Example 11 Given a sequence $a_{1}, a_{2}, \cdots$, where $a_{1}$ is a natural number, and $a_{n+1}=\left[1.5 a_{n}\right]+1$ (here $[x]$ denotes the greatest integer not exceeding $x$), can we choose the value of $a_{1}$ so that the first 100000 terms of this sequence are even, and the 100001st term is odd?
(1981 West Germany Competition Problem)
If $1 \leqslant n \leqslant 100000$ and $a_{n}$ is even, then $\left[1.5 a_{n}\right]=1.5 a_{n}$, thus
$$
a_{n+1}=1.5 a_{n}+1(1 \leqslant n \leqslant 100000) \text {. }
$$

By Theorem 9, we know that $a_{n}=\left(\frac{3}{2}\right)^{n-1}\left(a_{1}+2\right)-2(1 \leqslant n \leqslant 100000, f(x)=1.5 x+1$ has a fixed point at $x_{0}=-2$), so by taking $a_{1}=2^{100000} \cdot N-2$ (where $N$ is any odd number), we have
$$
a_{n}=3^{n-1} \cdot 2^{100000-n} \cdot N-2(1 \leqslant n \leqslant 100000) \text {, }
$$

at this time, $a_{1}, \cdots, a_{100000}$ are even, and $a_{10001}=3^{100000} \cdot N-2$ is odd.