\[a_n=n^2+3, n ge 1\]
. The first five terms are4, 7, 12, 19, 18
To find the greatest common denominator of consecutive terms,
\[a_n=n^2+3, \_{n+1}=(n+1)^2+3=n^2+2n+4\]
repeated subtract (or add) multiples of terms.\[\begin{equation} \begin{aligned} gcd(a_n,a_{n+1}) &= gcd(n^2+3, n^2+2n+4) \\&= gcd(n^2+3, (n^2+2n+4)-(n^2+3)=2n+1) \\ &= gcd(2n^2+6,2n+1) \\ &= gcd(n(2n+1)-(2n^2+6)=n-6,2n+1) \\ &= gcd(n-6,(2n+1)-2(n-6))=13 \end{aligned} \end{equation} \]
Hence
\[gcd(a_n,a_{n+1}\]
equals 1 or 13.General formula here..