to find the value v where prime p is a factor of 2^v-1
tempvalue = p
count = 0
while tempvalue != 0
{
if tempvalue is odd
{
shiftright tempvalue
count++
}
else
{
tempvalue+=p
}
}
if the count is a primenumber then p is thus a factor of a mersenne prime
if the count is not a primenunber it isn't
if p is a factor of 2^v-1 then it is also a factor of 2^(2v)-1
or just 2^(kv)-1 for all value of k are integers above 0
if kv is not prime them 2^(kv)-1 isn't also, so each prime can only be a factor of one
mersenne numer or 0 mersenne numbers
the first question is now simple to solve, just find the 2^v-1 where Mx is a factor of
*********** REPLY SEPARATOR ***********
On 20-3-02 at 0:21 Torben Schlüntz wrote:
>Just of curiosity:
>
>Has it ever happened that a factor for Mx later has proved to be a
>mersenne prime itself?
>
>Has the same factor been a factor for two different Mx and My?
>
>In my humble oppinion both questions answers No; but GIMPS could have
>proved otherwise.
>
>Anyway, it must exist a great deal of low primes; which by now never can
>become mersenne factors (by reason: 2kp+1). So with two types of primes,
>those that are mersenne factors and those that never can be, do we have
>any means of distinguish them?
>
>
>Happy hunting
>tsc
>
>Btw: (M29 mod 1 + M29 mod 2 +......+ M29 mod 32) = 233 which is 1.
>factor of M29
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