Logo
All Random Solved Random Open
SOLVED
Is there an infinite set of primes $P$ such that if $\{a_1<a_2<\cdots\}$ is the set of integers divisible only by primes in $P$ then $\lim a_{i+1}-a_i=\infty$?
Originally asked to Erdős by Wintner.

The limit is infinite for a finite set of primes, which follows from a theorem of Pólya [Po18], that if $P(n)$ is a quadratic integer polynomial without repeated roots then as $n\to \infty$ the largest prime factor of $f(n)$ also approaches infinity. Indeed, if $P$ is a finite set of primes and $(a_i)$ is the set of integers divisible only by primes in $P$, and $a_{i+1}-a_i$ is bounded, then there exists some $k$ such that $a_{i+1}=a_i+k$ infinitely often, which contradicts Pólya's theorem with $f(n)=n(n+k)$.

Tijdeman [Ti73] proved that, if $P$ is a finite set of primes, then \[a_{i+1}-a_i \gg \frac{a_i}{(\log a_i)^C}\] for some constant $C>0$ depending on $P$.

Tijdeman [Ti73] resolved this question, proving that, for any $\epsilon>0$, there exists an infinite set of primes $P$ such that, with $a_i$ defined as above, \[a_{i+1}-a_i \gg a_i^{1-\epsilon}.\]

See also [368].

Additional thanks to: Boris Alexeev, Dustin Mixon, Euro Vidal Sampaio, and Desmond Weisenberg