Example 2 For any natural number $k$, if $k$ is even, divide it by 2, if $k$ is odd, add 1 to it, this is called one operation. Let the number of numbers that become 1 after exactly $n$ operations be $a_{n}$, try to find $a_{15}$.
Solve: Represent the natural number $k$ with dots. If $k$ is transformed into $h$ after one operation, draw an arrow from $k$ to $h$. The resulting graph is called a directed graph. As shown in Figure 2.9.1 (this graph should have infinitely many points, but we only draw up to the 6th layer).

Clearly, $a_{1}=1$ (only the 2 on the 2nd layer becomes 1 after one operation), $a_{2}=1$ (only the 4 on the 3rd layer becomes 1 after two operations).

For $n \geqslant 2$, among the $a_{n}$ numbers on the $(n+1)$-th layer that become 1 after $n$ operations, each odd number $m$ has only $2m$ that becomes $m$ after one operation; each even number $m$ has both $2m$ and $m-1$ that become $m$ after one operation. Therefore, the number of additional numbers $a_{n+1}-a_{n}$ on the $(n+1)$-th layer compared to the $a_{n}$ numbers on the $n$-th layer is the number of even numbers among these $a_{n}$ numbers.

The even numbers on the $(n+1)$-th layer become the $a_{n-1}$ numbers on the $(n-1)$-th layer after one operation. Therefore,
$$
a_{n+1}-a_{n}=a_{n-1},
$$

which means
$$
a_{n+1}=a_{n}+a_{n-1}.
$$

From the recurrence relation (2) and the initial conditions $a_{1}=a_{2}=1$, it is not difficult to deduce step by step:
\begin{tabular}{c|ccccccccccccccc}
$n$ & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 \\
\hline$a_{n}$ & 1 & 1 & 2 & 3 & 5 & 8 & 13 & 21 & 34 & 55 & 89 & 144 & 233 & 377 & 610
\end{tabular}

Note 1: The sequence $\left\{a_{n}\right\}$ is the famous Fibonacci sequence. When the number of terms is not too large, using the recurrence formula to calculate $a_{n}$ is simpler than using the general term formula.

Note 2: "Every natural number can become 1 after a finite number of operations," this is known as the Collatz conjecture, which has not been proven to this day.