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^p1)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] 
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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