2^primenumber -1 is always a prime number ?

[Mastan]

Yes, Aketilander ,Thanks...

[Mastan]

Sorry , My question may be misunderstood for someone , my question in tcl scripting clearly is

set x 3
while { $x > 0} {
set x [expr POWER(2,$x) -1]
puts "$x is a prime number ? "
}

[Mastan]

Hi Aketilander,

Double mersenne number is only 2 levels of POWER of 2.
I am thinking for more than 2 (may be infinite levels) , it is always a prime number
MMp : Double mersenne number...
MMMMM...p is always a prime number ?


Regards,
Mastan

[Dubslow]

Originally Posted by Mastan

Sorry , My question may be misunderstood for someone , my question in tcl scripting clearly is

set x 3
while { $x > 0} {
set x [expr POWER(2,$x) -1]
puts "$x is a prime number ? "
}

No. 11 is prime, but 2^11-1 is not prime (=23*89, and yes I did have those factors memorized).

Originally Posted by Wikipedia/Mersenne prime

Though it was believed by early mathematicians that Mp is prime for all primes p, Mp is very rarely prime. In fact, of the 1,622,441 prime numbers p up to 25,964,951,[5] Mp is prime for only 42 of them. The smallest counterexample is the Mersenne number
M11 = 211 − 1 = 2047 = 23 × 89.

Looking at your Tcl, it looks like you are asking the following, much more specific question(s):
2^2-1 = M2 = 3 is prime.
2^3-1 = M3 = MM2 = 7 is prime.
2^7-1 = M7 = MMM2 = 127 is prime.
2^127-1 = M127 = MMMM2 is prime.

Is MM127 = MMMMM2 a prime?
For a natural number n and prime p, define M(p, n) = the nth iterated Mersenne number starting with p, such that M(p, 3) = MMMp.
Is M(2, n) prime for all n >= 1?

There is (as yet) no reason to believe that these numbers have a better chance to be prime than a "random" Mersenne number with a prime exponent. Since Mersenne primes are so few and far between, it is conjectured that MM127 (and M(2, n) for "almost all" n >4) are composite. With our current knowledge of mathematics and computational power though, we have no way to prove it one way or the other. aketilander is looking for a factor of MM127, which would undoubtedly prove its compositeness.

There has been much discussion of these questions -- try Googling "double mersenne numbers" or "iterated mersenne numbers".

Originally Posted by science_man_88

2^p-1 iff p is prime, however the reverse p is prime iff 2^p-1 is prime is false, first counter example is 2^11-1 =23*89.

Watch your 'f's there. You're saying contradictory things if those double 'f's are to be interpreted "properly".

[Prime95]

Originally Posted by Mastan

I am thinking for more than 2 (may be infinite levels) , it is always a prime number

You may think it, and we cannot disprove it, but there is little reason to believe it is true. In fact it is overwhelmingly believed by math experts to be false.

[Batalov]

Originally Posted by Mastan

I am thinking for more than 2 (may be infinite levels) , it is always a prime number
MMp : Double mersenne number...
MMMMM...p is always a prime number ?

It will take one factor to destroy this chain (proving MM127 composite and all of the rest of the chain), and this factor will in a few years be found (most experts agree).

[science_man_88]

Originally Posted by Dubslow

No. 11 is prime, but 2^11-1 is not prime (=23*89, and yes I did have those factors memorized).



Looking at your Tcl, it looks like you are asking the following, much more specific question(s):
2^2-1 = M2 = 3 is prime.
2^3-1 = M3 = MM2 = 7 is prime.
2^7-1 = M7 = MMM2 = 127 is prime.
2^127-1 = M127 = MMMM2 is prime.

Is MM127 = MMMMM2 a prime?
For a natural number n and prime p, define M(p, n) = the nth iterated Mersenne number starting with p, such that M(p, 3) = MMMp.
Is M(2, n) prime for all n >= 1?

There is (as yet) no reason to believe that these numbers have a better chance to be prime than a "random" Mersenne number with a prime exponent. Since Mersenne primes are so few and far between, it is conjectured that MM127 (and M(2, n) for "almost all" n >4) are composite. With our current knowledge of mathematics and computational power though, we have no way to prove it one way or the other. aketilander is looking for a factor of MM127, which would undoubtedly prove its compositeness.

There has been much discussion of these questions -- try Googling "double mersenne numbers" or "iterated mersenne numbers".



Watch your 'f's there. You're saying contradictory things if those double 'f's are to be interpreted "properly".

I did I think you need to reread.

[R.D. Silverman]

Originally Posted by Mastan

Ok, Thanks what about this .


2 ^ 2 ^127 -1 which is more clearly POWER(2,(POWER(2,127)) -1

Huh?? It's the difference of two squares! (x^2-y^2) = (x-y)(x+y)..........

[Prime95]

Originally Posted by Batalov

It will take one factor to destroy this chain (proving MM127 composite and all of the rest of the chain)

True.

and this factor will in a few years be found (most experts agree).

False.

[ewmayer]

Hey, George, I'll bet you never contemplated using Tcl or Python to implement your LL software, you dinosaur. No wonder it took so long to find the latest M-prime! Sheesh...

[Batalov]

I used 'a few' very, very generously.
I agree that it could be ...erhhhm... a while! (Found a better meaningless term! :-)