Baltic Way 2019

Answer: The greatest lighthouse number is 1260.

Denote by $p_n$ the $n$-th prime number. We first set out to prove that, for $n\ge 5$, the inequality $$p_{n+1}^3\le p_1\cdots p_n,n\ge 5,(\text{Eq-8})$$ holds. As is readily verified by computation, the inequality holds for $n=5,6$. Consider now the case $n\ge 7$. By Bertrand's Postulate, $p_k<2p_{k-1}$ for each $k$. Consequently, $$\begin{array}{rl}{p}_{n+1}^3& <2^3{p}_n^3=2^3{p}_n\cdot p_n^2\\ & <2^3{p}_n\cdot 2^2{p}_{n-1}^2=2^3{p}_n\cdot 2^2{p}_{n-1}\cdot p_{n-1}\\ & <2^3{p}_n\cdot 2^2{p}_{n-1}\cdot 2p_{n-2}=64p_n{p}_{n-1}{p}_{n-2}\\ & \le p_n{p}_{n-1}\cdots p_1,\end{array}$$ in so far as $64\le p_1{p}_2\cdots p_{n-3}$, which does indeed hold true when $n\ge 7$.

Consider now a number $N$ divisible by the primes $p_1,\dots ,p_n$, but not by $p_{n+1}$. Then $N$ is a lighthouse number precisely when $N<p_{n+1}^3$, for $p_{n+1}^3$ is the least number relatively prime to $N$ with more than two prime factors. But $p_1\cdots p_n\le N<p_{n+1}^3$, which, in conjunction with (Eq-8), enforces $1\le n\le 4$, so that the only possible prime divisors of $N$ are $2,3,5,7$. There remains to examine these cases in detail.

Suppose $n\le 3$. Then $$N<p_4^3=7^3=343.$$ Suppose now $n=4$. Writing $N=p_1{p}_2{p}_3{p}_{4}M$, we find $$210M=p_1{p}_2{p}_3{p}_{4}M=N<p_5^3=11^3=1331,$$ implying $M\le 6$, and $$N\le 210\cdot 6=1260.$$ Consequently, the lighthouse numbers are bounded by 1260, hence finite in number. Further, the number 1260 has the lighthouse property, for $$p_5^3=11^3=1331>1260.$$