# Are there infinitely many primes next to smooth numbers?

A side discussion over on this question has left me curious: is there any $B$ for which it's known that there are infinitely many primes adjacent to $B$-smooth numbers (i.e., for which there are infinitely many primes of the form $n\pm 1$, where all the prime factors of $n$ are $\leq B$)? Of course, the Lenstra-Pomerance-Wagstaff conjecture about the distribution of Mersenne primes (and specifically just the conjecture that there are infinitely many) implies this conjecture in the sharpest possible form (with $B=2$), but I'm wondering if results of this form have been proven for any value of $B$.

I'm pretty sure the answer is no, on the grounds that the $B$-smooth numbers are an exponentially thin set, and proofs of infinitely many primes are too much to expect in such circumstances. But I yield to anyone who can supply an actual reference to work on the topic.
• Hmm. Isn't it just 'pseudoexponentially thin' for larger $B$? For instance, between $n$ and $2n$ there's only one 2-smooth number, so assuming 'random' prime distribution the odds are small ($O(1/\log n)$) that there's a prime adjacent to it. But there are $O(\log n)$ 3-smooth numbers in that range, which means heuristically a constant chance of finding a prime next to one over any range $(n\ldots 2n)$, and with $O(\log^2 n)$ 5-smooth numbers in that range we start to hit high probabilities. It seems at least plausible that there could be some $B$ for which this could be made precise... – Steven Stadnicki Jun 9 '11 at 1:49
• You may be right. But just knowing that there are lots of candidates is rarely enough for proving there are lots of primes. E.g., between $n$ and $2n$ there are $O(\sqrt n)$ numbers of the form $x^2+1$, but we can't even prove that one of them is prime. – Gerry Myerson Jun 9 '11 at 5:07