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Source: arXiv v2, p. 2, Theorem 2 and its proof.

Statement

For every integer $j\ge5$, some minimal covering system consists of $j$ classes with distinct moduli, and these moduli, in increasing order, are:

$$2^1<2^2<\cdots<2^{j-4}<3\cdot2^{j-5}<2^{j-3} <3\cdot2^{j-4}<3\cdot2^{j-3}.$$

When $j=5$, the initial string $2^1,\ldots,2^{j-4}$ consists of the single term $2$.

Full proof

For $1\le i\le j-3$, let

$$D_i=2^{i-1}+2^i\mathbb Z.$$

For $k\in\{0,1,2\}$, let $A_k$ be the simultaneous congruence class

$$n\equiv k\pmod 3, \qquad n\equiv0\pmod {2^{j-5+k}}.$$

The Chinese remainder theorem makes $A_k$ one progression of modulus $3\cdot2^{j-5+k}$. Define

$$\mathcal C_j=\{D_1,\ldots,D_{j-3},A_0,A_1,A_2\}.$$

If $2^{j-3}\nmid n$, there is a unique $i\in\{1,\ldots,j-3\}$ with $2^{i-1}\mid n$ and $2^i\nmid n$. Then $n\in D_i$. If $2^{j-3}\mid n$, choose the unique $k\in\{0,1,2\}$ congruent to $n$ modulo $3$. The divisibility condition for $A_k$ is at most $2^{j-3}\mid n$, so $n\in A_k$. Hence $\mathcal C_j$ covers every integer.

The dyadic moduli are $2,2^2,\ldots,2^{j-3}$, and the other three are $3\cdot2^{j-5}$, $3\cdot2^{j-4}$, and $3\cdot2^{j-3}$. They are distinct. The inequalities $2<3<4$ put the first new modulus strictly between $2^{j-4}$ and $2^{j-3}$; the remaining two occur after $2^{j-3}$ in the displayed order. Thus the list has exactly $j$ entries in the claimed order.

It remains to prove minimality. The following give an integer covered by each class and by no other class.

• If $1\le i\le j-4$, the integer $2^{i-1}$ lies in $D_i$ and in no other $D_h$. For $i\le j-5$ its $2$-adic valuation is too small for every $A_k$. For $i=j-4$, it can meet only the divisibility condition of $A_0$, but the nonzero residue $2^{j-5}\pmod3$ excludes $A_0$.
• For $D_{j-3}$, choose an odd $u$ with $2^{j-4}u\equiv2\pmod3$. Then $2^{j-4}u$ has exact $2$-adic valuation $j-4$, so it lies in $D_{j-3}$, misses $A_2$, and its residue modulo $3$ excludes $A_0$ and $A_1$.
• For each $k\in\{0,1,2\}$, the Chinese remainder theorem supplies $n_k$ with $2^{j-3}\mid n_k$ and $n_k\equiv k\pmod3$. It lies in $A_k$, in no $D_i$, and in neither of the other two $A$-classes.

Every member therefore has a private witness. Deleting it leaves that witness uncovered, so $\mathcal C_j$ is minimal.

The source states minimality from its two partition statements but does not spell out the overlaps of $A_0,A_1$ with the last dyadic classes. The private witnesses above are the compilation's explicit completion of that step.

Bears on

• Problem 2: for each $j\ge5$ the system is a minimal distinct cover whose $j$-th smallest modulus is $3\cdot2^{j-3}$, which the paper presents as complementing Theorem 1. Its least modulus is $2$, so it says nothing new about the least-modulus question itself.
• Problem 1188, as structural information about minimal distinct covering systems rather than an estimate for their total number.