I consider two sequences of numbers $A=\{a_1,...,a_n\}$ and $B=\{k-a_1,...,k-a_n\}$, where $a_1 \le a_2 \le ... \le a_n \le k$.

**I am looking for such conditions under which**: $\gcd(a_1,...,a_n) = \gcd(k-a_1,...,k-a_n)=1$.

**In more general form**: $\gcd(a_1,...,a_n) = \gcd(k-a_1,...,k-a_n) \ge 1$.

It can be seen that the problem is solved only in a particular forms.

I found only four particular solutions.

If there is such a number $\exists a_s \in A: k-a_t=a_s$, where $a_t \in A$ then $\gcd(a_1,...,a_n) = \gcd(k-a_1,...,k-a_n)$.

Let $\gcd(a_1,...,a_n)=e$ and $\gcd(a_n-a_1,...,a_2-a_1)=E$. If $e=E$ and $e\mid k$, then $\gcd(a_1,...,a_n) = \gcd(k-a_1,...,k-a_n)$.

Let $P=p_1 \cdot ... \cdot p_n$ denotes the primorial equaling the product of the first $n$ prime numbers and $p_i$ is the $i^{th}$ prime number. Let $a_i=\frac{P}{p_i}$ and $k=P$, then $\gcd(a_1,...,a_n) = \gcd(k-a_1,...,k-a_n) = 1$.

Let $\gcd(k-a_1,...,k-a_n) = 1$ and $a_i\mid k, \forall a_i \in A$, then $\gcd(a_1,...,a_n) = 1$.

**I am convinced that there are other solutions, but I can not find them yet.
I will be grateful for any help.**