Simultaneous recurrence relations

Question

Currently working on solving this set of three simultaneous recurrences, but having some trouble. Tried various substitutions, but still cannot seem to make any progress. Also, none of the three recurrences are equal, so you cannot reduce this from a system of 3 to a system of 2. Any ideas or hints would be greatly appreciated (If this can even be solved)

For $n > 2$:

$f_n = 2g_{n-1}+h_{n-1}+3$

$g_n = g_{n-1}+2h_{n-1}+3$

$h_n = f_{n-1}+2h_{n-1}+2 $

Initial conditions are $f_1 = 2, g_1 = 1, h_1=2, f_2 = 6, g_2 = 6, h_2 = 6$

Answer 1

This means $$ h_{n-1} = \frac{1}{2}(g_n - g_{n-1} - 3) \quad (*) \\ f_{n-1} = h_n -2h_{n-1}-2 \quad (**) $$ So by applying $(**)$ we can have $$ f_n = 2g_{n-1}+h_{n-1}+3 \iff \\ h_{n+1} -2h_n - 2 = 2g_{n-1}+h_{n-1}+3 \iff \\ 2 g_{n-1} = h_{n+1} -2h_n - h_{n-1} - 5 $$ and by applying $(*)$ it should be possible to come up with a recurrence in $g$ terms only. This gives \begin{align} 2 h_{n-1} &= g_n - g_{n-1} - 3 \\ 2 h_n &= g_{n+1} - g_n - 3 \\ 2 h_{n+1} &= g_{n+2} - g_{n+1} - 3 \\ \end{align} and \begin{align} 4 g_{n-1} &= 2 h_{n+1} - 4 h_n - 2 h_{n-1} - 10 \\ &= g_{n+2} - g_{n+1} - 3 - 2 (g_{n+1} - g_n - 3)-(g_n - g_{n-1} - 3) - 10 \\ &= g_{n+2} - 3 g_{n+1} + g_n + g_{n-1} - 4 \iff \\ 4 &= g_{n+2} - 3 g_{n+1} + g_n - 3 g_{n-1} \end{align} This can be solved for $g$ (first turn into homogeneous version, then solve, no idea if one initial condition suffices here) then $(*)$ leads to $h$ and $(**)$ to $f$.

Answer 2

$$\begin{bmatrix} f_n \\ g_n \\ h_n \\ \end{bmatrix} = \begin{bmatrix} 0 & 2 & 1 \\ 0 & 1 & 2 \\ 1 & 0 & 2 \\ \end{bmatrix} \begin{bmatrix} f_{n - 1} \\ g_{n - 1} \\ h_{n - 1} \\ \end{bmatrix} + \begin{bmatrix} 3 \\ 3 \\ 2 \end{bmatrix}$$

This is an affine system that can be converted to a linear system with the usual trick:

$$\begin{bmatrix} f_n \\ g_n \\ h_n \\ 1 \\ \end{bmatrix} = \begin{bmatrix} 0 & 2 & 1 & 3 \\ 0 & 1 & 2 & 3 \\ 1 & 0 & 2 & 2 \\ 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} f_{n - 1} \\ g_{n - 1} \\ h_{n - 1} \\ 1 \\ \end{bmatrix}$$

So you have a system of the form $I_n = A~I_{n - 1} = A^{n}~I_0$ with the initial conditions given in the question.

If you wish to simplify the system further, you can diagonalize $A$.