<s>[INST] Simplify system of 3 equations of 3 recursive functions Is it possible to simplify this system into one equation of f only?
f(N) = f(N-1) + f(N-2) + 2*g(N) + h(N)
g(N) = f(N-2) + g(N-1)
h(N) = f(N-2) + h(N-2)
Initial values if needed:
f(1) = 2
f(2) = 5
g(2) = 1
g(3) = 2
h(2) = 1
h(3) = 1

I'm not sure it is possible. The simplification I have is: 
()=(−1)+5(−2)+(−3)−(−4)
But it maybe includes some additional inputs. 
 [/INST] We start with 
$$\begin{align}
&f(n) = f(n - 1) + f(n - 2) + 2·g(n) + h(n)
\\&-\\
&f(n - 2) = f((n - 2) - 1) + f((n - 2) - 2) + 2·g(n - 2) + h(n - 2)
\end{align}$$
And obtain 
$$
f(n) = - f(n - 4) - f(n - 3) + 2·f(n - 2) + f(n - 1) - 2·g(n - 2) + 2·g(n) - h(n - 2) + h(n)
$$
We can substitute $h(n)-h(n-2)$ in the equation and get
$$
f(n) = - f(n - 4) - f(n - 3) + 2·f(n - 2) + f(n - 1) - 2·g(n - 2) + 2·g(n)+f(n-2)
$$
Now all that's left to eliminate are the summands $- 2·g(n - 2) + 2·g(n)$.
Using the recurrence $g(n) = f(n - 2) + g(n - 1)$ this becomes:
$$- 2·g(n - 2) + 2·f(n - 2) + 2·g(n - 1)$$
Which, using the same recurrence (shifted by $n\mapsto n-1$) becomes
$$ 2·f(n - 2) + 2·f(n-3)$$
And so, we obtain the recurrence:
$$
f(n) = - f(n - 4) - f(n - 3) + 2·f(n - 2) + f(n - 1)+ 2·f(n - 2) + 2·f(n-3)+f(n-2)
$$
Or, simplified
$$f(n) = f(n - 1)+ 5·f(n - 2)+ f(n - 3) - f(n - 4)  $$
So, your reduction was correct.
</s>