Consider the set of all strictly decreasing sequences of $n$ natural numbers having the property that in each sequence no term divides any other term of the sequence. Let $A = (a_j)$ and $B = (b_j)$ be any two such sequences. We say that $A$ precedes $B$ if for some $k$, $a_k < b_k$ and $a_i = b_i$ for $i < k$. Find the terms of the first sequence of the set under this ordering.
To solve the problem, we need to find the first sequence of strictly decreasing natural numbers such that no term divides any other term in the sequence, and the sequence is minimal under the given ordering. Let's break down the solution step by step.

1. **Understanding the Ordering**:
   We define \( A \prec B \) if for some \( k \), \( a_k < b_k \) and \( a_i = b_i \) for all \( 1 \leq i < k \). This means that \( A \) precedes \( B \) if the first difference between the sequences occurs at position \( k \) and \( a_k < b_k \).

2. **Erdös Result**:
   According to Erdös, the largest subset \( A \) of \( \{1, 2, \ldots, N\} \) such that no two elements of \( A \) are in a divisibility relation has cardinality \( |A| = \lceil N/2 \rceil \). For \( |A| = n \), it follows that \( N \geq 2n - 1 \).

3. **Constructing the Sequence**:
   For \( n \) elements, we need to construct a sequence \( A_n \) such that no term divides any other term. One possible sequence is \( A = \{2n-1, 2n-2, \ldots, n+1, n\} \). However, another valid sequence is \( A' = \{2n-1, 2n-3, \ldots, n, n-1\} \), and \( A' \prec A \).

4. **Minimal Sequence**:
   To find the minimal sequence under the given ordering, we need to ensure that the sequence is strictly decreasing and no term divides any other term. For \( n = 1 \), the minimal set is \( A_1 = \{1\} \). For \( n = 2 \), the minimal set is \( A_2 = \{3, 2\} \).

5. **General Case**:
   For larger values of \( n \), we can construct the sequence by ensuring that each term is not divisible by any previous term. The sequence can be constructed as follows:
   - \( A_3 = \{5, 3, 2\} \)
   - \( A_4 = \{7, 5, 3, 2\} \)
   - \( A_5 = \{9, 7, 6, 5, 4\} \)
   - \( A_6 = \{11, 9, 7, 6, 5, 4\} \)
   - \( A_7 = \{13, 11, 9, 7, 6, 5, 4\} \)
   - \( A_8 = \{15, 13, 11, 10, 9, 8, 7, 6\} \)

6. **Conclusion**:
   Based on the examples computed by hand, the minimal sequence \( A_n \) for \( n \) elements is given by:
   \[
   A_n = \{2n-1, 2n-3, \ldots, n, n-1\}
   \]
   This sequence ensures that no term divides any other term and is minimal under the given ordering.

The final answer is \( \boxed{ A_n = \{2n-1, 2n-3, \ldots, n, n-1\} } \).