An estimate for the number of integers with no small factors Given integers $x > y$, I am interested in the number of positive integers  $\leq x$ and free of factors $\leq y$.  In my case $x$ and $y$ will be large. For a concrete example we can take $x=2^{100}$ and $y=2^{40}$.
Is there a computationally efficient method to estimate this number?  I was looking at Mertens' third theorem but I can't see how to use it make an efficient algorithm.
I also found On the number of positive integers ≦ x and free of prime factors > y but that is the opposite of what I want. I would like the number of positive integers $\leq x$ and free of factors $\leq y$.

If $y \gt \sqrt[3]x$ you are just looking for primes in the range from $y$ to $x$ plus the pairs of primes greater than $y$ that multiply to less than $x$.  Let $\pi(z)$ be the prime-counting function  For the primes we want $\pi(x)-\pi(y)\approx \operatorname{li}(x)-\operatorname{li}(y)$.  For the pairs of primes we get an estimate by saying that each number $z$ has $\frac 1{\ln(z)}$ chance of being prime.  The number of pairs is then about
$$\int_y^{\sqrt x} dt \frac 1{\ln(t)} \int _t^{\frac xt}du \frac 1 {\ln u}=\int_y^{\sqrt x} dt \frac 1{\ln(t)}\left(\operatorname{li}\left(\frac xt\right) - \operatorname{li}(t)\right)$$
which is an integral with a nice smooth integrand, so it should be easy to evaluate numerically.  If $y$ is between $\sqrt [4]x$ and $\sqrt[3]x$ you can do the same thing but add a triple integral for numbers with three large factors.
