Let the sequence $\{a_n\}_{n=0}^\infty$ be defined by $a_n=|n(n+1)-19|$. Show that for $n\neq 4$, if $a_n$ is relatively prime to $a_k$ for all $k<n$, then $a_n$ is prime.
The first few terms are $19, 17,13,7,1,11,23,37,53,71,91$, where $a_{10}=91$ is the first non-prime term (other than $a_4=1$), and indeed it is not relatively prime to $a_2=13$ and $a_3=7$.
Suppose $n>4$ and $a_n$ is composite. Let $p$ be the smallest prime dividing $a_n$, so $$ p^2\leq a_n=n(n+1)-19\leq 4n^2. $$ Hence $p\leq 2n$. Let $$ k=\begin{cases} n-p&\text{if }n\geq p,\\ p-1-n&\text{otherwise}. \end{cases} $$ Then $k<n$ and $k(k+1)\equiv n(n+1)\equiv19$ mod $p$. Thus $p|a_k$, so $\gcd(a_k,a_n)\geq p>1$.
$n\ge 4\Rightarrow 0<\overbrace{n^2\!+\!n\!-\!19}^{\large a(n)\ :=\ a_{\Large n}} \color{#c00}{< (n\!+\!1)^2}\,$ so a composite $\,a(n)\,$ has a proper factor $\,d\color{#c00}{\le n},\,$ thus $\,d\mid a(n\!-\!d)\,$ too, $ $ by $\ a(n\!-\!d)\equiv a(n)\equiv 0\,\pmod{\! d},\:$ by the Polynomial Congruence Rule.
