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Source: arXiv v2, p. 3, Theorem 3; proof on pp. 6--7.

Statement

There is an absolute constant $c>0$ such that every finite covering system $\mathcal A$ of multiplicity $s\ge1$ has smallest modulus at most

$$\exp\left(\frac{c\log^2(s+1)}{\log\log(s+2)}\right). \tag{1}$$

Full proof relative to the stated external inputs

If the family contains modulus $1$, (1) is immediate. Otherwise write its moduli as $1<d_1\le\cdots\le d_n$ and use the notation of the distortion setup.

First consider large $s$. Put $y=Cs^3$, where the absolute constant $C$ will be fixed below. Let $k$ be the largest index with $p_k\le y$, with the convention $k=0$ if no such prime divides $Q$. Set

$$\delta_i=0\quad(i\le k), \qquad \delta_i=\frac12\quad(i>k).$$

Using the first term in the minimum when $\delta_i=0$ and noting that the second denominator is $1$ when $\delta_i=1/2$, define

$$\eta_1=\sum_{i\le k}M_i^{(1)}, \qquad \eta_2=\sum_{k<i\le J}M_i^{(2)}. \tag{2}$$

By Lemma 3.3(b), followed by the standard Chebyshev upper bound $\pi(t)\ll t/\log t$ and partial summation,

$$\eta_2\ll s^2\sum_{p>y}\frac{(\log p)^6}{p^2} \ll\frac{s^2(\log y)^5}{y} =\frac{(\log(Cs^3))^5}{Cs}. \tag{3}$$

The supremum of the last expression over $s\ge1$ tends to zero as $C\to\infty$. Fix $C$ so large that $\eta_2<1/2$.

Choose a preliminary absolute constant $c_0$ and set

$$x_s=\exp\left(\frac{c_0\log^2(s+1)}{\log\log(s+2)}\right)$$

and suppose for a contradiction that $d_1>x_s$. Lemma 3.3(a) gives

$$\eta_1\le s\sum_{\substack{d>x_s\\P^+(d)\le y}}\frac1d. \tag{4}$$

For all sufficiently large $s$, one has $x_s\ge y$ and $y\ge(\log x_s)^3$. Moreover

$$u=\frac{\log x_s}{\log y} =\frac{c_0\log^2(s+1)} {\log(Cs^3)\log\log(s+2)} \sim\frac {c_0}3\frac{\log s}{\log\log s}. \tag{5}$$

Consequently $u\log u=(c_0/3+o(1))\log s$. Choose $c_0$ sufficiently large. The external smooth-number estimate in Lemma 3.4 then makes the sum in (4) less than $1/(2s)$ for all sufficiently large $s$. Thus $\eta_1<1/2$ and $\eta_1+\eta_2<1$.

Choose $S$ so that the preceding argument applies whenever $s>S$. For completeness, the finitely many multiplicities $s\le S$ can be handled by the same criterion without an asymptotic assertion. Choose $Y$ large, set

$$U=\left\lfloor\frac{Y^{1/4}}{\log Y}\right\rfloor, \qquad X_0=Y^U,$$

and take the cutoff $Y$. For large $Y$, $U\ge1$ and

$$Y\ge(\log X_0)^3,$$

because $U\log Y\le Y^{1/4}$. The right side of (3), with $s\le S$ and $y=Y$, tends to zero. At the same time, Lemma 3.4 bounds (4), with $x=X_0$, by

$$S\,O\!\left(\frac{\log Y}{U^U}\right),$$

which also tends to zero. Hence one common finite $X_0$ bounds the smallest modulus for every $1\le s\le S$. Since $\log^2(s+1)/\log\log(s+2)>0$ in this finite range, choose the final constant $c\ge c_0$ so that the right side of (1) is at least $X_0$ for all these $s$. For $s>S$, an assumed violation of (1) also gives $d_1>x_s$, so the preceding large-$s$ argument with the unchanged threshold $x_s$ still applies.

We have therefore arranged, for every $s\ge1$ and under an assumed violation of (1),

$$\sum_{j=1}^J\min\left\{M_j^{(1)}, \frac{M_j^{(2)}}{4\delta_j(1-\delta_j)}\right\}<1.$$

The exact external distortion criterion says that $\mathcal A$ then fails to cover $\mathbb Z$, a contradiction. Thus (1) holds.

The printed proof gives the asymptotic large-$s$ calculation and leaves the finite range implicit; the common-$X_0$ paragraph supplies that routine closure. Also, its final asymptotic for $u$ omits the factor $1/3$ coming from $\log(Cs^3)\sim3\log s$. The corrected relation (5) has the same consequence after enlarging the unspecified absolute constant $c$.

Bears on

• Problem 2. At $s=1$ the theorem bounds the least modulus of every covering system with distinct moduli by an unspecified absolute constant; it is not a new explicit improvement on the density paper's bound.