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f(n + 1)f(n - 1) = nf(n)f(n - 1) + (f(n))^2 and f(0)=f(1)=1

Source: 2023 OLCOMA Costa Rica National Olympiad, Final Round, 3.1

March 20, 2024
Functional Equationsalgebranonnegative integers

Problem Statement

Let Z≥0\mathbb Z^{\geq 0} be the set of all non-negative integers. Consider a function f:Z≥0→Z≥0f:\mathbb Z^{\geq 0} \to \mathbb Z^{\geq 0} such that f(0)=1f(0)=1 and f(1)=1f(1)=1, and that for any integer n≥1n \geq 1, we have f(n+1)f(n−1)=nf(n)f(n−1)+(f(n))2.f(n + 1)f(n - 1) = nf(n)f(n - 1) + (f(n))^2. Determine the value of f(2023)/f(2022)f(2023)/f(2022).
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