\documentclass{amsart} \def\KK{\mathbb K} \begin{document} Setting $X=x+\frac{1}{x}$, we have $X^3 + C_n X^2 - (C_n-6) X -(C_n^2+2C_n-16)=0,$ and with $X=Y-C_n/3$, we obtain $Y^3+(6-C_n-C_n^2/3)Y+(16-4C_n-2C_n^2/3+2C_n^3/27)=0.$ Put $Y=u+v$, $p=6-C_n-C_n^2/3$ and $q=16-4C_n-2C_n^2/3+2C_n^3/27$. Thus $u^3$ and $v^3$ are solutions of $Z^2+qZ-p^3/27=0,$ and so $u=\sqrt[3]{-27q/2\!+\!3\sqrt{3\Delta}/2},\;a_1=\frac{u\!+\!\overline{u}\!-\!C_n}{3},\; a_2=\frac{ju\!+\!\overline{ju}\!-\!C_n}{3},\;a_3=\frac{j^2u\!+\!\overline{j^2u}\!-\!C_n}{3},$ where $\overline{u}$ is the complex conjugate of $u$. Therefore $\nu_1=\frac{a_2-\sqrt{a_2^2-4}}{2},\;\; \nu_2=\frac{a_1+\sqrt{a_1^2-4}}{2},\;\; \nu_3 =\frac{a_3+\sqrt{a_3^2-4}}{2},$ $\nu_4=\frac{a_2+\sqrt{a_2^2-4}}{2},\;\; \nu_5=\frac{a_1-\sqrt{a_1^2-4}}{2},\;\; \nu_6 =\frac{a_3-\sqrt{a_3^2-4}}{2},$ whereby $\varepsilon=|\nu_1|,\; \varepsilon'=|\nu_2|.$ We determine the constants $X_0,\, X_1,\, X_2,\, X_3$, $c_1,\dots , c_{12}$, $\delta_i$ and $\lambda_i$: $X_0=1, \, c_1=\frac{16\theta(\theta-1)(\theta+1)(\theta+2)(2\theta+1)^5}{9(\theta^2+\theta+1)^5},\, c_2=\frac{\theta^2+\theta+1}{(2\theta+1)(\theta+1)},$ $X_1=\max(X_0, (7c_1/c_2)^{1/6}),\; c_3=\frac{7}{6(\theta^2+\theta+1)}c_1, \; c_4=1.39c_3, X_2=\max(X_1, (2c_3)^{1/6}),$ $\begin{array}{lcl} \vspace{5mm} c_5 = \left\{ \begin{array}{ll} \vspace{5mm} \left|\frac{\log(\varepsilon_3^2 \varepsilon_3')}{U}\right| & if\; n=0,\\ \left|\frac{\log(\varepsilon^2/ \varepsilon')}{V}\right| & \mbox{otherwise}, \end{array} \right. \end{array}$ $\begin{array}{lcl} \vspace{5mm} c_6 \!=\! \!\left\{ \!\begin{array}{l} \vspace{5mm} -\frac{\log(\varepsilon_3^2\varepsilon_3')}{6U}\!+\!\left|\frac{-\log(\varepsilon_3)\log(T_2T_5)-\log(\varepsilon_3\varepsilon_3')\log(T_3T_6)}{2U}\right| \;\;\mbox{if } n=0,\\ \frac{\log(\varepsilon^2/\varepsilon')}{6V}\!+\!\left|\frac{\log((\varepsilon')^2/\varepsilon)\log(T_2)\!+\!3\log(\varepsilon'/\varepsilon)\log(T_3)\!+\!2\log(\varepsilon'/\varepsilon^2)\log(T_4)\!-\!3\log(\varepsilon)\log(T_5)\!-\!\log(\varepsilon\varepsilon')\log(T_6)}{6V}\right|, \end{array} \right. \end{array}$ where $T_j=\theta-\theta_j,\;\; j\neq 1,\;\; U=\log^2(\varepsilon_3)+\log^2(\varepsilon_3')+\log(\varepsilon_3)\log(\varepsilon_3'),$ $V=\log^2(\varepsilon)+\log^2(\varepsilon')-\log(\varepsilon)\log(\varepsilon'),$ $c_7=c_5,\;\;\; c_8=c_6,\;\;\; c_9=c_4 \exp(6c_6/c_5),\;\;\; c_{10}=6/c_5$ because $\KK_t$ is real; $c_{11}=7\times 5\times 3^{14}\times 