(22) For the sequence $\left\{a_{n}\right\}$, we define $\left\{\Delta a_{n}\right\}$ as the first-order difference sequence of $\left\{a_{n}\right\}$, where $\Delta a_{n} = a_{n+1} - a_{n} \left(n \in \mathbf{N}^{*}\right)$. For a positive integer $k$, we define $\left\{\Delta^{k} a_{n}\right\}$ as the $k$-th order difference sequence of $\left\{a_{n}\right\}$, where $\Delta^{k} a_{n} = \Delta^{k-1} a_{n+1} - \Delta^{k-1} a_{n} = \Delta\left(\Delta^{k-1} a_{n}\right)$.
(1) If the sequence $\left\{a_{n}\right\}$ has the first term $a_{1} = 1$, and satisfies
$$
\Delta^{2} a_{n} - \Delta a_{n+1} + a_{n} = -2^{n},
$$

find the general term formula of the sequence $\left\{a_{n}\right\}$;
(2) For the sequence $\left\{a_{n}\right\}$ in (1), does there exist an arithmetic sequence $\left\{b_{n}\right\}$ such that
$$
b_{1} \mathrm{C}_{n}^{1} + b_{2} \mathrm{C}_{n}^{2} + \cdots + b_{n} \mathrm{C}_{n}^{n} = a_{n}
$$

holds for all positive integers $n \in \mathbf{N}^{*}$? If it exists, find the general term formula of the sequence $\left\{b_{n}\right\}$; if not, explain the reason;
(3) Let $c_{n} = (2 n - 1) b_{n}$, and set
$$
T_{n} = \frac{c_{1}}{a_{1}} + \frac{c_{2}}{a_{2}} + \cdots + \frac{c_{n}}{a_{n}},
$$

if $T_{n} < M$ always holds, find the smallest positive integer $M$.
(22) (1) Since $\Delta^{2} a_{n}-\Delta a_{n+1}+a_{n}=-2^{n}$, and
$$
\Delta^{2} a_{n}=\Delta a_{n+1}-\Delta a_{n},
$$

we can get
$$
\Delta a_{n}-a_{n}=2^{n}.
$$

Therefore,
$$
a_{n+1}-2 a_{n}=2^{n}, \frac{a_{n+1}}{2^{n+1}}-\frac{a_{n}}{2^{n}}=\frac{1}{2},
$$

so $\left\{\frac{a_{n}}{2^{n}}\right\}$ is an arithmetic sequence with the first term $\frac{1}{2}$ and common difference $\frac{1}{2}$, thus
$$
\frac{a_{n}}{2^{n}}=\frac{1}{2}+(n-1) \cdot \frac{1}{2},
$$

which means $a_{n}=n \cdot 2^{n-1}\left(n \in \mathbf{N}^{*}\right)$.
(2) $b_{1} \mathrm{C}_{n}^{1}+b_{2} \mathrm{C}_{n}^{2}+\cdots+b_{n} \mathrm{C}_{n}^{n}=a_{n}$, that is
$$
b_{1} \mathrm{C}_{n}^{1}+b_{2} \mathrm{C}_{n}^{2}+\cdots+b_{n} \mathrm{C}_{n}^{n}=n \cdot 2^{n-1},
$$

and since $k \mathrm{C}_{n}^{k}=n \cdot \mathrm{C}_{n-1}^{k-1}$, we have
$$
\begin{aligned}
\mathrm{C}_{n}^{1}+2 \mathrm{C}_{n}^{2}+\cdots+n \mathrm{C}_{n}^{n} & =n \mathrm{C}_{n-1}^{0}+n \mathrm{C}_{n-1}^{1}+\cdots+n \mathrm{C}_{n-1}^{n-1} \\
& =n\left(\mathrm{C}_{n-1}^{0}+\mathrm{C}_{n-1}^{1}+\cdots+\mathrm{C}_{n-1}^{n-1}\right) \\
& =n \cdot 2^{n-1},
\end{aligned}
$$

thus $b_{n}=n$.
Therefore, there exists an arithmetic sequence $\left\{b_{n}\right\}, b_{n}=n$ such that
$$
b_{1} \mathrm{C}_{n}^{1}+b_{2} \mathrm{C}_{n}^{2}+\cdots+b_{n} \mathrm{C}_{n}^{n}=a_{n}
$$

holds for all positive integers $n \in \mathbf{N}^{*}$.
(3) From (2) we know
$$
\begin{aligned}
T_{n} & =\frac{1}{1}+\frac{3}{2}+\frac{5}{2^{2}}+\cdots+\frac{2 n-1}{2^{n-1}}, \\
\frac{1}{2} T_{n} & =\frac{1}{2}+\frac{3}{2^{2}}+\frac{5}{2^{3}}+\cdots+\frac{2 n-3}{2^{n-1}}+\frac{2 n-1}{2^{n}},
\end{aligned}
$$
(1) minus (2) gives:
$$
\begin{aligned}
\frac{1}{2} T_{n} & =1+1+\frac{1}{2}+\frac{1}{2^{2}}+\cdots+\frac{1}{2^{n-2}}-\frac{2 n-1}{2^{n}} \\
& =3-\frac{1}{2^{n-2}}-\frac{2 n-1}{2^{n}}
\end{aligned}
$$

Therefore,
$$
T_{n}=6-\frac{1}{2^{n-3}}-\frac{2 n-1}{2^{n-1}}5.
$$

Thus, the smallest positive integer $M$ that satisfies the condition is 6.