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Old 2012-11-09, 19:49   #1
aketilander
 
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"Åke Tilander"
Apr 2011
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Default Number of distinct prime factors of a Double Mersenne number

I am trying to figure out a way to estimate the number of distinct prime factors of a Double Mersenne number. If I understand it rightly, for a specific number n the number of distinct prime factors x are:

ω (n) which is asymptotically equal to ln (ln n)

for a Double Mersenne number n=MMp:

ln(ln (2^(2^p-1)-1))

ignoring both "-1" since those parts will be infinitesimally small with growing p.

ln(2^p * ln(2)) =
ln(ln(2)) + p*ln(2) =
-0.367 + p*ln(2) =
-0.367 + 0.693*p

i.e. for MM127 (i.e. p=127) x=87.64

Maybe it may be argued that since both p and Mp are prime x may be a little smaller?

Have I understood this rightly or have I done something wrong?

If this is right the nice thing is that the estimated number of distinct prime factors of a MMp are directely proportional to p.

Last fiddled with by aketilander on 2012-11-09 at 20:09
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Old 2012-11-09, 21:16   #2
ewmayer
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Any such estimate needs to take into account the special "restricted" form of Mersenne factors, which in general will cause M(p) (and by extension M(M(p) for M(p) prime) to have a lower expected number of factors than a general odd number of similar size. Here is a paper I found via cursory online search - the paper itself is not so much of interest in the present context as are the references, several of which appear to have investigated the question you ask.
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