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).




m = 15
n = 1
For j = 2 To m
    If j = 2 Then
        s = 0
    Else
        s = 4
    EndIf
    n = (n + 1) * 2 - 1
    For i = 1 To j - 2
        s = (s * s - 2) % n
    Next i
    If s = 0 Then Print "M";j
Next