Expression of $x^n+\frac1{x^n}$ by $x+\frac1{x}$ where $n$ is a positive odd number.

I don't see a clear pattern in the coefficients but there is a recurrence: $$ y_{n+2} = y_2 y_n -y_{n-2} = (y^2-2)y_n -y_{n-2} $$ where $$ y_n = x^n+\dfrac{1}{x^n} $$

Let $x=e^{iz}$. We want to express $x^n+\frac{1}{x^n}=2\cos(nz)$ in terms of $x+\frac{1}{x}=2\cos(z)$, which can be done through Chebyshev polynomials of the first kind:

$$ x^n+\frac{1}{x^n}=2\cos(nz) = 2\,T_n(\cos z) = 2\, T_n\left(\frac{x+\frac{1}{x}}{2}\right).$$

Let $s_1:=x+\dfrac1x$. Then

$$x=\frac{s_1\pm\sqrt{s_1^2-4}}2$$ and

$$s_n=\left(\frac{s_1+\sqrt{s_1^2-4}}2\right)^n+\left(\frac{s_1-\sqrt{s_1^2-4}}2\right)^n.$$

Then by the Binomial development, after cancellation of the odd terms

$$s_n=\frac1{2^{n-1}}\sum_{2k=0}^n\binom n{2k} s_1^{n-2k}\left(s_1^2-4\right)^{k}.$$

Then, developing again,

$$s_n=\frac1{2^{n-1}}\sum_{2k=0}^n\sum_{j=0}^k\binom n{2k}\binom kj(-4)^{k-j} s_1^{n-2k+2j}$$

and you can regroup the terms by equal powers of $s_1$.