MathDB
2018 MMATHS Tiebreaker p4 - s_n = cs_{n-1} + ds_{n-2}, t_n=s_n mod 2018

Source:

October 8, 2023
recurrence relationalgebraMMATHS

Problem Statement

A sequence of integers fsng is defined as follows: fix integers aa, bb, cc, and dd, then set s1=as_1 = a, s2=bs_2 = b, and sn=csn1+dsn2s_n = cs_{n-1} + ds_{n-2} for all n3n \ge 3. Create a second sequence {tn}\{t_n\} by defining each tnt_n to be the remainder when sns_n is divided by 20182018 (so we always have 0tn20170 \le t_n \le 2017). Let N=(20182)!N = (2018^2)!. Prove that tN=t2Nt_N = t_{2N} regardless of the choices of aa, bb, cc, and dd.