Thus for the exponent 1165 bits long, if it only has 2 factors , then the first
factor must be between 2 and 3413 bits long, and the second factor must be between
3413 and 1164 bits long. Note that the bit lengths of the 2 factors added
together must equal the bit length of the Prime
Dave Mullin wrote:
Result = MODPOW(2,MersenneExponent,TrialFactor)
where TrialFactor is the MersenneExponent * 2 * (some k in range 1 to 2^16)
+ 1.
If Result = 1 then TrialFactor divides the Mersenne Prime. As UBASIC can
handle around 2600
decimal digits, in theory (and with a lot of time), I
Sorry, I'm no mathematician, and new to the
Mersenne field.
No, in the x-y bit range (remember that n bit integers
are about 2^n) thefirst factor could be x/2 to y/2 bits long (powers of a
power multiply).
What I was trying to say in my disjointed way was
...
(Example) M11 = 2047 (11
