## double mersenne prime

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• thanks andersen for your information (links). my algo supose that : for double prime mersenne : p = (2^n)-1 : is mersenne numbre : if (p - (n+1)) /n = prime
Message 1 of 3 , Jun 28, 2009

my algo supose that : for double prime mersenne :

p = (2^n)-1 : is mersenne numbre :

if (p - (n+1)) /n = prime then (2^p)-1 is probable double mersenne prime

exempe : p = (2^5)-1

= 31 is mersenne numbre

test : (p-(n+1) / n

= 31-(5+1)/5

= 5 is prime so (2^p)-1 is probable prime = (2^31)-1

exempe : p = (2^7)-1

= 127 is mersenne numbre

test : (127-(7+1) / 7

= 17 is prime so (2^p)-1 is probable prime = (2^127)-1

exempe : p = (2^13)-1

= 8191 is mersenne numbre

test : (p-(n+1) / n

= 8191-(13+1)/13

= 629 is not prime so (2^p)-1 is not probable prime = (2^8191)-1

exempe : p = (2^17)-1

= 131071 is mersenne numbre

test : (p-(n+1) / n

= 131071-(17+1)/17

= 7709 is not prime so (2^p)-1 is not probable prime = (2^131071)-1

ex .........

for the mersenne numbre p = (2^521)-1

my be (p-(n+1) / n is probable prime then

2^ (2^521)-1 is a probable double mersenne prime if (p-(n+1)/n =prime

thanks

rachid baih

> From: jens.k.a@...
> To: sta_staf@...
> Subject: Re: sorry
> Date: Sat, 27 Jun 2009 01:16:10 +0200
>
> There are infinitely many composite n not of form 2^p-1 such that 2^n - 2
> mod n = 0.
> See for example http://www.research.att.com/~njas/sequences/A001567
> and even n in http://www.research.att.com/~njas/sequences/A006935
>
> --
> Jens Kruse Andersen (off list)
>
> ----- Original Message -----
> From: "sta staf" <sta_staf@...>
> Sent: Friday, June 26, 2009 11:56 PM
>
>
>
> sory for all because the 2 numbre are not the mersenne numbre .
> the great probleme is : my 2^n - 2 mod n = 0 if n is prime
> this formula work with al integer else 2^p-1
> example 2^p - 1 = 2^11 - 1 = 2047
> 2^2047 - 2 mod 2047 = 0 but 2047 is not prime
> so the formul 2^n - 2 mod n = 0 is corecte if n # 2p -1
> for n = 99000313
> 2^n-2 mod n = 0
>

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• ... sta staf wrote, ... (2^521-521-2)/521 has factors: 19 (2^607-607-2)/607 has factors: 13 (2^1279-1279-2)/1279 has factors: 11489
Message 2 of 3 , Jun 28, 2009
sta staf <sta_staf@...> wrote,
with careless brackets:

> for the mersenne numbre p = (2^521)-1
> my be (p-(n+1) / n is probable prime

(2^521-521-2)/521 has factors: 19
(2^607-607-2)/607 has factors: 13
(2^1279-1279-2)/1279 has factors: 11489
(2^2203-2203-2)/2203 has factors: 4409
(2^2281-2281-2)/2281 has factors: 19
(2^3217-3217-2)/3217 has factors: 11
(2^4253-4253-2)/4253 has factors: 31
(2^4423-4423-2)/4423 has factors: 37
(2^9689-9689-2)/9689 has factors: 13
(2^9941-9941-2)/9941 has factors: 17
(2^11213-11213-2)/11213 has factors: 7477
(2^19937-19937-2)/19937 has factors: 11
(2^21701-21701-2)/21701 has factors: 13
(2^23209-23209-2)/23209 has factors: 29
(2^44497-44497-2)/44497 has factors: 139
(2^86243-86243-2)/86243 has factors: 73
(2^110503-110503-2)/110503 has factors: 13183
(2^132049-132049-2)/132049 has factors: 79
(2^216091-216091-2)/216091 has factors: 5
(2^756839-756839-2)/756839 has factors: 13

I appreciate that it is difficult to write in a foreign
language, but that is no excuse for your frequent abuse of
brackets, inordinate numbers of blank lines and failures to
find trivially small factors.

Please note that it takes less than a second to find all of
the factors above, using PFGW. You might try to learn to use
this simple tool, so as not to ask for easily found factors,
in future?

David
• ... But that s ludicrous, even by your low standards of coherence, since Fermat s little theorem shows that (2^n - 1 - (n+2))/n is not an integer when n is
Message 3 of 3 , Jun 29, 2009
sta staf <sta_staf@...> wrote:

> sory i mean n+1 is the second odd after n
> is personel expersion (sory)

> p = (2^n)-1 : is mersenne numbre :
> if (p - (n+2)) /n = prime ...

But that's ludicrous, even by your low
standards of coherence, since Fermat's
little theorem shows that
(2^n - 1 - (n+2))/n
is not an integer when n is prime.

David
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