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).




lucas_lehmer(p) := block([s, n, i],
   if not primep(p) then false elseif p = 2 then true else
   (s: 4,
   n: 2^p - 1,
   for i: 2 thru p - 1 do s: mod(s*s - 2, n),
   is(s = 0))
)$
 
sublist(makelist(i, i, 1, 200), lucas_lehmer);
/* [2, 3, 5, 7, 13, 17, 19, 31, 61, 89, 107, 127] */