Growth of a linear recurrent sequence

Question

Consider the sequence defined by $a_0=a_1=1$ and $a_n=2a_{n-1}-3a_{n-2}$ for $n\geq 2$. This is the sequence https://oeis.org/A087455.

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 few 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.

Answer 1

First note that $\frac{1+i\sqrt{2}}{1-i\sqrt{2}}\in\mathbb{Q}(\sqrt{-2})$ is not a root of unity, because it does not equal $\pm 1$. Therefore Baker's famous theorem shows that, for some effectively computable constant $c>0$, $$\left|\left(\frac{1+i\sqrt{2}}{1-i\sqrt{2}}\right)^m-1\right|>m^{-c},\qquad m\geq 2.\tag{$*$}$$ (The variables $m$ and $n$ are integers in this post.) Applying this for $m=2n$, we get $$\left|\left(\frac{1+i\sqrt{2}}{1-i\sqrt{2}}\right)^n-1\right|\cdot\left|\left(\frac{1+i\sqrt{2}}{1-i\sqrt{2}}\right)^n+1\right|>(2n)^{-c},\qquad n\geq 1.$$ The first factor is less than $2$, hence multiplying both sides by the absolute value of $(1-i\sqrt{2})^n$, we infer $$\Bigl|(1+i\sqrt{2})^n+(1-i\sqrt{2})^n\Bigr|>\frac{1}{2}\cdot\frac{3^{n/2}}{(2n)^c},\qquad n\geq 1.$$ This means that the sequence $(a_n)$ grows exponentially: $$ |a_n|>\frac{1}{4}\cdot\frac{3^{n/2}}{(2n)^c},\qquad n\geq 1.$$ In particular, there are only finitely many $n$'s with $|a_n|\leq 100$, and these can be effectively bounded. Then, up to that bound, one can check with a computer (at least in theory) which $n$'s satisfy $|a_n|\leq 100$.

Added. In ($*$), the exponent $c=5\times 10^9$ is admissible by Corollary 2.3 in Matveev: An explicit lower bound for a homogeneous rational linear form in logarithms of algebraic numbers. II (Izv. Math. 64 (2000), 1217-1269). This means that $|a_n|\leq 100$ implies $n<3\times 10^{11}$.

Answer 2

Besides the archimedean technique from Baker's theorem in another answer, this type of question can be handled both qualitatively and quantitatively using $p$-adic methods. This was illustrated, for the sequence you asked about, on https://math.stackexchange.com/questions/873147/finding-non-negative-integers-m-such-that-1-sqrt-2m-has-real-part/873529 with the task being a determination of when $a_n = \pm 1$. I wrote up an account of this in another $p$-adic field at http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/strassmannapplication.pdf.