Five. (20 points) The increasing sequence $1,3,4,9,10,12$, $13, \cdots$ consists of some positive integers, which are either powers of 3 or the sum of several different powers of 3. Find the value of the 2014th term.
Let the sequence be $\left\{a_{n}\right\}$. Then
$$
\begin{array}{l}
a_{1}=1, a_{2}=3, a_{3}=4, a_{4}=9, a_{5}=10, \\
a_{6}=12, a_{7}=13 .
\end{array}
$$

Represent the index $n$ in binary and the term $a_{n}$ in ternary:
$$
\begin{array}{l}
1=(1)_{2}, 2=(10)_{2}, 3=(11)_{2}, \\
4=(100)_{2}, 5=(101)_{2}, 6=(110)_{2}, \\
7=(111)_{2}, \\
a_{1}=1 \times 3^{0}=(1)_{3}, \\
a_{2}=1 \times 3^{1}+0 \times 3^{0}=(10)_{3}, \\
a_{3}=1 \times 3^{1}+1 \times 3^{0}=(11)_{3}, \\
a_{4}=1 \times 3^{2}+0 \times 3^{1}+0 \times 3^{0}=(100)_{3}, \\
a_{5}=1 \times 3^{2}+0 \times 3^{1}+1 \times 3^{0}=(101)_{3}, \\
a_{6}=1 \times 3^{2}+1 \times 3^{1}+0 \times 3^{0}=(110)_{3}, \\
a_{7}=1 \times 3^{2}+1 \times 3^{1}+1 \times 3^{0}=(111)_{3} .
\end{array}
$$

From the problem, we know that the ternary representation of $\left\{a_{n}\right\}$ can only have the digits $0$ and $1$ and is monotonically increasing. Therefore, we can conjecture that when $i=\left(b_{i} b_{i-1} \cdots b_{1} b_{0}\right)_{2}$, then $a_{i}=\left(b_{i} b_{i-1} \cdots b_{1} b_{0}\right)_{3}$.
Next, we prove this by mathematical induction.
Assume that the proposition holds for $i \leqslant t$.
Then, when $t+1=\left(c_{s} c_{s-1} \cdots c_{1} c_{0}\right)_{2}$, where $c_{s} \neq 0$ and $c_{j} \in\{0,1\}(j=0,1, \cdots, s)$, there must be an integer $k \in \mathbf{N}$ such that $c_{k}=1$ and $c_{j}=0(1 \leqslant j \leqslant k-1)$.
Thus, $t+1=(c_{s} c_{s-1} \cdots c_{k} \underbrace{00 \cdots 0}_{k \uparrow})_{2}$.
Therefore, $t=(c_{s} c_{s-1} \cdots c_{k+1} 0 \underbrace{11 \cdots 1}_{k \uparrow})_{2}$.
By the assumption, $a_{t}=(c_{s} c_{s-1} \cdots c_{k+1} 0 \underbrace{11 \cdots 1}_{k \uparrow})_{3}$, which is the sum of several different powers of 3 and is the maximum value less than $a_{t+1}$.
Since $\left\{a_{n}\right\}$ is monotonically increasing, we have
$$
\begin{array}{l} 
a_{t+1}>a_{t} \\
= 3^{s} c_{s}+3^{k-1} c_{s-1}+\cdots+3^{k+1} c_{k+1}+3^{k-1}+ \\
3^{k-2}+\cdots+1,
\end{array}
$$

where $c_{j} \in\{0,1\}(j=k+1, k+2, \cdots, s)$.
$$
\begin{array}{l}
\text { Then } a_{k+1}-\left(3^{s} c_{s}+3^{s-1} c_{s-1}+\cdots+3^{k+1} c_{k+1}\right) \\
>3^{k-1}+3^{k-2}+\cdots+1,
\end{array}
$$
$$
\text { and } a_{t+1}-\left(3^{s} c_{s}+3^{s-1} c_{s-1}+\cdots+3^{k+1} c_{k+1}\right) \text { is still }
$$

a term $a_{u}$ in $\left\{a_{n}\right\}$ (of course, $\left.u(c_{s} c_{s-1} \cdots c_{k+1} 0 \underbrace{11 \cdots 1}_{k \uparrow})_{3}=a_{t},
\end{array}
$$

of course, $(c_{s} c_{s-1} \cdots c_{k+1} 1 \underbrace{00 \cdots 0}_{k \uparrow})_{3}$ can be expressed as the sum of several different powers of 3, so there exists $v \in \mathbf{Z}_{+}$ such that $a_{v}=(c_{s} c_{s-1} \cdots c_{k+1} 1 \underbrace{00 \cdots 0}_{k \uparrow})_{3}$.
Thus, $a_{t+1} \geqslant a_{v}$.
Since $a_{v}>a_{t}$, we have $v>t$, i.e., $v \geqslant t+1$.
Therefore, $a_{0} \geqslant a_{t+1}$.
This implies $a_{t+1}=a_{v}=(c_{s} c_{s-1} \cdots c_{k+1} \underbrace{00 \cdots 0}_{k \uparrow})_{3}$.
Hence, when $n=t+1$, the proposition still holds.
Since $2014=(11111011110)_{2}$, we have
$$
\begin{array}{l}
a_{2014}=(11111011110)_{3} \\
=3^{10}+3^{9}+\cdots+3^{6}+3^{4}+3^{3}+3^{2}+3 \\
=88329 .
\end{array}
$$