Does anyone have an idea in finding common terms of two following sequences? \begin{matrix} x_0=2,x_1=12, x_{n+1}=6x_n-x_{n-1} \\ x'_0=8,x'_1=144,x'_{n+1}=18x'_n-x'_{n-1}\end{matrix} What is the most general method to find the common terms of two given sequences?
Thanks in advanced.
Sorry. Edited.
Under the given initial values, the recurrence relation $x_{n+1}=6x_n-x_{n-1}$ has the solution $$\color{Blue}{x_n=\dfrac{1}{2\sqrt{2}}(3+2\sqrt{2})^{n+1}-\dfrac{1}{2\sqrt{2}}(3-2\sqrt{2})^{n+1}=\dfrac{1}{\sqrt{2}}\sinh(2(n+1) \ln(1+\sqrt{2})).}$$ Similarly, the recurrence relation $y_{n+1}=18y_n-y_{n-1}, y_0=8, y_1=144$ has the solution $$\color{Purple}{y_n=\dfrac{1}{\sqrt{5}}(9+4\sqrt{5})^{n+1}-\dfrac{1}{\sqrt{5}}(9-4\sqrt{5})^{n+1}=\dfrac{2}{\sqrt{5}}\sinh(2(n+1) \ln(2+\sqrt{5})).}$$ It is an unfortunate that we don't have a notation for $\sinh$ in a base different from $e$ :). Observe that the second sequence grows quite faster than first one. Therefor there will be no positive integer $n$ such that $x_n=y_n.$ In fact, even there are no such reals except $n=-1.$
However, (as @GerryMyerson pointed out in a comment) finding pairs $(m, n)$ such that $x_m=y_n$ is a bit difficult problem.