3. Given a sequence $a_{1}, a_{2}, \cdots$, where $a_{1}$ is a natural number, and $a_{n+1}=\left[1.5 a_{n}\right]+1$ (where $[x]$ denotes the greatest integer not exceeding the real number $x$). Can we choose the value of $a_{1}$ such that the first 100000 terms of this sequence are even, and the 100001st term is odd?
(1981, German Mathematics Competition)
If $1 \leqslant n \leqslant 100000$ and $a_{n}$ is even, then
$$
\begin{array}{l}
{\left[1.5 a_{n}\right]=1.5 a_{n}} \\
\Rightarrow a_{n+1}=1.5 a_{n}+1(1 \leqslant n \leqslant 100000) .
\end{array}
$$

Since the fixed point of $f(x)=1.5 x+1$ is $x_{0}=-2$, combining with conclusion 3 we have
$$
a_{n}=\left(\frac{3}{2}\right)^{n-1}\left(a_{1}+2\right)-2(1 \leqslant n \leqslant 100000) .
$$

Therefore, by taking $a_{1}=2^{100000} N-2$ (where $N$ is any odd number), we get $a_{n}=3^{n-1} \times 2^{100000-n} N-2$. In this case, $a_{1}$, $a_{2}, \cdots, a_{100000}$ are even, while $a_{100001}=3^{100000} N-2$ is odd.