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If $n-1$ is $f(n)$-smooth, $n$ is prime infinitely often. What is the best $f$?

For all primes $p$, $p-1$ is $\frac{p}{2}$-smooth, so $f(n) = \frac{n}{2}$ works.

If $q$ is a Fermat prime, then $q-1$ is $2$-smooth. Since we cannot prove there are a finite number of Fermat primes it is possible that $f(n) = 2$ works.

Can anything in between be established with certainty?

I considered $f(n) = \sqrt{n}$. The simplest kind of $\sqrt{n}$-smooth number is a square, and since we don't know if there are an infinite number of primes of the form $k^2+1$ we are in the same situation as with Fermat primes. I'm not sure how to approach the case when $n-1$ is $\sqrt{n}$-smooth but not square.

I feel like it should be possible to prove this for $f(n)=\frac{n}{\log{n}}$ (there are an infinite number of primes $p$ where $p-1$ is $\frac{p}{\log p}$-smooth). How can this be shown? What is the best that is known?

Related:

Is there any infinite set of primes for which membership can be decided quickly?

Are there infinitely many primes next to smooth numbers?

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There has been many works on that problem. A result of Baker and Harman asserts that $p-1$ is $p^{0.2961}$-smooth infinitely often, and seems to be the current published record. It uses very strong results about primes in arithmetic progression. An earlier result of M. Goldfeld has $f(n)=\sqrt{n}$ using the theorems of Bombieri-Vinogradov and Brun-Titchmarsh.

It's not obvious to me that there should be an easy proof for $f(n)=n/\log n$, since this relates to good estimates for the number of primes of size $\approx x$ in a congruence class to a modulus between $x/\log x$ and $x$.

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