We call a subset $A = \{a_1, a_2, a_3, \dots\}$ of $\mathbb N$ with $a_1 < a_2 < \dots $ transparent if $a_{k+1} - a_k$ goes to $\infty$ as $k \rightarrow \infty$. Is the following true? For every finite set $P$ of prime numbers, the set $A_P := \{p_1 \cdots p_s : p_i \in P\}$ is transparent.
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asked Apr 9, 2011 at 9:58
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$\begingroup$ How can the limit approach infinity if $P$ is finite? $\endgroup$– J.C. OttemCommented Apr 9, 2011 at 10:12
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1$\begingroup$ @J.C. Ottem: primes can appear more than once in the product. @Jens Reinhold: yes, it is true. Your set $A_P$ is a subset of the set of $(\max P)$-smooth numbers, which thin out. $\endgroup$– felixCommented Apr 9, 2011 at 10:15
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$\begingroup$ (If this would not be true, factoring large integers would be much easier...) $\endgroup$– felixCommented Apr 9, 2011 at 10:16
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$\begingroup$ @Felix: the (max P)-smooth numbers form a very thin set, yes, but that does not preclude the possibility that it also contains pairs of consecutive integers infinitely often. Thus "thin out" is not a proof, indeed it is a restatement of the problem. $\endgroup$– Greg MartinCommented Apr 12, 2011 at 22:03
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$\begingroup$ by the way, "friable" is a better term than "smooth" in this context, despite the latter's head start in the literature $\endgroup$– Greg MartinCommented Apr 12, 2011 at 22:04
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Proving that $a_{k+1}-a_k\to \infty$ reduces to proving that $$a_{k+1}-a_k=n$$ has finitely many solutions for every $n$. If you let $S$ be the set of prime divisors of $n$ union $P$, then this follows from the S-unit theorem.
answered Apr 9, 2011 at 11:18