jmc

Let $\frac{a_2}{a_1}=\frac{b}{a},$ where $a$ and $b$ are relatively prime positive integers, and $a<b.$ Then $a_2=\frac{b}{a}\cdot a_1,$ and $$a_3=\frac{a_2^2}{a_1}=\frac{(b/a\cdot a_1{)}^2}{a_1}=\frac{b^2}{a^2}\cdot a_1.$$This implies $a_1$ is divisible by $a^2.$ Let $a_1=c{a}^2$; then $a_2=cab,$ $a_3=c{b}^2,$ $$\begin{array}{rl}{a}_4& =2a_3-a_2=2c{b}^2-cab=cb(2b-a),\\ a_5& =\frac{a_4^2}{a_3}=\frac{[cb(2b-a){]}^2}{(c{b}^2)}=c(2b-2a{)}^2,\\ a_6& =2a_5-a_4=2c(2b-a{)}^2-cb(2b-a)=c(2b-a)(3b-2a),\\ a_7& =\frac{a_6^2}{a_5}=\frac{[c(2b-a)(3b-2a){]}^2}{c(2b-a{)}^2}=c(3b-2a{)}^2,\\ a_8& =2a_7-a_6=2c(3b-2a{)}^2-c(2b-a)(3b-2a)=c(3b-2a)(4b-3a),\\ a_9& =\frac{a_8^2}{a_7}=\frac{[c(3b-2a)(4b-3a){]}^2}{[c(3b-2a{)}^2}=c(4b-3a{)}^2,\end{array}$$and so on.

More generally, we can prove by induction that $$\begin{array}{rl}{a}_{2k}& =c[(k-1)b-(k-2)a][kb-(k-1)a],\\ a_{2k+1}& =c[kb-(k-1)a{]}^2,\end{array}$$for all positive integers $k.$

Hence, from $a_{13}=2016,$ $$c(6b-5a{)}^2=2016=2^5\cdot 3^2\cdot 7=14\cdot 12^2.$$Thus, $6b-5a$ must be a factor of 12.

Let $n=6b-5a.$ Then $a<a+6(b-a)=n,$ and $$n-a=6b-6a=6(b-a),$$so $n-a$ is a multiple of 6. Hence, $$6<a+6\le n\le 12,$$and the only solution is $(a,b,n)=(6,7,12).$ Then $c=14,$ and $a_1=14\cdot 6^2=\boxed{504}.$