14. Given $O(0,0), A(1,0), B(0,2)$. Let $P_{1}, P_{2}, P_{3}$ be the midpoints of $AB, OB, OP_{1}$ respectively. For any positive integer $n$, $P_{n+3}$ is the midpoint of the segment $P_{n} P_{n+1}$. Let $P_{n}(x_{n}, y_{n})$, and set
$$
\begin{array}{l}
a_{n}=\frac{x_{n}}{2}+x_{n+1}+x_{n+2}, \\
b_{n}=a_{n} y_{n}+y_{n+1}+y_{n+2} .
\end{array}
$$
(1) Find $a_{1}, a_{2}, a_{3}$;
(2) Prove: $y_{n+4}=1-\frac{y_{n}}{4}$;
(3) Let $c_{n}=1-\frac{5}{4} y_{4 n}$, prove that $\{c_{n}\}$ is a geometric sequence, and find $c_{n}$.
14. (1) From the problem, we know
$$
x_{1}=\frac{1}{2}, x_{2}=0, x_{3}=\frac{1}{4}, y_{1}=1, y_{2}=1, y_{3}=\frac{1}{2} \text {. }
$$

From $x_{n+3}=\frac{x_{n}+x_{n+1}}{2}$, we get $x_{4}=\frac{1}{4}, x_{5}=\frac{1}{8}$.
Therefore, $a_{1}=\frac{1}{2}, a_{2}=\frac{1}{2}, a_{3}=\frac{1}{2}$.
(2) From $a_{n+1}=\frac{x_{n+1}}{2}+x_{n+2}+x_{n+3}$
$$
\begin{array}{l}
=\frac{x_{n+1}}{2}+x_{n+2}+\frac{x_{n}+x_{n+1}}{2} \\
=\frac{x_{n}}{2}+x_{n+1}+x_{n+2}=a_{n},
\end{array}
$$

we get $\quad a_{n}=a_{1}=\frac{1}{2}$.
Similarly, from $b_{n}=\frac{y_{n}}{2}+y_{n+1}+y_{n+2}$, we get
$$
b_{n}=b_{1}=2 \text {, }
$$

i.e., $\frac{y_{n}}{2}+y_{n+1}+y_{n+2}=2$.
Thus, $y_{n+1}+y_{n+2}=2-\frac{y_{n}}{2}$,
$$
y_{n+4}=\frac{y_{n+1}+y_{n+2}}{2}=1-\frac{y_{n}}{4} \text {. }
$$
(3) From $c_{n}=1-\frac{5 y_{4 n}}{4}$
$$
\begin{array}{l}
=\frac{y_{4 n}}{4}+\frac{y_{4 n+1}+y_{4 n+2}}{2}-\frac{5 y_{4 n}}{4} \\
=y_{4 n+4}-y_{4 n},
\end{array}
$$

we get $\frac{c_{n+1}}{c_{n}}=\frac{y_{4 n+8}-y_{4 n+4}}{y_{4 n+4}-y_{4 n}}=\frac{1}{4} \cdot \frac{y_{4 n}-y_{4 n+4}}{y_{4 n+4}-y_{4 n}}=-\frac{1}{4}$. Hence, $\left\{c_{n}\right\}$ is a geometric sequence.
Also, $c_{1}=1-\frac{5 y_{4}}{4}=-\frac{1}{4}$, then
$$
c_{n}=\left(-\frac{1}{4}\right)\left(-\frac{1}{4}\right)^{n-1}=(-1)^{n} \frac{1}{4^{n}} \text {. }
$$