Proving an exponential bound for a recursively defined function I am working on a function that is defined by
$$a_1=1, a_2=2, a_3=3, a_n=a_{n-2}+a_{n-3}$$
Here are the first few values:
$$\{1,1,2,3,3,5,6,8,11,\ldots\}$$
I am trying to find a good approximation for $a_n$. Therefore I tried to let Mathematica diagonalize the problem,it seems to have a closed form but mathematica doesn't like it and everytime I simplify it gives:
a_n = Root[-1 - #1 + #1^3 &, 1]^n Root[-5 + 27 #1 - 46 #1^2 + 23 #1^3 &, 1] + 
 Root[-1 - #1 + #1^3 &, 3]^n Root[-5 + 27 #1 - 46 #1^2 + 23 #1^3 &, 2] + 
 Root[-1 - #1 + #1^3 &, 2]^n Root[-5 + 27 #1 - 46 #1^2 + 23 #1^3 &, 3]

I used this to get a numerical approximation of the biggest root: $$\text{Root}\left[x^3-x-1,1\right]=\frac{1}{3} \sqrt[3]{\frac{27}{2}-\frac{3 \sqrt{69}}{2}}+\frac{\sqrt[3]{\frac{1}{2}
   \left(9+\sqrt{69}\right)}}{3^{2/3}}\approx1.325$$
Looking at the function I set 
$$g(n)=1.325^n$$
and plotted the first 100 values of $\ln(g),\ln(a)$ ($a_n=:a(n)$) in a graph (blue = $a$, red = $g$):

It seems to fit quite nicely, but now my question:
How can I show that $a \in \mathcal{O}(g)$, if possible without using the closed form but just the recursion. If there would be some bound for $f$ thats slightly worse than my $g$ but easier to show to be correct I would be fine with that too.

As another angle to approach things, note that once you have the characteristic equation it's easy to see that growth is exponential: we're looking for a root $x=\alpha$ of $f(x) = x^3-x-1=0$ and $f(1)\lt 0$ while $f(2)\gt 0$, so there's at least one root $\alpha\gt 1$, implying exponential growth by the usual theory of linear recurrences.
On the flip side, there's an even easier (if looser) exponential bound than QED's on the high end: $a_n\lt 2^n$ for $n\in\{1,2,3\}$, and assuming the inductive hypothesis we get $$\begin{align}a_n &= a_{n-2}+a_{n-3} \\
&\lt 2^{n-2}+2^{n-3} \\
&\lt 2^{n-2}+2^{n-2} \\
&= 2^{n-1} \\
&\lt 2^n.\end{align}$$
