- Hi to all,

By popular request, a new Yahoo Groups number. This one could be

BIG.

It is 3*2^n 1 n is prime. Note a certain similarity to the

one and only 2^n 1 (Mersenne). Note also the 3,2,1. So here are

the first 3Mn numbers of note.

(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 stands

at n=827

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) Proth/NewPGen

have found a 3Mn (n = 827) which is not a million miles away from

Mn . It took all of us until 1952 (Robinson) to find 2^607 1. I

mean, this is PROGRESS! I think I can wait another 10 years and

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 10000

so go for it!

Regards,

Paul Mills

England - --- In primenumbers@y..., paulmillscv@y... wrote:
> Hi to all,

be

> By popular request, a new Yahoo Groups number. This one could

> BIG.

the

> 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 here

are

> 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=827

Proth/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 from

I

> 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 and

10000

> 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

ABC2 3*2^$a-1

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):

3*2^43-1

3*2^103-1

3*2^827-1

3*2^7559-1

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).

http://groups.yahoo.com/group/primeform

You might want to add this program to your prime searching arsenal.

Jim. > 3*2^43-1

... and few weeks later, we obtain:

> 3*2^103-1

> 3*2^827-1

> 3*2^7559-1

>

> 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

Yves- 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 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

decimal expansion.

Sounds fun? Good luck!

regards,

Paul Mills

Kenilworth, England > > 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.

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!

> >

> > 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: primenumbers-unsubscribe@egroups.com

> > > The Prime Pages : http://www.primepages.org

> > >

> > >

> > >

> > > Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/

> > >

> > >

> >

> >

>

>