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[a_{2020}] =? a_{n+1}=\sqrt{2020+a_n}

Source: Mathematics Regional Olympiad of Mexico Northeast 2020 P1

September 7, 2022
floor functionalgebrarecurrence relationSequence

Problem Statement

Let a1=2020a_1=2020 and let an+1=2020+ana_{n+1}=\sqrt{2020+a_n} for n≥1n\ge 1. How much is ⌊a2020⌋\left\lfloor a_{2020}\right\rfloor?
Note: ⌊x⌋\lfloor x\rfloor denotes the integer part of a number, that is that is, the immediate integer less than xx. For example, ⌊2.71⌋=2\lfloor 2.71\rfloor=2 and ⌊π⌋=3\lfloor \pi\rfloor=3.
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