Products of sparse sets of primes Let $S$ be a set of primes such that $\prod_{p \in S} (1 - 1/p)^{-1}$ converges, so the sum of the reciprocals of the products of these primes converges.
If if $n_S$ is the largest factor of $n$ that is a product of elements of $S$, then this condition is $$\sum_{n_S = n} \frac{1}{n} = O(1)$$
In "On the periods of the linear congruential and power generators" (2005), Pomerance and Kurlberg claim that this implies that $n_S < \log n$ for almost all $n$.
I don't see why this follows. I've tried adding other primes slowly to the Euler product, but this doesn't seem to help.
The paper is can be found at https://math.dartmouth.edu/~carlp/PDF/par13.pdf (proof of Lemma 7, bottom of page 6)

I think the $\log n$ here is a bit of a red herring; I suspect it was made deliberately weaker than necessary in order to retain the strong analogy with Lemma 14.  (As we'll see in a moment, it would still be true with the RHS replaced by $\log \log \log n$.)
Let $S^*$ denote the set of values of $n_S$ (in other words $S^* = \{n : n_S = n\}$).
Pick your favourite $\epsilon > 0$.  Since the reciprocal sum of $S^*$ is convergent, there exists a $C := C(\epsilon)$ such that $$\sum_{\substack { n \in S^* \\n > C}} \frac1n < \epsilon.$$
This easily implies the number of $n\le x$ that are divisible by any large element of $S^*$ is $< \epsilon x$.  For any other $n$, we must have $n_S \le C$, and there are only finitely many $n$ for which $\log n \le C$ so then $n_S < \log n$ for large enough $n$.
Therefore $n_S < \log n$ for all $n \le x$ with at most $\epsilon x + O_\epsilon(1)$ exceptions.  Since we can take $\epsilon$ arbitrarily small, this is $o(x)$.
