Question Polynomial
$$
(1-z)^{b_{1}}\left(1-z^{2}\right)^{b_{2}} \cdots\left(1-z^{32}\right)^{b_{32}}\left(b_{i} \in \mathbf{N}_{+}\right)
$$

has the following property: when expanded, if terms with powers of $z$ higher than 32 are ignored, what remains is $1-2 z$. Find $b_{32}$.
(17th United States of America Mathematical Olympiad)
The solution to this problem is relatively straightforward:
$$
\begin{array}{l}
(1-z)^{b_{1}}\left(1-z^{2}\right)^{b_{2}} \cdots\left(1-z^{32}\right)^{b_{32}} \\
\equiv 1-2 z\left(\bmod z^{33}\right) .
\end{array}
$$

By comparing the coefficients of $z$, we get $b_{1}=2$.
Let the polynomial on the left side of the above equation be denoted as $f(z)$. Then
$$
\begin{array}{l}
f(z) f(-z) \\
=\left(1-z^{2}\right)^{b_{1}+2 b_{2}}\left(1-z^{4}\right)^{2 b_{4}}\left(1-z^{6}\right)^{b_{3}+2 b_{6} \ldots}\left(1-z^{32}\right)^{2 b_{32}} \\
\equiv 1-4 z^{2}\left(\bmod z^{34}\right) \text {, } \\
\end{array}
$$

i.e.,
$$
\begin{array}{l} 
g(w) \\
=(1-w)^{b_{1}+2 b_{2}}\left(1-w^{2}\right)^{2 b_{4}}\left(1-w^{3}\right)^{b_{3}+2 b_{6}} . \\
\left(1-w^{4}\right)^{2 b_{8}} \cdots\left(1-w^{16}\right)^{2 b_{32}} \\
\equiv 1-4 w\left(\bmod w^{17}\right) .
\end{array}
$$

By comparing the coefficients of $w$, we get $b_{1}+2 b_{2}=4$.
$$
\begin{array}{l}
\equiv 1-16 w^{2}\left(\bmod w^{18}\right) \\
\Rightarrow b_{1}+2 b_{2}+4 b_{4}=16 \Rightarrow b_{4}=\frac{16-4}{4}=3 \text {. } \\
\end{array}
$$

Continuing in this manner, we sequentially obtain
$$
b_{8}=30, b_{16}=2^{12}-2^{4}=4080 \text {. }
$$

Finally, from the equation
$$
b_{1}+2 b_{2}+4 b_{4}+8 b_{8}+16 b_{16}+32 b_{32}=2^{32} \text {, }
$$

we get $b_{32}=\frac{2^{32}-2^{16}}{32}=2^{27}-2^{11}$.
The question now is, how can we obtain a general expression for $b_{n}$? If we use the famous Möbius function $\mu(d)$ from number theory, the general formula can be expressed as
$$
b_{n}=\frac{1}{n} \sum_{d \mid n} \mu(d) 2^{\frac{n}{d}}(1 \leqslant n \leqslant 32),
$$

where “$\sum_{d \mid n}$” denotes the sum over all positive divisors $d$ of $n$.