THE Fourteenth IMAR MATHEMATICAL COMPETITION

The answer is in the affirmative. If $n$ is odd, set $a_1=1$ and $a_2=2$, and if $n>3$ define the other $a_k$ recursively by $a_k{a}_{k-1}=a_0+a_1+\cdots +a_{k-1}$, $k=3,\dots ,n$. It is easily seen that $a_{k+1}=a_{k-1}+1$, $k=3,\dots ,n-1$, so $a_{2k}=k+1$ and $a_{2k+1}=k+(n+1)/2$, $k=1,\dots ,(n-1)/2$. By construction, the integers $a_k$ satisfy the divisibility condition (even at $k=n$). The $a_{2k}$ and $a_{2k+1}$, $k=1,\dots ,(n-1)/2$, form strictly increasing sequences of integers greater than 1, and no $a_{2i}$ equals an $a_{2j+1}$, $i,j=1,\dots ,(n-1)/2$, so the $a_k$, $k=1,\dots ,n-1$, form an injective sequence of positive integers. Since they all lie in the range $1,2,\dots ,n-1$, they form indeed a permutation of the latter.

Similarly, if $n$ is even, set $a_1=2$ and define $a_k$, $k=2,\dots ,n-2$, by the same recursive relation, to get $a_{2k-1}=k+1$ and $a_{2k}=k+n/2$, $k=1,\dots ,n/2-1$. Setting $a_{n-1}=1$ settles the case as above and completes the proof.

Remark. For an odd $n>3$, we may equally well start by setting $a_1=1$ and $a_2=(n+1)/2$, and define the remaining $a_k$ by the recursive relation in the solution, to obtain another sequence of positive integers satisfying the divisibility condition. Explicitly, this time $a_{2k}=k+(n-1)/2$, $k=1,\dots ,(n-1)/2$, and $a_{2k+1}=k+2$,