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 1log(2).