F5 (20-3, UK) Let $f, g: \mathbf{N}^{*} \rightarrow \mathbf{N}^{*}$ be strictly increasing functions, and
$$
f\left(\mathbf{N}^{*}\right) \cup g\left(\mathbf{N}^{*}\right)=\mathbf{N}^{*}, f\left(\mathbf{N}^{*}\right) \cap g\left(\mathbf{N}^{*}\right)=\varnothing, g(n)=f(f(n))+1 .
$$

Find $f(240)$.
Let $A=\{f(k)\}, B=\{g(k)\}, k \in \mathbf{N}^{*}$.
Each positive integer is either in $A$ or in $B$, but not in both. When $k \neq 1$, $f(1)$ is less than $f(k)$ and $g(k)$, so $f(1)=1$.

From the problem, the $n$-th positive integer not in $A$ is $g(n)=f(f(n))+1$, hence in the segment of positive integers $1 \sim g(n)$, there are $n$ values not in $B$ and $f(n)$ values in $A$, thus we have

or
$$
\begin{array}{c}
f(n)+n=g(n), \\
f(f(n))=f(n)+n-1 .
\end{array}
$$

Using this recurrence relation and $f(1)=1$, we get
$$
\begin{array}{c}
f(1)=1, g(1)=2, f(2)=3, \\
f(f(2))=f(3)=f(2)+2-1=4, \\
f(f(3))=f(4)=f(3)+3-1=6, \\
f(f(4))=f(6)=f(4)+4-1=9, \\
f(f(6))=f(9)=f(6)+6-1=14, \\
f(f(9))=f(14)=f(9)+9-1=22, \\
f(f(14))=f(22)=f(14)+14-1=35, \\
f(f(22))=f(35)=f(22)+22-1=56, \\
f(f(35))=f(56)=f(35)+35-1=90, \\
g(35)=f(f(35))+1=91, \\
f(57)=92, \\
f(f(57))=f(92)=f(57)+57-1=148, \\
f(f(92))=f(148)=f(92)+92-1=239, \\
f(f(148))=f(239)=f(148)+148-1=386, \\
g(148)=f(f(148))+1=f(239)+1=387,
\end{array}
$$

Therefore,
$$
f(240)=388 \text{. }
$$

Note: In fact, the expressions for these two functions are
$$
f(n)=\left[\frac{\sqrt{5}+1}{2} n\right], g(n)=\left[\frac{\sqrt{5}+3}{2} n\right] .
$$

This is a special case of the following Betty's Theorem.
Betty's Theorem If two irrational numbers $\alpha, \beta$ are both greater than 1, and satisfy $\frac{1}{\alpha}+\frac{1}{\beta}=1$, then the functions $f(n)=[n \alpha], g(n)=[n \beta]$ satisfy
(1) $f(n), g(n)$ are increasing functions from $\mathbf{N}^{*} \rightarrow \mathbf{N}^{*}$;
(2) $f\left(\mathbf{N}^{*}\right) \cup g\left(\mathbf{N}^{*}\right)=\mathbf{N}^{*}$;
(3) $f\left(\mathbf{N}^{*}\right) \cap g\left(\mathbf{N}^{*}\right)=\varnothing$.
In this problem, $\alpha=\frac{\sqrt{5}+1}{2}, \beta=\frac{\sqrt{5}+3}{2}$.