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Statement

Corollary 1.3 (p. 5, quoted). "There are infinitely many positive integers $n$ such that among the subsets of $D_n=\{d\geqslant2:\,d\mid n\}$ only $D_n$ can be the set of all the moduli in a cover of $\mathbb Z$ with distinct moduli."

The proof shows this for every $n=2^{p-1}p$ with $p$ an odd prime, and that for these $n$ the set $D_n$ itself is the set of moduli of a cover: every minimal cover of $\mathbb Z$ whose moduli $1<n_1<\cdots<n_k$ have least common multiple $2^{p-1}p$ has $\{n_1,\ldots,n_k\}=D_n$, and $2^{p-1}p$ is a covering number by Theorem 1.4 (i). The paper presents the corollary as an affirmative answer to a question in Erdős's 1980 survey (Ann. Discrete Math. 6, 89--115).

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 was read on the page image of the print, and the proof was followed. Simpson's theorem is cited in the paper, not proved. Nothing here is independently reviewed.

Proof pointer

P. 5. Take a minimal cover with distinct moduli greater than one and least common multiple $N=2^{p-1}p$. Simpson's theorem, a conjecture of Znám, gives $k\ge1+f(N)$ with $f(\prod p_t^{\alpha_t})=\sum\alpha_t(p_t-1)$, here $1+(p-1)+(p-1)=2p-1$, while $D_N$ has exactly $2p-1$ elements. So the moduli are all of $D_N$. Primitivity of $N$ (Theorem 1.4 (i)) makes every minimal cover with moduli in $D_N$ have least common multiple exactly $N$.

Dependencies

Theorem 1.4 (i); R. J. Simpson, Regular coverings of the integers by arithmetic progressions, Acta Arith. 45 (1985), 145--152, whose card is Simpson 1985.

Bears on

Problem 1189: for each odd prime $p$ the divisors of $2^{p-1}p$ greater than one form a covering set of which no proper subset is a covering set, so the problem's last question, whether infinitely many $n$ have their divisors above one forming an irreducible covering set, has the answer yes. The paper does not address the problem's other questions.