5. (BRA 1) 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 $A=\left(a_{j}\right)$ and $B=\left(b_{j}\right)$ 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.
5. Each natural number $p$ can be written uniquely in the form $p=2^{0}(2 r-1)$. We call $2 r-1$ the odd part of $p$. Let $A_{n}=\left(a_{1}, a_{2}, \ldots, a_{n}\right)$ be the first sequence. Clearly the terms of $A_{n}$ must have different odd parts, so those parts must be at least $1,3, \ldots, 2 n-1$. Being the first sequence, $A_{n}$ must have the numbers $2 n-1,2 n-3, \ldots, 2 k+1$ as terms, where $k=[n+1 / 3]$ (then $3(2 k-1)<2 n-1<3(2 k+1))$. Smaller odd numbers $2 s+1$ (with $s<k$ ) obviously cannot be terms of $A_{n}$. In this way we have obtained the $n-k$ odd numbers of $A_{n}$. The other $k$ terms must be even, and by the same reasoning as above they must be precisely the terms of $2 A_{k}$ (twice the terms of $A_{k}$ ). Therefore $A_{m}$ is defined recursively as
$$
\begin{array}{c}
A_{0}=0, \quad A_{1}=\{1\}, \quad A_{2}=\{3,2\} ; \\
A_{m}=\{2 n-1,2 n-3, \ldots, 2 k+1\} \cup 2 A_{k} .
\end{array}
$$