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a_{k+1} = 3a_k -[ 2a_k]- [ a_k ]

Source: 2024 Czech and Slovak Olympiad III A p5

May 18, 2024
Sequencealgebrarecurrence relationfloor function

Problem Statement

Let (ak)k=0∞(a_k)^{\infty}_{k=0} be a sequence of real numbers such that if kk is a non-negative integer, then ak+1=3ak−⌊2ak⌋−⌊ak⌋.a_{k+1} = 3a_k - \lfloor 2a_k \rfloor - \lfloor a_k \rfloor. Definitely all positive integers nn such that if a0=1/na_0 = 1/n, then this sequence is constant after a certain term.
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