How to find recurrence relation for $n$-digits number where $1$ and $3$ occur an odd number of times 
Let $h(n)$ be the number of $n$-digits number with each digit odd, where the digits $1$ and $3$ occur an odd number of times. Find a recurrence relation for $h(n)$.

I have found the recurrence relation:
\begin{equation}
\begin{aligned}h(n+2) &= 6h(n+1) - 5h(n) + 2.3^n \\
h(0) &= h(1) = 0
\end{aligned} \tag{1}
\end{equation}
But the solution I tried was not quite satisfying. I got $(1)$ by using a system of recurrence.
Let $a(n),b(n),g(n)$ respectively be the number of $n$-digits number with even $3$s and odd $1$s, odd $3$s and even $1$s, and both even. Thus,
$a(n) :$ even $3$s odd $1$s
$b(n) :$ odd $3$s even $1$s
$h(n) :$ both odd
$g(n) :$ both even
I got some recurrences by considering the last digit of the number, whether it is $1,3$ or other than them and making connection between the $n$-digits and $(n-1)$digits number.
The system is
\begin{align*}
h(n) &= a(n-1) + b(n-1) + 3h(n-1)\\
b(n) &= g(n-1) + h(n-1) + 3b(n-1)\\
a(n) &= g(n-1) + h(n-1) + 3a(n-1)\\
g(n) &= a(n-1) + b(n-1) + 3g(n-1)\\
\end{align*}
with $a(0) = b(0) =h(0) = 0, g(0) = 1$. Note that $a(n)$ is, ofcourse, equal $b(n)$ since it is only swapping $1$ and $3$. From this system, I got the recurrence $(1)$. It is quite tedious.
Question: How can we proceed directly to get $(1)$ using some combinatorial argument?
The term $-5h(n)$ in $(1)$ suggests that we have to use the inclusion-exclusion principle but I just don't see how. The term $2.3^n$ also seems mysterious.
 A: The following does not derive $(1)$ from your recurrence but instead finds an explicit formula for $h_n$, which is perhaps the motivation for $(1)$.  First replace $b$ with $a$:
\begin{align}
h_n &= 2a_{n-1} + 3h_{n-1}\\
a_n &= g_{n-1} + h_{n-1} + 3a_{n-1}\\
g_n &= 2a_{n-1} + 3g_{n-1}\\
\end{align}
With generating functions $H(z)=\sum_{n \ge 0} h_n z^n$, $A(z)=\sum_{n \ge 0} a_n z^n$, and $G(z)=\sum_{n \ge 0} g_n z^n$, we obtain
\begin{align}
H(z) - 0 &= 2z A(z) + 3z H(z) \\
A(z) - 0 &= z G(z) + z H(z) + 3z A(z) \\
G(z) - 1 &= 2 z A(z) + 3z G(z) \\
\end{align}
Solving this system yields
\begin{align}
H(z) &= \frac{2 z^2}{1-9z+23z^2-15 z^3} \tag2\\
A(z) &= \frac{z}{1-6z+5 z^2} \\
G(z) &= \frac{1-6z+7 z^2}{1-9z+23z^2-15 z^3} \\
\end{align}
Now $(2)$ implies that
\begin{align}
H(z) &= \frac{2 z^2}{(1-z)(1-3z)(1-5z)} \\
&= \frac{1/4}{1-z}-\frac{1/2}{1-3z}+\frac{1/4}{1-5z} \\
&= \frac{1}{4}\sum_{n\ge 0}z^n-\frac{1}{2}\sum_{n\ge 0}(3z)^n+\frac{1}{4}\sum_{n\ge 0}(5z)^n
\end{align}
which immediately yields explicit formula
$$h_n = \frac{1}{4}-\frac{1}{2}\cdot 3^n+\frac{1}{4}\cdot 5^n = \frac{1-2\cdot 3^n+5^n}{4}.$$
If a recurrence relation is required, note that the denominator of $(2)$ implies that $$h_n = 9h_{n-1}-23h_{n-2}+15h_{n-3}.$$
