I have two very related questions:


If $f(N)$ is the number of square-free integers in the interval $[1, N]$, it is well known that $$f(N) \sim \frac{6}{\pi^{2}} N.$$ 


My first question is, if we impose the additional condition that the integer is not divisible by any prime smaller than $ N^{1/k}$, for some fixed integer $k \geq 2$, then what is the precise asymptotic? Clearly it is at least $\frac{N}{\log{N}}$ by the prime number theorem. Furthermore, by an asymptotic for squarefree integers divisible by precisely $c$ prime factors [here](https://mathoverflow.net/questions/160963/asymptotics-of-special-square-free-numbers/160974#160974) we see that it is at most $\frac{k\_{1}N(\log \log{N})}{\log{N}}$ with $k\_{1}$ a constant depending on $k$. 


Secondly, there's a connection with a product over all primes. That is, analogously to finding the density of square-free integers, we have that the density of squarefree integers not divisible by any prime less than $N^{1/k}$ is


$$ \prod\_{p < N^{1/k}} \left( 1 - \frac{1}{p} \right) \prod\_{p > N^{1/k}} \left( 1 - \frac{1}{p^2} \right). $$ Now the second term converges to some constant, and the first one decreases, by Mertens' theorem, like $\frac{c\_{2}}{\log{N}}$. This looks similar to what I mentioned before, with the $\log{N}$ appearing in the denominator, but this doesn't a priori tell us anything about what happens inside the interval $[1, N]$ - only an interval $[1, M]$ where $M$ is sufficiently large (and this might be much larger than $N$). So, is there a way to formulate this so that one can find the asymptotic up to $N$ by looking at this product (or perhaps, see how large relative to $N$ that $M$ must be)?



Square-free integers not divisible by any "small" primes
Forgetting the squarefree condition for a moment, the number of integers up to $N$ that are not divisible by any primes less than $N^{1/k}$ is asymptotic to
$$
\omega(k) \frac N{\log N} \sim e^\gamma \omega(k) N \prod\_{p\le N^{1/k}} \bigg( 1-\frac1p \bigg),
$$
where $\omega$ is the [Buchstab function](http://mathworld.wolfram.com/BuchstabFunction.html). (In particular, the first half of the density you derive in your answer is not correct: there is a correction factor of the form $e^\gamma \omega(k)$.)


Now the number of integers up to $N$ that are divisible by the square of a particular prime $p$ is at most $N/p^2$. So the number of integers up to $N$ that are divisible by the square of a prime greater than $N^{1/k}$ is at most
$$
\sum\_{p>N^{1/k}} \frac N{p^2} < N \sum\_{n>N^{1/k}} \frac1{n^2} < N \cdot \frac1{N^{1/k}}.
$$
Therefore the above asymptotic also holds for squarefree numbers not divisible by small primes.


