Question

Let $m_1,\ldots, m_n$ be pairwise coprime natural numbers $\geq 1$. We consider the product $$G(m_1,\ldots,m_n) := \prod_{i=1}^{n} \mathbb{Z} / m_i \mathbb{Z}.$$ We define $M(n)$ as the set $n$-tuple of natural numbers $\geq 1$ with the property that the entrys are pairwise coprime. We define $l : M(n) \rightarrow \mathbb{N}_0$ by $$l(m_1,\ldots,m_n) = \max_{(x_1,\ldots,x_n) \in G(m_1,\ldots,m_n)} \min \{ k \in \mathbb{N}_0 | x_i+k \neq 0 \forall 1 \leq i \leq n \},$$ hence $l(m_1,\ldots,m_n)$ may be regarded as the maximal distance of the elements of $G$ from the elements which have no identity in their entrys. It is clear, that $l(m_1,\ldots,m_n)$ is always a natural number.

The question is now: Does there exist a real number $r$, such that $$r \cdot n \cdot \ln (n) \geq \sup_{(m_1,\ldots,m_n) \in M(n)} l(m_1,\ldots,m_n)$$ holds for all $n >> 0$ and is it possible to take $r=2$ ?

Answer 1

We have $l(m_1,\cdots,m_n)$ is at least equal to the maximum length of a sequence of consecutive integers each of which is divisible by some $m_i$. Let's suppose for a moment that $m_i$ is the $i$th prime. In "On the problem of Jacobsthal", Iwaniec proves that this number is $\ll (n\log n)^2$, while your bound would be much stronger, and if you see the lower bound I mention in this equvalent question ($\sim\frac{n(\log n)^2 \log \log \log n}{(\log \log n)^2}$), it is actually false. In any case the question reduces to finding the best upper bound on Jacobsthal's function (In the comments of the question I linked to above there is an argument for why it suffices to reduce to the $m_i$ prime case).