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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 1.4 (p. 4).

• (i) An integer $n>1$ with at most two distinct prime divisors is a primitive covering number if and only if $n=2^{p-1}p$ for some odd prime $p$.
• (ii) A positive integer $n\equiv0\pmod3$ with exactly three distinct prime divisors is a primitive covering number if and only if $n=2\cdot3^{(p-1)/2}p$ for some prime $p>3$.
• (iii) For every prime $p>5$, both $2^35^{\lfloor(p-1)/4\rfloor}p$ and $2\cdot3\cdot5^{\lfloor(p-1)/4\rfloor}p$ are primitive covering numbers. For every prime $p>7$, $2\cdot3^27^{\lfloor(p-1)/6\rfloor}p$ is a primitive covering number, and so is $2^57^{\lfloor(p-1)/6\rfloor}p$ provided $p\ne13,19$.

Remark 1.3 (p. 4). The two excluded numbers $2^57^2\cdot13$ and $2^57^3\cdot19$ are covering numbers by Theorem 1.1; the paper does not know whether they are primitive.

The proof also records that no prime power is a primitive covering number (p. 9, from Lemma 2.1).

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 1.3 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. 9--10. The sufficiency halves of (i) and (ii) and all of (iii) are cases of Theorem 1.3 ($r=2$; $r=3$ with chain $2,3,p$; $r=3$ with chains $2,5,p$ and $2,7,p$ and $r=4$ with $2,3,5,p$ and $2,3,7,p$), with the primes $p\in\{11,13,17,19\}$ (for $2^57^{\lfloor(p-1)/6\rfloor}p$ only $11$ and $17$), which fall below the bound $(7-2)(7-3)$, handled by rerunning that theorem's argument. For necessity in (i), Lemma 2.1 excludes prime powers; for $n=p_1^{\alpha_1}p_2^{\alpha_2}$ the necessary condition $\sigma(n)/n>2$ (the paper's (1.1), p. 2) forces $p_1=2$, and Lemma 2.1 then gives $\alpha_1+1\ge p_2$, so $2^{p_2-1}p_2\mid n$ and primitivity forces equality. For (ii), the necessary condition $\sum_{d\mid n,\ d\text{ composite}}1/\varphi(d)\ge1$ (the paper's (1.2), p. 3) forces $p_1=2$, so $p_2=3$ as $3\mid n$; part (i) keeps $2^2\cdot3$ from dividing $n$, so $n=2\cdot3^{\alpha}p$, Lemma 2.1 gives $\alpha\ge(p-1)/2$, and primitivity forces $n=2\cdot3^{(p-1)/2}p$.

Dependencies

Theorem 1.3; Lemma 2.1 (p. 6), stated on the Theorem 1.2 page; the necessary conditions (1.1) and (1.2) of pp. 2--3, which the paper derives from its author's earlier results (1996, Theorem I(iv); 2001, Theorem 5(ii)).

Bears on

Problem 1189: part (i) makes $2^{p-1}p$ a primitive covering number for every odd prime $p$, the input from which Corollary 1.3 answers the problem's last question.