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2019-08-21, 10:29   #4
R. Gerbicz

"Robert Gerbicz"
Oct 2005
Hungary

2×3×5×47 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).