267 digits) and 3*2^5134 1. (1546 digits)
> Not an ounce of theory so far!
> Also, can we stand back and appreciate just what Proth and NewPGen
> have accomplished? In a few seconds (OK, 10 seconds)
> have found a 3Mn (n = 827) which is not a million miles away fromI
> Mn . It took all of us until 1952 (Robinson) to find 2^607 1.
> mean, this is PROGRESS! I think I can wait another 10 years and10000
> type in 2^10^9 1 to find (is PRIME!?) in just a few seconds. I
> can't wait!
> OK, so here is the puzzle of the moment. Can anyone improve on
> n=827 n must be prime. I haven't checked the range n= 5000
> so go for it!Paul,
I created a simple file in the new ABC2 format (usable by PFGW
development version). The file "aaa" is simply
a: primes from 2 to 10000
I ran this using the command pfgw -f aaa
This took a couple of minutes to run. The -f adds trial factoring to
a pfgw search, which speeds up this unfactored search. It processed
all numbers which n was prime from 2 to 10000 and these numbers came
up as 3Mn primes (where n is prime):
So 7559 is the next prime satisfying the 3*2^n-1 with n prime.
Note pfgw only did a PRP check when used as above. I still had to
do a -tp primality test to validate that this number was prime:
pfgw -tp "-q3*2^7559-1"
pfgw can be found in the primeform egroup (or now Yahoo group).
You might want to add this program to your prime searching arsenal.
... ... and few weeks later, we obtain: rank description digits who year ... 212 3*2^164987-1 49667 gb 1999 266 3*2^155930-1Message 1 of 5 , Feb 1, 2001View Source
> 3*2^43-1... and few weeks later, we obtain:
> So 7559 is the next prime satisfying the 3*2^n-1 with n prime.
rank description digits who year
---- -------------------- ------ --- ----
212 3*2^164987-1 49667 gb 1999
266 3*2^155930-1 46941 gb 1999
568 3*2^123630-1 37217 gb 1999
1153 3*2^97063-1 29220 gb 1999
1373 3*2^88171-1 26543 gb 1999
1507 3*2^85687-1 25795 gb 1998
1797 3*2^80330-1 24183 gb 1998
2746 3*2^71783-1 21610 gb 1998
Hi to all, Many thanks for the info about the 3Mn number and (aMn numbers) from various people. Yves Gallot points out that the record for 3Mn stands atMessage 1 of 5 , Feb 5, 2001View SourceHi to all,
Many thanks for the info about the 3Mn number and (aMn
numbers) from various people. Yves Gallot points out that the record
for 3Mn stands at 3*2^164987 -1 is prime. 164987 is prime!
Not one to stray from theory too long I think I have come up
with a way to use the Proth/NewPGen combo to arrive at an interesting
equation/conjecture. The equation is
n*X = p + 1 n is an integer, n>= 1. X is an integer >=1
and p is any prime.
Some simple solutions are
n= 1 X= 3 p = 2
n= 2 X= 2 p = 3
n= 3 X= 1 p = 3
n= 4 X= 1 p = 3
n =5 X= 4 p = 19 etc
We see that if X = 1 then n = p + 1 so we can write down a solution
for every integer n equal to a prime value p plus 1. I.e the
equation has a solution for n = 3,4,6,8,12 etc (all p + 1) . Note
how the minimal solution for n= 5 jumped to p = 19 from p = 3 so this
is not a trival series.
I will be working on solutions to this equation for n <= 100,000 at
first, to see how we get on. A minimal solution to this equation for
any n will be the smallest X and p. The use of this equation will be
evident because it connects the 'hats off, pause' integers n, with
primes. So if anyone wants to leap in and start finding solutions,
go ahead. I will compile the results. Note that solutions to this
equation can be enjoyably found with Proth/NewPGen set to values such
as A*2^25 - 1 = p and just listing the values of A which give the 'is
prime!' value. Then A*2^25 = p + 1 and we have a solution for n = A
which is X = 2^25 and p is some value. The point is that we leave
the prime number as part of the equation so we don't have to know its
Sounds fun? Good luck!
... Now,if n=m what do we have? 2^n=1+n*p This implies n can never be even...except of course n=0...then it is trivial. If n is not even...cann t say offhand!Message 1 of 5 , May 30, 2001View Source
> > Hi!Now,if n=m what do we have?
> > First of all...the two n's on either side do not seem to be the same.So
> > you basically say
> > 2^n-1=m*p.Right?We'll say about n=m in a while.
> > Now it seems trivial.
> > 2^n-1 is odd and we know any odd number is factorisable into odd integers.
> > Now taking one of the prime factors as p and the rest as m the result
> > follows.
> > Did I clear the point or did I miss it altogether?
> > Again,for Mersenne primes m=1.
This implies n can never be even...except of course n=0...then it is
If n is not even...cann't say offhand!
> > On Wed, 30 May 2001 paulmillscv@... wrote:
> > > Hi to all,
> > > Apologies for not being on the list recently but the local TV
> > > station did a rerun of the Pink Panther movies. So, I wish to
> > > have "speaks" with the group.
> > >
> > > I have 'good reason to believe' that
> > > 2^n - 1 = n*p for some integer n, p a prime.
> > >
> > > n is odd, "I know that, I know that.."
> > >
> > > Can you prove me wrong, right!
> > >
> > > regards
> > > Paul Mills
> > > Keniworth,
> > > England.
> > >
> > >
> > > Unsubscribe by an email to: email@example.com
> > > The Prime Pages : http://www.primepages.org
> > >
> > >
> > >
> > > Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
> > >
> > >