Three, (50 points) Find the number of all sequences that satisfy the following conditions:
(1) Each term is an integer no less than 2;
(2) The sum of all terms equals a fixed value $m$.


Three, the sequence that satisfies conditions (1) and (2) is called an "$f(m)$ sequence". Below, we calculate the number of $f(m)$ sequences.
Since for a sequence $\left\{a_{n}\right\}$ with $k$ terms, there is
$$
\begin{array}{l}
a_{1}+a_{2}+\cdots+a_{k} \\
=m\left(a_{i} \geqslant 2, i=1,2, \cdots, k\right),
\end{array}
$$

The number of sequences $\left\{a_{n}\right\}$ is the number of integer solutions to equation (1) that satisfy $a_{i} \geqslant 2$, which is the number of non-negative integer solutions to the equation
$$
x_{1}+x_{2}+\cdots+x_{k}=m-2 k
$$

Since the number of non-negative integer solutions to equation (2) is $\mathrm{C}_{m-k-1}^{k-1}$, the number of $f(m)$ sequences is
$$
\begin{aligned}
|f(m)|= & \mathrm{C}_{m-2}^{0}+\mathrm{C}_{m-3}^{1}+\mathrm{C}_{m-4}^{2}+\mathrm{C}_{m-5}^{3}+\cdots \\
= & \mathrm{C}_{m-2}^{0}+\left(\mathrm{C}_{m-4}^{1}+\mathrm{C}_{m-4}^{0}\right)+\left(\mathrm{C}_{m-5}^{2}+\mathrm{C}_{m-5}^{1}\right)+ \\
& \left(\mathrm{C}_{m-6}^{3}+\mathrm{C}_{m-6}^{2}\right)+\cdots \\
= & \left(\mathrm{C}_{m-3}^{0}+\mathrm{C}_{m-4}^{1}+\mathrm{C}_{m-5}^{2}+\mathrm{C}_{m-6}^{3}+\cdots\right)+ \\
& \left(\mathrm{C}_{m-4}^{0}+\mathrm{C}_{m-5}^{1}+\mathrm{C}_{m-6}^{2}+\cdots\right) \\
= & |f(m-1)|+|f(m-2)| .
\end{aligned}
$$

Given $|f(2)|=F_{1}=1,|f(3)|=F_{2}=1$, we have
$$
|f(m)|=F_{m-1} \text {, }
$$

where the sequence $\left\{F_{n}\right\}$ is the Fibonacci sequence:
$$
F_{1}=F_{2}=1, F_{n+2}=F_{n+1}+F_{n} \text {. }
$$

Solving $\lambda^{2}=\lambda+1$, we get $\lambda=\frac{1 \pm \sqrt{5}}{2}$,
$$
F_{n}=A\left(\frac{1+\sqrt{5}}{2}\right)^{n-1}+B\left(\frac{1-\sqrt{5}}{2}\right)^{n-1} \text {. }
$$

Solving $\left\{\begin{array}{l}A+B=1, \\ \frac{1+\sqrt{5}}{2} A+\frac{1-\sqrt{5}}{2} B=1,\end{array}\right.$, we get
$$
A=\frac{1+\sqrt{5}}{2 \sqrt{5}}, B=\frac{1-\sqrt{5}}{2 \sqrt{5}} \text {. }
$$

Thus, $F_{n}=\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{n}-\left(\frac{1-\sqrt{5}}{2}\right)^{n}\right]$.
Therefore, $|f(m)|=F_{m-1}$
$$
=\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{m-1}-\left(\frac{1-\sqrt{5}}{2}\right)^{m-1}\right] .
$$