Algebra or Computer Science?

Let's generalize $x^n+\frac{1}{x^n}$ when $x+\frac{1}{x}=a$ and $n$ is a power of $2$ as follows;

$x+\frac{1}{x}=a$ , where $n=1$ .

$x^2+\frac{1}{x^2}=\left( x+\frac{1}{x} \right)^2-2=a^2-2$ , where $n=2$ .

$x^4+\frac{1}{x^4}=\left( x^2+\frac{1}{x^2} \right)^2-2=\left( a^2-2 \right)^2-2$ , where $n=4$ .

$x^8+\frac{1}{x^8}=\left( x^4+\frac{1}{x^4} \right)^2-2=\left( \left( a^2-2 \right)^2-2 \right)^2-2$ , where $n=8$ .

$\vdots$

$x^n+\frac{1}{x^n}=\left( x^{\frac{n}{2}}+\frac{1}{x^{\frac{n}{2}}} \right)^2-2$ , where $n=2^m$ and $m>0$ .

Therefore;

$x^n+\frac{1}{x^n} = \left\{ \begin{array}{ll} a & &\text{if } n = 1\\ {\left( x^{\frac{n}{2}}+\frac{1}{x^{\frac{n}{2}}} \right)}^2-2 & &\text{if } n=2^m, m \ge 1. \end{array} \right.$

We can simply write a recursive function in any programing language, say Python3, as follows;

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def f(n, a):
    if n == 1:
        return a
    return (f(n//2, a)**2-2)

• In the question, $n=2048$ and $a=2018$ ; simply call f(2048, 2018) . The result is a long number, and $2018$ occurs at $4313^{th}$ position.