For the real sequence
$$V_n = |a^n+b^n| \quad\text{ with }\quad a=\frac12(1+\sqrt{-7}),\ \ b=\frac12(1-\sqrt{-7}) \tag1$$
I'd like to derive some estimates on the growth rate of $V_n$ and $V_{2^n}$.
It's clear that $$|V_n| = |a^n+b^n| = |a|^n|1+(b/a)^n)| \leqslant 2^{1+n/2}\tag2$$ because $|a|=|b|=\sqrt2$. What's missing is a estimate from below like:
$$|V_{2^n}| > C\cdot 2^{n/2}\quad\text{for infinitely many }n \text{ and any }C.\tag 3$$
$V_n=V_n(P=1,Q=2)$ is OEIS A002249 which is a Lucas Sequence of the 2nd Kind with eigenvalues $a$ and $b$ which satisfy $ab=Q$ and $a+b=P$. $V_n$ satisfies many (recurrence) relations, for example
$$V_{2n} = V_n^2-2^{n+1}\tag4$$
which might be useful to estimate $V_{2^n}$. Where I am stuck is the relation
$$V_n=a^n(1+\varepsilon^n)\quad\text{ where }\quad \varepsilon=\frac{\sqrt{-7}-3}4~\text{ and }~|\varepsilon| = 1\tag 5$$
so that I don't see how to bound $1+\varepsilon^n$ from below like $|1+\varepsilon^{2^n}|>\mathit{const}>0$ (or similar) infinitely many times. It seems to be related to diophantine approximations of $\varepsilon$, but I don't see how to proceed.
Purpose of the estimate is a conclusion in this answer where I used the growth rate to show that a specific representation of $V_{2^n}$ does not exist. I thought it was obvious, but the more I think about it, the less obvious it becomes...
Using Euler's formula, $a = \sqrt{2} e^{i\theta}$ and $b = \sqrt{2} e^{-i\theta}$ where $\theta = \tan^{-1}(\sqrt{7})$.
$$ a^n + b^n = 2^{n/2} e^{in \theta} + 2^{n/2} e^{-in \theta} = 2^{1+n/2} \cos(n\theta)$$
$$ V_n = 2^{1+n/2} \left|\cos(n\theta)\right| $$
If $\left|\cos(n\theta)\right| < \frac{1}{2}$ then $$\cos(2n\theta) = 2\cos^2 (n\theta) - 1 < -\frac{1}{2}$$ $$\left|\cos(2n\theta)\right| > \frac{1}{2} $$
So for at least one of $n=2^k$ or $n=2^{k+1}$, and therefore for infinitely many $n=2^k$,
$$ V_n \geq 2^{n/2} $$
(This is also another way to show $V_n \leq 2^{1+n/2}$ for all $n$.)