Anyhow bringing 11 in a problem ?

Given a recurrence relation a0=0a_0=0, a1=1a_1=1 and for n2n\geq 2 , an=6an19an2a_n=6a_{n-1} -9a_{n-2} find the remainder when a20a_{20} is divided by 1111 ?


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

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