Let:

$\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 \text { } \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)&=\left{{\begin{array}{lc} x(k) && \text{if } k \ge n \ 2^{60} - 1 - max(y_n(2k),y_n(2k+1)) && \text{if } k

You are given:

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

Find $A(10^{12})$.