$a_i=i*m+n+1-i,i \leq n$ $(a_i,a_j)=(i*m+n+1-i,j*m+n+1-j)=((j-i)(m-1), i*(m-1)+n+1)$ If $n+1=ab$ where $1<a \leq b$ and $b>2$ then for $i=a,j=2a$ we have $j=2a<ab \leq ab+1=n$ and $(a_i,a_j)=(j-i,i(m-1)+n+1) =(a,a(m-1)+ab)=a>1$ If $n+1=4,n=3$ then for $m=2$ we have $a_1=5,a_2=6,a_3=7$ If $n=p-1$ where $p$ is prime than for $m=(p-2)!+1$ we have $(a_i,a_j)=((j-i)(p-2)!,i* (p-2)!+p)=1$ because $j-i|(p-2)!$