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PAMO Problem 3: Index of 2018 in sequence

Source: 2018 Pan-African Mathematics Olympiad

July 3, 2018
algebranumber theorySequence

Problem Statement

For any positive integer xx, we set g(x)= largest odd divisor of x, g(x) = \text{ largest odd divisor of } x, f(x)={x2+xg(x) if x is even;2x+12 if x is odd. f(x) = \begin{cases} \frac{x}{2} + \frac{x}{g(x)} & \text{ if } x \text{ is even;} \\ 2^{\frac{x+1}{2}} & \text{ if } x \text{ is odd.} \end{cases}
Consider the sequence (xn)n∈N(x_n)_{n \in \mathbb{N}} defined by x1=1x_1 = 1, xn+1=f(xn)x_{n + 1} = f(x_n). Show that the integer 20182018 appears in this sequence, determine the least integer nn such that xn=2018x_n = 2018, and determine whether nn is unique or not.
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