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.

Your function $a_n$ is a classical recurrence relation. It is well-known that
$$a_n = \sum_{i=1}^3 A_i \alpha_i^n,$$
where $\alpha_i$ are the roots of the equation $x^3 = x + 1$. You can find the coefficients $A_i$ by solving a system of linear equations. In your case, one of the roots is real, and the other two are complex conjugates whose norm is less than $1$, so their contribution to $a_n$ tends to $0$. So actually
$$a_n = A_1 \alpha_1^n + o(1).$$
Mathematica diligently found for you the value of $A_1$, and this way you can obtain your estimate (including the leading constant).
