16、令$n=2^{m}+t,0\leq t < 2^{m}$，即$n=({1b_{m-1}b_{m-2}...b_{2}b_{1}b_{0}})_{2}$.令$g(n)=A_{n}\alpha +B_{n}\gamma +C_{n}\beta _{0}+D_{n}\beta _{1}$
(1)设$\alpha =1,\beta _{0}=\beta _{1}=\gamma =0$,那么可以得到$g(1)=1,g(2n)=g(2n+1)=3g(n)$,所以$g(n)=3^{m}=A_{n}\alpha +B_{n}\gamma +C_{n}\beta _{0}+D_{n}\beta _{1}=A_{n}$
(2)令$g(n)=n$可以得到$g(1)=\alpha = 1,2n=3n+\gamma n + \beta _{0},2n+1=3n+\gamma (n+1) + \beta _{1}$,所以$\alpha = 1, \gamma = -1, \beta_{0}=0,\beta_{1}=1$,可以得到$n=A_{n}-B_{n}+D_{n}$,所以$B_{n}-D_{n}=3^{m}-n$
(3)令$\alpha =1,\beta _{0}=1,\beta _{1}=\gamma =0$,即$g(1)=1, g(2n)=3g(n)+1,g(2n+1)=3g(n)$，可以得到$g(n)=3^{m}+\sum_{i=0}^{m-1}3^{i}(1-b_{i})=A_{n}+C_{n}$，所以 $C_{n}=\sum_{i=0}^{m-1}3^{i}(1-b_{i})$
(4)令$\alpha =1,\beta _{1}=1,\beta _{0}=\gamma =0$,即$g(1)=1, g(2n)=3g(n),g(2n+1)=3g(n)+1$，同理可以得到 $D_{n}=\sum_{i=0}^{m-1}3^{i}b_{i}$
所以$g(n)=3^{m}\alpha +\left (3^{m}-n+\sum_{i=0}^{m-1}3^{i}b_{i} \right )\gamma +\left (\sum_{i=0}^{m-1}3^{i}(1-b_{i}) \right )\beta _{0}+\left (\sum_{i=0}^{m-1}3^{i}b_{i} \right )\beta _{1}$