Ireland

Define $b_n(x)=a_{n+1}x+a_n$ for $n\ge 0$ and any integer $x$. We have to find all $n\ge 1$ for which $\gcd (b_n(n),b_{n-1}(n))>1$. By definition, for $k\ge 1$ we have $b_{k+1}(x)=b_k(x)+b_{k-1}(x)$, hence $\gcd (b_{k+1}(x),b_k(x))=\gcd (b_k(x),b_{k-1}(x))$ and so, using induction and substituting $x=n$, $$\begin{array}{rl}\gcd (b_n(n),b_{n-1}(n))& =\gcd (b_1(n),b_0(n))=\gcd (a_{2}n+a_1,a_{1}n+a_0)\\ & =\gcd (4n+3,3n+1)=\gcd (n+2,3n+1)\\ & =\gcd (n+2,-5)\in \{1,5\}.\end{array}$$ Hence, $n{a}_{n+1}+a_n$ and $n{a}_n+a_{n-1}$ are not coprime iff $5\mid n+2$, and this is equivalent to $n\equiv 3(\mathrm{mod}5)$.