, 2024

It is easy to check that for $n = 2, 4, 10$ the sequences are $(1), (1, 3), (1, 3, 7, 9)$ respectively. Thus these numbers are solution of the given problem. Considering $n \geq 3$, if $n$ is odd then $\text{gcd}(n, 2) = 1$ so clearly $a_1 = 1, a_2 = 2$ and their sum is divisible by $3$, which not satisfy. Hence, $n$ is even. It is easy to check that $n = 6$ or $n = 8$ also do not work. Now we consider $n \geq 12$ and separate in the following cases:

1. If $n = 12k$ for $k \geq 1$ then consider $a = 6k + 1, b = 6k - 1$. Since $a$ is odd then $\text{gcd}(n, a) \mid \text{gcd}(n, 2a) = \text{gcd}(12k, 12k + 2) = 2$ so $\text{gcd}(n, a) = 1$. Similarly, we have $\text{gcd}(n, b) = 1$ so $a, b$ are two consecutive terms in the sequence of $n$, but $a + b = 12k$ which is divisible by $3$. This case does not give any solution.

2. If $n = 12k + 2$ for $k \geq 1$ then consider $a = 6k + 3, b = 6k + 5$. Let denote $d = \text{gcd}(n, a)$ then $d|2a = 12k + 6$ so $d|4$, but $d$ is odd then $d = 1$, thus $\text{gcd}(n, a) = 1$. Similarly, denote $d' = \text{gcd}(n, b)$ then $d'|2b = 12k + 10$ so $d'|8$, but $d'$ is odd then $d' = 1$, thus $\text{gcd}(n, b) = 1$. Now $a, b$ are consecutive terms in the sequence and $3 \mid a + b$, not satisfies.

3. If $n = 12k + 4$ for $k \ge 1$ then consider $a = 6k + 1, b = 6k - 1$.

4. If $n = 12k + 6$ for $k \ge 1$ then consider $a = 6k + 1, b = 6k - 1$.

5. If $n = 12k + 8$ for $k \ge 1$ then consider $a = 6k + 3, b = 6k + 1$.

6. If $n = 12k + 10$ for $k \ge 1$ then consider $a = 6k + 1, b = 6k - 1$.