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 aketilander 2012-11-09 19:49

Number of distinct prime factors of a Double Mersenne number

[SIZE=3][FONT=Times New Roman]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 [B]n[/B] the number of distinct prime factors [B]x[/B] are:[/FONT][/SIZE]

[SIZE=3][FONT=Times New Roman]ω (n) which is asymptotically equal to ln (ln n)[/FONT][/SIZE]

[SIZE=3][FONT=Times New Roman]for a Double Mersenne number n=MMp:[/FONT][/SIZE]

[SIZE=3][FONT=Times New Roman]ln(ln (2^(2^p-1)-1))[/FONT][/SIZE]

[SIZE=3][FONT=Times New Roman]ignoring both "-1" since those parts will be infinitesimally small with growing p.[/FONT][/SIZE]

[FONT=Times New Roman][SIZE=3]ln(2^p * ln(2)) =[/SIZE][/FONT]
[FONT=Times New Roman][SIZE=3]ln(ln(2)) + p*ln(2) =[/SIZE][/FONT]
[FONT=Times New Roman][SIZE=3]-0.367 + p*ln(2) =[/SIZE][/FONT]
[FONT=Times New Roman][SIZE=3]-0.367 + 0.693*p[/SIZE][/FONT]

[SIZE=3][FONT=Times New Roman]i.e. for MM127 (i.e. p=127) x=87.64[/FONT][/SIZE]

[SIZE=3][FONT=Times New Roman]Maybe it may be argued that since both p and Mp are prime x may be a little smaller?[/FONT][/SIZE]

[SIZE=3][FONT=Times New Roman]Have I understood this rightly or have I done something wrong?[/FONT][/SIZE]

[FONT=Times New Roman][SIZE=3]If this is right the nice thing is that the estimated number of distinct prime factors of a MMp are directely proportional to p. :smile:[/SIZE][/FONT]

 ewmayer 2012-11-09 21:16

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 [url=http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.131.5361&rep=rep1&type=pdf]paper[/url] 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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