$a_0 = 2$ $a_1 = 5$ $a_2 = 29$ $a_3 = 869$... If $\gcd(a_{n-1}, 2n-1) = r$ then $a_{n}(a_{n}+1) \equiv 1 \pmod{r}$. Further $2n+1 = a_{n-1} k/r + 2$ so we'll prove $\gcd(a_n, a_{n-1}\cdot k/r + 2) = 1$. [deleted because it was incorrect, incomplete post.] Starting with $r=1$ for $n=0$, we get that all gcd's are 1.