让：

$\begin{array}{ll}x(0)& =0\\ x(1)& =1\\ x(2k)& =(3x(k)+2x(\lfloor \frac{k}{2}\rfloor ))\text{ mod }{2}^{60}\text{ for }k\ge 1\text{, where }\lfloor \rfloor \text{ is the floor function}\\ x(2k+1)& =(2x(k)+3x(\lfloor \frac{k}{2}\rfloor ))\text{ mod }{2}^{60}\text{ for }k\ge 1\\ y_n(k)& =\{\begin{array}{lcc}x(k)& & \text{if }k\ge n\\ 2^{60}-1-max(y_n(2k),y_n(2k+1))& & \text{if }k<n\end{array}\\ A(n)& =y_n(1)\end{array}$

您将获得：

$\begin{array}{ll}x(2)& =3\\ x(3)& =2\\ x(4)& =11\\ y_4(4)& =11\\ y_4(3)& =2^{60}-9\\ y_4(2)& =2^{60}-12\\ y_4(1)& =A(4)=8\\ A(10)& =2^{60}-34\\ A({10}^3)& =101881\end{array}$

求$A({10}^{12})$。