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From the property given in the problem, $$a_n=a_{n-1}{a}_{n+1}-1.$$Isolating $a_{n+1},$ we find $$a_{n+1}=\frac{a_n+1}{a_{n-1}}.$$Let $a=a_1$ and $b=a_2.$ Then $$\begin{array}{rl}{a}_3& =\frac{b+1}{a},\\ a_4& =\frac{(b+1)/a+1}{b}=\frac{a+b+1}{ab},\\ a_5& =\frac{(a+b+1)/(ab)+1}{(b+1)/a}=\frac{a+1}{b},\\ a_6& =\frac{(a+1)/b+1}{(a+b+1)/(ab)}=a,\\ a_7& =\frac{a+1}{(a+1)/b}=b.\end{array}$$Note that $a_6=a=a_1$ and $a_7=b=a_2.$ Since each term depends only on the two previous terms, the sequence is periodic from here on, and the length of the period is 5. Therefore, $$a_{2003}=a_3=\frac{b+1}{a}=\frac{a_2+1}{a_1}=\boxed{\frac{1777}{1492}}.$$