Problem Statement

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}

.

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

Let's start with the solution:

see "Mersenne Primes :" + nl
for p = 2 to 18
    if lucasLehmer(p) see "M"  + p + nl ok
next
 
func lucasLehmer p
     i = 0 mp = 0 sn = 0
     if p = 2 return true ok
     if (p and 1) = 0 return false ok
     mp = pow(2,p) - 1
     sn = 4
     for i = 3 to p
         sn = pow(sn,2) - 2
         sn -= (mp * floor(sn / mp))
     next
     return (sn=0)