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).




(de prime? (N)
   (or
      (= N 2)
      (and
         (> N 1)
         (bit? 1 N)
         (let S (sqrt N)
            (for (D 3  T  (+ D 2))
               (T (> D S) T)
               (T (=0 (% N D)) NIL) ) ) ) ) )
 
(de mersenne? (P)
   (or
      (= P 2)
      (let (MP (dec (>> (- P) 1))  S 4)
         (do (- P 2)
            (setq S (% (- (* S S) 2) MP)) )
         (=0 S) ) ) )