## a new way to proof Mersenne numbers , but faster ???

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• Hello members, years ago, I had have an idea to proof Mersenne numbers only with the exponent. So the largest number of M(45981737) we use is smaller than 93M.
Message 1 of 1 , Jun 28, 2009
Hello members,

years ago, I had have an idea to proof Mersenne numbers only with the exponent. So the largest number of M(45981737) we use is smaller than
93M.

The problem, is it faster? I use also the Lucas-Lehmer Test.

Here a confuse example:

n=2^5-1 ,p=5

start with 14: Dual = 2^3 + 2^2+2^1 :short 3 2 1

Calculate 14^2 : we get matrix

6 5 4, 5 4 3, 4 3 2 : optimize with p=5
1 0 4, 0 4 3, 4 3 2 : so 4 + 4 =5 , I mean 2^4+2^4 =2^5 , 2^5 mod 2^5-1=1

so 4+4=0 mod (p=5)

after all we get : 3 1 -> 2^3+2^1=10 : 10-2=8 , here only 3 , because (14^2-2) = 8 mod 31

3+3=6 : 6 mod 5 = 1 -> 2^6 mod 31 = 2 (2=2^"1"): 1-1=0 : Rest 0 , 2^5-1 is prime.

Best

Norman

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@get2net. dk

> To: sta_staf@hotmail. com

> 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@hotmail. com>

> 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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