Note: I changed the terminology somewhat; this sequence starts with $a_0$ rather than $a_1$.
Suppose we have a sequence $a_0,a_1,a_2,\dots$ whose generating function is
$$
f_a(x)=a_0+a_1x+a_2x^2+\cdots
$$
satisfying the recurrence relation
$$
a_n-P\,a_{n-1}-Q\,a_{n-2}-R\,a_{n-3}=0\iff\\
a_n=P\,a_{n-1}+Q\,a_{n-2}+R\,a_{n-3}
$$
Multiply $f_a(x)$ by the polynomial $1-Px-Qx^2-Rx^3$ to get the polynomial
$$
g(x) = b_0+b_1 x+ b_2 x^2+\cdots
$$
where for $n\geq 3$, $b_n=a_n-P\,a_{n-1}-Q\,a_{n-2}-R\,a_{n-3}$. By our recurrence relation, this means that $b_n=0$ whenever $n\geq 3$. So, we have
$$
(1-Px-Qx^2-Rx^3)f_a(x)=b_0+b_1 x+ b_2 x^2
$$
Which is to say that
$$
f_a(x)=\frac{b_0+b_1 x+ b_2 x^2}{1-Px-Qx^2-Rx^3}
$$
Where
$$
b_0 = a_0\\
b_1 = a_1 - P\,a_0\\
b_2 = a_2 - P\,a_1 - Q\,a_0
$$
Can you take it from there?
So in order to bring this back to the characteristic equation, we just need to use another little trick. Instead of writing this as a function of $x$, write it as a function of $\frac1x$. You could do this by making a substitution like $x=\frac1\omega$, but I prefer a more direct approach.
We have:
$$
f_a(x)=\frac{b_0+b_1 x+ b_2 x^2}{1-Px-Qx^2-Rx^3}
$$
With $b_1,b_2,b_3$ as defined above. From there, just divide the top and bottom by $x^3$ to get
$$
f_a(x)=\frac{b_0\left(\frac1{x}\right)^3+b_1 \left(\frac1{x}\right)^2
+ b_2 \left(\frac1{x}\right)}{
\left(\frac1{x}\right)^3-P\left(\frac1{x}\right)^2-
Q\left(\frac1{x}\right)-R}
$$
Now, suppose we have one repeated root. That is, $t^3 - Pt^2 - Qt - R=(t-r_1)(t-r_2)^2$ for roots $r_1,r_2$. We then can write the above as
$$
f_a(x)=\frac{b_0\left(\frac1{x}\right)^3+b_1 \left(\frac1{x}\right)^2
+ b_2 \left(\frac1{x}\right)}{
\left(\left(\frac1{x}\right)-r_1\right)
\left(\left(\frac1{x}\right)-r_2\right)^2}
$$
Where would you go from there? For the case of a triply repeated root, we have $t^3 - Pt^2 - Qt - R=(t-r)^3$ for the repeated root $r$. We then can write the generating function as
$$
f_a(x)=\frac{b_0\left(\frac1{x}\right)^3+b_1 \left(\frac1{x}\right)^2
+ b_2 \left(\frac1{x}\right)}{
\left(\left(\frac1{x}\right)-r\right)^3}
$$
Where would you go from there?
I started your problem from scratch and arrived to something different (I have not been able to find where there has been a difference). First, I arrived to $$A(z)=\frac{-616 z^5+1540 z^4-1198 z^3+267 z^2+55 z-23}{(z-1)^2 (2 z-1)^3 (4 z+1)}$$ which decomposes as $$A(z)=-\frac{19}{z-1}-\frac{2}{(1-2 z)^2}+\frac{5}{(z-1)^2}-\frac{2}{(2 z-1)^3}-\frac{1}{4
z+1}$$ Reworking from here, I arrived to quite messy expressions for $g_n$ which, fortunately, simplify to $$g_n=2^n (n+1) n+5 n-(-4)^n+24$$
Best Answer
Rewriting $\,2n+5 = (n+3)^2 - (n+2)^2\,$:
$$ \quad\quad a_{n+1} = \frac{(n+3)^2 - (n+2)^2}{n+2} a_{n} + \frac{(n+1)(n+3)^2}{n+2} a_{n-1} \\ \iff\quad\quad a_{n+1} + (n+2) a_n = \frac{(n+3)^2}{n+2} \big(a_n + (n+1) a_{n-1}\big) $$
With $\,b_n = a_{n} + (n+1) a_{n-1}\,$, this telescopes to:
$$ \require{cancel} b_{n+1} = \frac{(n+3)^2}{n+2} \,b_n = \frac{(n+3)^2}{\bcancel{n+2}} \,\frac{(n+2)^{\bcancel{2}}}{n+1} \, b_{n-1} = \dots = (n+3)\,\frac{(n+3)!}{2} \, b_0 $$
Once $\,b_n\,$ is determined, what's left to solve is the simpler recurrence:
$$ a_{n} + (n+1) a_{n-1} = b_n \quad\iff\quad \frac{a_n}{(n+1)!} = -\frac{a_{n-1}}{n!} + \frac{b_n}{(n+1)!} $$