Consider the sequence defined by $a_0=a_1=1$ and $a_n=2a_{n-1}-3a_{n-2}$ for $n\geq 2$.

I would like to prove that $|a_n|>100$ when $n>10$. How can we do it ? (Also, is there an explicit non-trivial lower bound for the sequence $|a_n|$?)

The first terms :

1, 1, -1, -5, -7, 1, 23, 43, 17, -95, -241, -197, 329, 1249, 1511, -725, -5983, -9791, -1633, 26107, 57113, 35905, -99529, -306773, -314959, 290401, 1525679, 2180155, -216727, -6973919, -13297657, -5673557, 28545857, 74112385, 62587199, -97162757, -382087111, -472685951, 200889431, 1819836715, 3037005137, 614500129, -7882015153, -17607530693, -11569015927, 29684560225, 94076168231, 99098655787, -84031193119, -465358353599, -678623127841, 38828805115, 2113526993753, 4110567572161, 1880554163063, -8570594390357, -22782851269903, -19853919368735, 28640715072239, 116843188250683, 147764231284649, -55001102182751, -553294898219449, -941586489890645, -223288285122943, 2378182899426049, 5426230654220927, 3717912610163707, -8842866742335367, -28839471315161855, -31150342403317609, 24217729138850347

Explicit formulas :

$$\begin{align*}a_n&=\frac{(1+i\sqrt2)^n+(1-i\sqrt2)^n}{2}\\&=(\sqrt{3})^n\cdot \cos (n\cdot \theta)\end{align*}$$

where $\theta=\tan^{−1}(\sqrt2)$.

Thank you.