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$a_{n+1}=\frac{1}{2\lfloor a_n \rfloor -a_n+1}$

Source: Philippine Mathematical Olympiad 2024 P6

February 20, 2024

floor functionnumber theory

Problem Statement

The sequence

\{a_n\}_{n\ge 1}

of real numbers is defined as follows: a_1=1, \text{and} a_{n+1}=\frac{1}{2\lfloor a_n \rfloor -a_n+1} \text{for all} n\ge 1 Find

a_{2024}