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 2019-08-20, 03:02 #1 carpetpool     "Sam" Nov 2016 1010001102 Posts Probability N has a prime factor > sqrt(N) For some integer N chosen at random, what is the probability that N has a prime factor > sqrt(N)? This also includes when N is prime, therefore the probability is greater than 1/ln(N). Furthermore, what's the probability that N has a prime factor > N^(1/k) ? Thanks for any new leads.
2019-08-21, 01:33   #2
wblipp

"William"
May 2003
New Haven

23·5·59 Posts

Quote:
 Originally Posted by carpetpool For some integer N chosen at random, what is the probability that N has a prime factor > sqrt(N)?
See Dickmanâ€“de Bruijn function in Wikipedia, Wolfram and elsewhere.

 2019-08-21, 10:16 #3 fivemack (loop (#_fork))     Feb 2006 Cambridge, England 143518 Posts It takes a long time to reach an asymptote ... Code: c=0;for(t=10^12,10^12+10^6,F=factor(t);lp=F[matsize(F)[1],1];if(lp*lp<=t,c=1+c)); c The normal suggestion is that it's about k^-k, so 1/4. But sampling ranges of 10^6 at different places suggests that the count (and so the implied probability) goes up perceptibly for larger N Code: 1e6 270639 1e8 276912 1e10 282202 1e12 286014 1e14 288791 1e16 291417 1e18 293196 1e20 294238 Assuming that we're dealing with a Poisson process with a fixed probability at each sampling point, which I don't think is unfair, those observations are not consistent with the probability being the same at 1e6 as at 1e18. Last fiddled with by fivemack on 2019-08-21 at 10:21
2019-08-21, 10:29   #4
R. Gerbicz

"Robert Gerbicz"
Oct 2005
Hungary

1,429 Posts

Quote:
 Originally Posted by fivemack Code: c=0;for(t=10^12,10^12+10^6,F=factor(t);lp=F[matsize(F)[1],1];if(lp*lp<=t,c=1+c)); c The normal suggestion is that it's about k^-k, so 1/4. But sampling ranges of 10^6 at different places suggests that the count (and so the implied probability) goes up perceptibly for larger N
For k=2 the exact result is known, it is 1-log(2).

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