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).




define Lucas_Lehmer_Test(p);
   lvars mp = 2**p - 1, sn = 4, i;
   for i from 2 to p - 1 do
       (sn*sn - 2) rem mp -> sn;
   endfor;
   sn = 0;
enddefine;
 
lvars p = 3;
printf('M2', '%p\n');
while p < 1000 do
   if Lucas_Lehmer_Test(p) then
       printf('M', '%p');
       printf(p, '%p\n');
   endif;
   p + 2 -> p;
endwhile;