20121109, 19:49  #1 
"Åke Tilander"
Apr 2011
Sandviken, Sweden
1066_{8} Posts 
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^p1)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 20121109 at 20:09 
20121109, 21:16  #2 
∂^{2}ω=0
Sep 2002
República de California
2·3^{3}·5·43 Posts 
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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