MathDB

3

Part of 1984 Czech And Slovak Olympiad IIIA

Problems(1)

a_{n + 2} =4a_{n + 1}-3a_n, b_n= [ a_{n+1} / a_{n-1} ]

Source: Czech and Slovak Olympiad 1984, National Round, Problem 3

9/11/2024
Let the sequence {an}\{a_n\} , n≥0n \ge 0 satisfy the recurrence relation an+2=4an+1−3an,  (1)a_{n + 2} =4a_{n + 1}-3a_n, \ \ (1) Let us define the sequence {bn}\{b_n\} , n≥1n \ge 1 by the relation bn=[an+1an−1]b_n= \left[ \frac{a_{n+1}}{a_{n-1}} \right] where we put bn=1b_n =1 for an−1=0a_{n-1}=0. Prove that, starting from a certain term, the sequence also satisfies the recurrence relation (1). Note: [x][x] indicates the whole part of the number xx.
algebrarecurrence relationfloor function