15. For natural numbers $n$ greater than 0, define an operation “ $G$ ”:
(1) When $n$ is odd, $G(n)=3 n+1$;
(2) When $n$ is even, $G(n)$ equals $n$ continuously divided by 2 until the quotient is odd;

The $k$-th “ $G$ ” operation is denoted as $G^{k}$, for example, $G^{1}(5)=3 \times 5+1=16, G^{2}(5)=G^{1}(16)=16 \div 2 \div 2$ $\div 2 \div 2=1, G^{3}(5)=3 \times 1+1=4, G^{4}(5)=4 \div 2 \div 2=1$.

Calculate:
(1) $G^{1}$ (2016) value;
(2) $G^{5}$ (19) value;
(3) $G^{2017}$ (19) value.

【Solution】According to the problem, we have:
(1) $G^{1}$ (2016) $=2016 \div 2 \div 2 \div 2 \div 2 \div 2=63$
(2) 
$$
\begin{array}{l}
G^{1}(19)=3 \times 19+1=58 . \\
G^{2}(19)=58 \div 2=29 . \\
G^{3}(19)=3 \times 29+1=88 . \\
G^{4}(19)=88 \div 2 \div 2 \div 2=11 . \\
G^{5}(19)=3 \times 11+1=34 .
\end{array}
$$
(3) 
$$
\begin{array}{l}
G^{6}(19)=17 \\
G^{8}(19)=13 . \\
G^{9}(19)=40 . \\
G^{10}(19)=5 . \\
G^{11}(19)=16 . \\
G^{12}(19)=1 . \\
G^{13}(19)=4 . \\
G^{14}(19)=1 . \\
G^{15}(19)=4 . \\
G^{16}(19)=1 . \\
\ldots
\end{array}
$$

Summary of the periodic pattern: For numbers greater than 11, the result of odd terms is 4, and the result of even terms is 1. Therefore, $G^{2017}(19)=4$.

Answer: $G^{1}(2016)=63, G^{5}(19)=34, G^{2017}(19)=4$.
15. For natural numbers $n$ greater than 0, define an operation “ $G$ " as follows:
(1) When $n$ is odd, $G(n)=3 n+1$;
(2) When $n$ is even, $G(n)$ equals $n$ continuously divided by 2 until the quotient is odd;
The $k$-th “ $G$ ” operation is denoted as $G^{k}$, for example, $G^{1}(5)=3 \times 5+1=16, G^{2}(5)=G^{1}(16)=16 \div 2 \div 2 \div 2 \div 2=1, G^{3}(5)=3 \times 1+1=4, G^{4}(5)=4 \div 2 \div 2=1$.
Calculate:
(1) The value of $G^{1}(2016)$;
(2) The value of $G^{5}(19)$;
(3) The value of $G^{2017}(19)$.