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\(a_1=3, a_2=3\)

\(n\geq2,a_{n+1} a_{n-1}={a^2_n}+2007\)

Find \(\lfloor \frac{{a^2_{2007}}+{a^2_{2006}}}{{a_{2007}}{a_{2006}}}\rfloor\)

 Mar 5, 2021
 #1
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The answer is 210.

 May 2, 2021

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