I have a complicated looking function
$$ f_n^{i,l}\left(z\right) = \sum _{k=0}^{2 l} \sum _{m=0}^{k/2} \binom{2 n}{k-2 m} \binom{l-n}{m} y^{k-m} x^{l+n-(k-m)} z^{k-m} \sum _{j=0}^i \frac{\binom{2 i}{2 j} \binom{i+j-1/2}{2 i}}{j+m+n+1/2}\\ \times\,_2F_1\left((1-k)/2,-k/2;-j-m-n+1/2;-(1-z)/z\right), $$ where I know that $i<l$ and $n<l$. Investigating this function it turns out I can write it simply as polynomial in $z$;
$$ f_n^{i,l}\left(z\right) = \sum_{m=0}^{n+i} c^{i,l}_{n, m} z^{m}. $$
I know I can get the coefficients $c^{i,l}_{n, m}$ by taking the derivatives of $f_n^{i,l}\left(z\right)$, for example I get:
$$ c^{0,0}_{0,0} = 2, ~~~ c^{0,1}_{0,0} = \frac{2}{3} (3 x+y),~~~ c^{0,1}_{1,0} = \frac{2 x^2}{3}, ~~~ c^{0,1}_{1,1} = \frac{2}{3} y (2 x+y), \\ c^{1,1}_{0,0} = -\frac{y}{30},~~~ c^{1,1}_{1,0} = \frac{x^2}{15}, ~~~ c^{1,1}_{1,1} = \frac{1}{30} y (4 x-y), $$
I would like to get the closed expression for $c^{i,l}_{n, m}$. I tried by just guessing from first several terms and I was able to get close (up to numerical factors) in the case when $i=0$, $c^{0,l}_{n, m}$. I got something of the form
$$ c^{0,l}_{n,m} = (y/x)^m \underset{k_1+k_2=l+n}{\sum_{k_1=0}^{l+n+m} \sum_{k_2=0}^{l-n+m}} n^{l,n,m}_{k_1,k_2} x^{k_1}y^{k_2}, $$
where $n^{l,n,m}_{k_1,k_2}$ are numeircal factors. I'd like to generalize this for $c^{i,l}_{n,m}$. Is that doable?