Lucas-Lehmer Test:

for 



p


{\displaystyle p}

 an odd prime, the Mersenne number 




2

p


−
1


{\displaystyle 2^{p}-1}

 is prime if and only if 




2

p


−
1


{\displaystyle 2^{p}-1}

 divides 



S
(
p
−
1
)


{\displaystyle S(p-1)}

 where 



S
(
n
+
1
)
=
(
S
(
n
)

)

2


−
2


{\displaystyle S(n+1)=(S(n))^{2}-2}

, and 



S
(
1
)
=
4


{\displaystyle S(1)=4}

.



Task

Calculate all Mersenne primes up to the implementation's
maximum precision, or the 47th Mersenne prime   (whichever comes first).




def mersenne( p ) =
  if p == 2 then return true
 
  var s = 4
  var M = 2^p - 1
 
  repeat p - 2
    s = (s*s - 2) mod M
 
  s == 0
 
import integers.primes
 
for p <- primes().filter( mersenne ).take( 20 )
  println( 'M' + p )