Five, given $P_{n}(x)=\sum_{1 \leqslant 2 k+1 \leq n} C_{n}^{2 k+1} x^{n-2 k-1}\left(x^{2}-1\right)^{k-1}$. $S_{n}$ is the sum of the absolute values of the coefficients of the polynomial $P_{n}(x)$. For any positive integer $n$, find the exponent of 2 in the prime factorization of $S_{n}$.
Five
$$
\begin{aligned}
P_{n}(x) & =\sum_{k} C_{n}^{2 k+1} x^{n-2 k-1} \sum_{h}(-1)^{h} C_{k+1}^{h} x^{2(k+1-h)} \\
& =\sum_{h}(-1)^{h}\left(\sum_{k} C_{n}^{2 k+1} C_{k+1}^{h}\right) x^{n+2-2 h}
\end{aligned}
$$

Therefore, $S_{n}$ is the sum of the coefficients of the polynomial
$$
\sum_{h}\left(\sum_{k} C_{n}^{2 k+1} C_{k+1}^{h}\right) x^{n+2-2 h}=\sum_{2 k+1 \leqslant n} C_{n}^{2 k+1} x^{n-2 k-1}\left(x^{2}+1\right)^{k+1}
$$

which is $\sum_{2 k+1 \leq n} C_{n}^{2 k+1} \cdot 2^{k+1}$.
$$
\text{When } n=1, S_{1}=2. \text{ When } n=2, S_{2}=2^{2}.
$$

Assume $n>2$, and in the factorization of $n$, the exponent of 2 is $t_{n}$, then
$$
S_{n}=2 n+\sum_{1 \leqslant k} \frac{n}{2 k+1} C_{n-1}^{2 k} \cdot 2^{k+1}
$$

Therefore, in the factorization of $S_{n}$, the exponent of 2 is $t_{n}+1$ (this also holds for $n=1,2$).