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Statement

Setting (pp. 2--4). A positive integer $n$ is a covering number (Definition 1.1, p. 2) if some cover of $\mathbb Z$ by finitely many residue classes has its moduli distinct, greater than one and dividing $n$; a covering number is primitive (Definition 1.2, p. 4) if none of its proper divisors is a covering number.

Theorem 1.3 (p. 4). Let $p_1=2<p_2<\cdots<p_r$, $r>1$, be distinct primes such that $p_t-1\mid p_{t+1}-1$ for all $0<t<r-1$ and $p_r\ge(p_{r-1}-2)(p_{r-1}-3)$. Then

$$p_1^{\frac{p_2-1}{p_1-1}-1}\cdots p_{r-2}^{\frac{p_{r-1}-1}{p_{r-2}-1}-1}\,p_{r-1}^{\left\lfloor\frac{p_r-1}{p_{r-1}-1}\right\rfloor}\,p_r$$

is a primitive covering number. For $r=2$ the number is $2^{p_2-1}p_2$.

Remark 1.2 (p. 4). The theorem makes $2\cdot3\cdot5\cdot7=210$ a primitive covering number; the paper adds that Erdős constructed a cover of $\mathbb Z$ whose moduli are "all the 14 proper divisors of 210", citing Guy's book and Guo and Sun. These are the divisors of $210$ other than $1$ and $210$.

Source. Zhi-Wei Sun, On covering numbers, Integers 7 (2007), no. 2, A33, also printed in Combinatorial Number Theory (de Gruyter, Berlin, 2007), 443--453. Labels and pages here are those of arXiv:math/0601017v2 (9 September 2006), the edition read, which is named on the source card.

Read depth. Claims checked: the statement and Remark 1.2 were read clause by clause on the page images of the print, and the proof was followed. Nothing here is independently reviewed.

Proof pointer

Pp. 7--9. With $\alpha_t$ the exponents displayed, the products $\prod_{t<s}(\alpha_t+1)$ telescope to $p_s-1$ for $s<r$ and exceed $p_r-1$ at $s=r$, so Theorem 1.1 makes $n$ a covering number. For primitivity, let $d$ be the least covering number $>1$ dividing $n$. Lemma 2.1 (p. 6) forces the largest prime of $d$ to be $p_r$, then forces the exponent of $p_{r-1}$ in $d$ to be full, and then, using $p_r\ge(p_{r-1}-2)(p_{r-1}-3)$, rules out a smaller exponent at any $p_j$, $j\le r-2$: the divisor count would be at most an integer $m$ with $m<p_r+1$ and $m\ne p_r$, so below $p_r$. So $d=n$.

Dependencies

Theorem 1.1; Lemma 2.1 (p. 6), stated on the Theorem 1.2 page.

Bears on

Problem 1189, through Theorem 1.4 (i), whose sufficiency half is the case $r=2$, and Corollary 1.3. For $r\ge3$ the theorem gives primitive covering numbers $n$; the paper does not say whether the divisors of such $n$ greater than one form an irreducible covering set.