What is the series $ \dfrac{1}{\left( x-{y}^{2} \right) \left(1- yx \right) \left( {x}^{2}-y \right)} $ expansion? Consider the series expansion
$$
\frac{1}{\left( x-{y}^{2} \right)  \left(1- yx \right)  \left( {x}^{2}-y \right)}=P_0(y)+P_1(y)x+P_2(y)x^2+\cdots+P_n(y) x^n+\cdots.
$$
For small $n$ we have
\begin{gather*}
P_0(y)=\frac{1}{y^3},\\
P_1(y)={\frac {{y}^{3}+1}{{y}^{5}}},\\
P_2(y)={\frac {{y}^{6}+2\,{y}^{3}+1}{{y}^{7}}},\\
P_3(y)={\frac {{y}^{9}+2\,{y}^{6}+2\,{y}^{3}+1}{{y}^{9}}},\\
P_4(y)=\frac{{y}^{12}+2\,{y}^{9}+3\,{y}^{6}+2\,{y}^{3}+1}{y^{11}},\\
P_5(y)={\frac {{y}^{15}+2\,{y}^{12}+3\,{y}^{9}+3\,{y}^{6}+2\,{y}^{3}+1}{{y}^{
13}}}.
\end{gather*}
Question: What is a close expression for $P_n(y)$?
My attempt.  The $P_n(y)$  has  a form
$$
 P_n(y)=\frac{1}{y^{2n+3}}(c_{n,1}+c_{n,2}y^3+c_{n,3}y^{6}+\cdots+с_{n,k} y^{3(k-1)}+\cdots+c_{n,n+1}y^{3n})
$$
for some  unknown sequence $c_{n,k}.$
By empirical way I have found that
$$
c(n,k)= \left \{
\begin{array}{l}
\min(n{-}k+2,k) , 1 \leq    k \leq n+1, \\
\\
0,  \text{ otherwise, }
\end{array}
\right.
$$
but I can't find a rigorous proof.
Any help?
 A: Your expression is a product of three geometric series (first two multiplied by $-1$ which does not affect the final product):
$$
\frac{1}{y-x^2}=\frac{1}{y}\frac{1}{1-\frac{x^2}{y}}=\frac{1}{y}+\frac{1}{y^2}x^2+\dots+\frac{1}{y^{n+1}}x^{2n}+\dots\\
\frac{1}{y^2-x}=\frac{1}{y^2}\frac{1}{1-\frac{x}{y^2}}=\frac{1}{y^2}+\frac{1}{y^4}x+\dots+\frac{1}{y^{2n+2}}x^{n}+\dots\\
\frac{1}{1-xy}=1+xy+x^2y^2+\dots+x^{n}y^n+\dots\\
$$
The product of the first two is
\begin{align}
\sum_{n=0}^{\infty}(\sum_{k=0}^{\lfloor n/2 \rfloor}\frac{1}{y^{k+1}}\frac{1}{y^{2(n-2k)+2}})x^n
&=\sum_{n=0}^{\infty}(\sum_{k=0}^{\lfloor n/2 \rfloor}\frac{1}{y^{2n-3k+3}})x^n
\end{align}
Multiplying with the third one we get
\begin{align}
\frac{1}{y-x^2}\frac{1}{y^2-x}\frac{1}{1-xy}
&=\sum_{n=0}^{\infty}\sum_{i=0}^{n}\sum_{k=0}^{\lfloor i/2 \rfloor}\frac{1}{y^{2i-3k+3}}y^{n-i} x^n\\
\end{align}
So the value you want is
\begin{align}
P_n(y)=\sum_{i=0}^{n}\sum_{k=0}^{\lfloor i/2 \rfloor}\frac{1}{y^{2i-3k+3}}y^{n-i}
&=y^{n-3}\sum_{i=0}^{n}\sum_{k=0}^{\lfloor i/2 \rfloor}\frac{1}{y^{3i-3k}}.
\end{align}
For fixed $t$, how many ways we can choose $i,k$ to get $t=i-k$, subject to $0\leq i \leq n$, $0 \leq k \leq i/2$? Since $k=i-t$ we have $0 \leq i-t \leq i/2$ we get $i/2 \leq t \leq i \leq n$. This works in the other direction as well, for any $i$ with $i/2 \leq t \leq i$ we set $k=i-t$ and the original conditions are satisfied. So the question reduces to how many $i$ exist such that $i/2 \leq t \leq i \leq n$? We can see that if $t \geq n/2$, then any $t \leq i \leq n$ will work, hence there is $n-t+1$ of these. If on the other hand $t < n/2$, then any $i/2 \leq t \leq i$ will work, so $t\leq i \leq 2t$ and hence we have $t+1$ possibilities. We can combine these as $\min(t+1,n-t+1)=\min(t,n-t)+1$. So the value simplifies to
\begin{align}
\boxed{P_n(y)=\sum_{t=0}^{n}\frac{\min(t,n-t)+1}{y^{3t-n+3}}}
\end{align}
