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 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)?