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Is there any way to calculate the number of squarefree numbers with three distinct prime factors below given $N$? I.e. how many numbers below $N$ can be factored to the form $$ prq $$ where $p$,$r$ and $q$ are distinct prime numbers?

Thanks!

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The number of possibilities will depend on the primes less than $N$. Taking the primes to be in ascending size $p<r<q$, we can see that $p<\sqrt[3]N, r<\sqrt{N/2}$ and $q<N/6$. Of course the restriction that the product is less than $N$ will also reduce the options significantly - for example, for $p=5$ and $r=13$, $q$ has a much lower maximum, $N/65$.

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  • $\begingroup$ Thanks! Is there any combinatorial approach to know exact number of such triples? $\endgroup$
    – user144765
    Feb 9 '15 at 7:48
  • $\begingroup$ Well, it would be difficult because of the shifting limits of which primes are permitted into the combination. So I would think probably there is not a pure combinatorial answer. $\endgroup$
    – Joffan
    Feb 9 '15 at 7:50

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