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.
 A: Induction would work. 
Notice that the characteristic equation has a unique real positive root $\lambda \approx 1.325 \cdots$. We will show that $a_n = O(\lambda^n)$. Fix a constant $c > 0$ such that $a_n \leqslant c \lambda^n$ for $n \leqslant 3$; such a constant clearly exists. We will show that $a_n \leqslant c \lambda^n$ for all $n$. 
Fix $n \geqslant 4$, and assume that $a_k \leqslant c \lambda^k$ for all $k < n$. Then we have
$$
\begin{align*}
a_n 
&= a_{n-2} + a_{n-3}
\\ &\leqslant c \lambda^{n-2} + c \lambda^{n-3}
\\ &= c \lambda^{n-3} (\color{Purple}{\lambda+1})
\\ &= c \lambda^{n-3} \cdot \color{Purple}{\lambda^3} 
\\ &= c \lambda^n,
\end{align*}
$$
where we have used the fact that $\lambda^3 = \lambda+1$. Thus the induction hypothesis is verified for $n$.
A: 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).
A: It might not be exactly what you are asking for but you can dominate the series by a rough exponential estimate without finding any closed forms like so:
Given $$a_n = a_{n-2} + a_{n-3} = a_{n-3} + a_{n-4} + a_{n-5}$$ it can be seen the sequence is increasing (i.e. $a_{n-1} < a_n$) when $0 < a_{n-5}$, which holds when $n > 5$ and it can be checked directly that we can strengthen this to $n \ge 5$.
This implies that $$a_n < 3 a_{n-3}$$ and on this basis split the sequence into three parts:


*

*$b_n = a_{3n}$

*$b'_n = a_{3n+1}$

*$b''_n = a_{3n+2}$


which are all smaller than $3^n$ for $n \ge 2$ by induction, we find then (by dividing $n$ by $3$) that for $n \ge 4$:
$$a_n < \sqrt[3]{3}^n.$$
which is roughly $1.442\ldots$
P.S. Thanks to Srivatsan for the idea behind this estimate!
A: 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 $$\begin{align}a_n &= a_{n-2}+a_{n-3} \\
&\lt 2^{n-2}+2^{n-3} \\
&\lt 2^{n-2}+2^{n-2} \\
&= 2^{n-1} \\
&\lt 2^n.\end{align}$$
