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a_{n+4} = (a_n + 2a_{n+1} + 3a_{n+2} + 4a_{n+3}$) mod 9

Source: KJMO 2010 p4

May 3, 2019
number theory with sequencesSequencenumber theoryrecurrence relation

Problem Statement

Let there be a sequence ana_n such that a1=2,a2=0,a3=1,a4=0a_1 = 2,a_2 = 0, a_3 = 1, a_4 = 0, and for n≥1,an+4n \ge 1, a_{n+4} is the remainder when an+2an+1+3an+2+4an+3a_n + 2a_{n+1} + 3a_{n+2} + 4a_{n+3} is divided by 99. Prove that there are no positive integer kk such that ak=0,ak+1=1,ak+2=0,ak+3=2.a_k = 0, a_{k+1} = 1, a_{k+2} = 0,a_{k+3} = 2.
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