2^{59} \log(72) \log\left(\varepsilon_2\right) \log^2(\varepsilon_3^{-2}(\varepsilon_3')^{-1}) \log^2(\varepsilon) \log[9(\theta+2)],$ $B_0=\max\left(e,2c_{10}^{-1}c_{11}\log\left(c_{10}^{-1}c_{11}c_9^{1/c_{11}}\right)\right),$ $i_1=\left\{ \begin{array}{ll} 2 & if\; n=0,\\ 4 & \mbox{otherwise}, \end{array} \right. \;\;\; i_2=1,$ $\delta=\left\{ \begin{array}{ll} \vspace{3mm} \delta_1/\delta_2 & if\; n=0,\\ \delta_1/\delta_4 & \mbox{otherwise}, \end{array} \right. \;\;\; \lambda=\left\{ \begin{array}{ll} \vspace{3mm} (\delta_1\lambda_2-\delta_2\lambda_1)/\delta_2 & if\; n=0,\\ (\delta_1\lambda_4-\delta_4\lambda_1)/\delta_4 & \mbox{otherwise}, \end{array} \right.$ $c_{14}=c_{12} \exp(6c_6/c_5), c_{15}=6/c_5,$ $\left\{ \begin{array}{lll} \vspace{3mm} \delta_1=-\frac{1}{\log(\varepsilon_2)}, & \delta_2=-\frac{\log(\varepsilon_3^2 \varepsilon_3')}{U}, & \delta_3=-\frac{\log(\varepsilon_3 (\varepsilon_3')^2)}{U},\\ \delta_4=-\frac{\log(\varepsilon^2/\varepsilon')}{V}, & \delta_5=\frac{\log(\varepsilon/(\varepsilon')^2)}{V}, & \end{array} \right.$ and $\left\{ \!\begin{array}{l} \vspace{3mm} \lambda_1=-\frac{\log(T_2T_4T_6)}{3\log(\varepsilon_2)},\\ \vspace{3mm} \lambda_2=\frac{-\log(\varepsilon_3)\log(T_2T_5)-\log(\varepsilon_3\varepsilon_3')\log(T_3T_6)}{2U},\\ \vspace{3mm} \lambda_3=\frac{-\log(\varepsilon_3\varepsilon_3')\log(T_2T_5)-\log(\varepsilon_3')\log(T_3T_6)}{2U},\\ \vspace{3mm} \lambda_4\!=\!\frac{\log((\varepsilon')^2/\varepsilon)\log(T_2)+3\log(\varepsilon'/\varepsilon)\log(T_3)+2\log(\varepsilon'/\varepsilon^2)\log(T_4)-3\log(\varepsilon)\log(T_5)-\log(\varepsilon\varepsilon')\log(T_6)}{6V},\\ \lambda_5\!=\!\frac{-\log(\varepsilon\varepsilon')\log(T_2)-3\log(\varepsilon')\log(T_3)-2\log((\varepsilon')^2/\varepsilon)\log(T_4)-3\log(\varepsilon'/\varepsilon)\log(T_5)-\log(\varepsilon'/\varepsilon^2)\log(T_6)}{6V}. \end{array} \right.$ Using the following system $\left\{ \begin{array}{l} \gamma_1=x-\theta y=\varepsilon_2^{b_1} \varepsilon_3^{b_2} (\varepsilon_3')^{b_3} \varepsilon^{b_4} (\varepsilon')^{b_5},\\ \gamma_2=x-\theta_2 y=\varepsilon_2^{-b_1} (\varepsilon_3')^{b_2} \left(\varepsilon_3\varepsilon_3'\right)^{-b_3} (\varepsilon')^{b_4} \left(-\varepsilon^{-1}\varepsilon'\right)^{b_5}, \end{array} \right.$ we obtain $\left\{ \begin{array}{l} \vspace{5mm} x= \frac{-(\theta-1)\gamma_1 + \theta(\theta+2)\gamma_2}{\theta^2+\theta+1},\\ y= \frac{(\theta+2)(-\gamma_1+\gamma_2)}{\theta^2+\theta+1}. \end{array} \right.$ The computations are done with a SUN PARC ULTRA1, and for each value of $n$, the time of computation is roughly 9 seconds. \end{document}