Anyhow bringing 11 in a problem ?

Given a recurrence relation \(a_0=0\), \(a_1=1\) and for \(n\geq 2\) , \[a_n=6a_{n-1} -9a_{n-2}\] find the remainder when \(a_{20}\) is divided by \(11\) ?


This is a part of the set 11≡ awesome (mod remainders)

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