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From the given recursion, $$a_{n+1}=\frac{a_n}{a_{n-1}}.$$Let $a=a_1$ and $b=a_2.$ Then $$\begin{array}{rl}{a}_3& =\frac{a_2}{a_1}=\frac{b}{a},\\ a_4& =\frac{a_3}{a_2}=\frac{b/a}{b}=\frac{1}{a},\\ a_5& =\frac{a_4}{a_3}=\frac{1/a}{b/a}=\frac{1}{b},\\ a_6& =\frac{a_5}{a_4}=\frac{1/b}{1/a}=\frac{a}{b},\\ a_7& =\frac{a_6}{a_5}=\frac{a/b}{1/b}=a,\\ a_8& =\frac{a_7}{a_6}=\frac{a}{a/b}=b.\end{array}$$Since $a_7=a=a_1$ and $a_8=b=a_2,$ and each term depends only on the two previous terms, the sequence is periodic from here on. Furthermore, the length of the period is 6. Therefore, $a_6=a_{1776}=13+\sqrt{7}$ and $a_{2009}=a_5.$ Also, $a_7=a_1,$ and $$a_7=\frac{a_6}{a_5}.$$Hence, $$a_5=\frac{a_6}{a_7}=\frac{13+\sqrt{7}}{1+\sqrt{7}}=\frac{(13+\sqrt{7})(\sqrt{7}-1)}{(1+\sqrt{7})(\sqrt{7}-1)}=\frac{-6+12\sqrt{7}}{6}=\boxed{-1+2\sqrt{7}}.$$