Coefficients of the polynomial? I have the following function
$$ p_n(x,y) = \sum_{i=0}^n c_{n}(i)~x^{n-i}y^i . $$
I know that first couple of $n$'s have a form
$$
n=1, ~~~ p_1(x,y) = 3 x + y\, , \\
n=2, ~~~ p_2(x,y) = 15 x^2 + 10 x y + 3 y^2\, , \\
n=3, ~~~ p_3(x,y) = 105 x^3 + 105 x^2 y + 63 x y^2 + 15 y^3\, , \\
n=4, ~~~ p_4(x,y) = 945 x^4 + 1260 x^3 y + 1134 x^2 y^2 + 540 x y^3 + 105 y^4\, ,\\
n=5, ~~~ p_5(x,y) = 10395 x^5 + 17325 x^4 y + 20790 x^3 y^2 + 14850 x^2 y^3 + 
 5775 x y^4 + 945 y^5\, .
$$
How do I figure out the general expression for $c_{n}(i)$ for arbitrary $n$ and $i$?
I managed to figure out edge cases, when $i=0$ and $i=n$, where
$$ c_n(0) = (2 n + 1)!!,~~{\rm while}~~ c_n(n) = (2 n - 1)!! ~.$$

Thanks to the information provided in OPs comment we can show
\begin{align*}
\color{blue}{c_n(i)=\frac{2n+1}{2i+1}\,\frac{(2n)!}{2^nn!}\binom{n}{i}\qquad\qquad 0\leq i\leq n}\tag{1}
\end{align*}
We are looking according to OPs comment for the coefficients $c_i(n)$ of the bivariate polynomial
\begin{align*}
p_n(x,y)=\sum_{i=0}^nc_n(i)x^{n-i}y^i:=(2n+1)!!\lim_{p\to 0}\left(x-y\frac{d^2}{dp^2}\right)^n\left(\frac{\sin p}{p}\right)\tag{2}
\end{align*}

We obtain from (2)
\begin{align*}
\color{blue}{\lim_{p\to 0}}&\color{blue}{\left(x-y\frac{d^2}{dp^2}\right)^n\left(\frac{\sin p}{p}\right)}\\
&=\lim_{p\to 0}\sum_{j=0}^n\binom{n}{j}x^{n-j}(-1)^j\left(y\frac{d^2}{dp^2}\right)^{j}\frac{\sin p}{p}\tag{3.1}\\
&=\lim_{p\to 0}\sum_{j=0}^n\binom{n}{j}x^{n-j}y^j(-1)^j\frac{d^{2j}}{dp^{2j}}\frac{\sin p}{p}\tag{3.2}\\
&=\lim_{p\to 0}\sum_{j=0}^n\binom{n}{j}x^{n-j}y^j(-1)^j
\frac{d^{2j}}{dp^{2j}}\sum_{k=0}^{\infty}\frac{p^{2k}}{(2k+1)!}(-1)^k\tag{3.3}\\
&=\lim_{p\to 0}\sum_{j=0}^n\binom{n}{j}x^{n-j}y^j(-1)^j
\sum_{k=0}^{2j}\frac{(2k)^{\underline{2j}}p^{2k-2j}}{(2k+1)!}(-1)^k\tag{3.4}\\
&=\sum_{j=0}^n\binom{n}{j}x^{n-j}y^j(-1)^j\frac{(2j)^{\underline{2j}}}{(2j+1)!}(-1)^j\tag{3.5}\\
&\,\,\color{blue}{=\sum_{j=0}^n\binom{n}{j}\frac{1}{2j+1}}x^{n-j}y^j\tag{3.6}
\end{align*}

Comment:

*

*In (3.1) we apply the binomial theorem.


*in (3.2) we can separate the factor $y^j$ since it does not affect the derivation of the variable $p$.


*In (3.3) we use the series expansion of $\sin x$.


*In (3.4) we differentiate $2j$ times and use the falling factorial notation $q^{\underline{j}}=q(q-1)\cdots(q-j+1)$. We also set the upper limit of the sum to $2j$ since other terms cancel.


*In (3.5) we do the limit operation which leaves us with the term with index $k=j$ only and we make some simplifications in the last step noting that $q^{\underline{q}}=q!$.

Denoting with $[y^i]$ the coefficient of $y^i$ in a series we finally obtain from (2) and (3.6)
\begin{align*}
\color{blue}{c_n(i)}&=[x^{n-i}y^i]p_n(x,y)\\
&=[x^{n-i}y^i](2n+1)!!\sum_{j=0}^n\binom{n}{j}\frac{1}{2j+1}x^{n-j}y^j\\
&=(2n+1)\frac{(2n)!}{(2n)!!}\binom{n}{i}\frac{1}{2i+1}\\
&\,\,\color{blue}{=\frac{2n+1}{2i+1}\,\frac{(2n)!}{2^nn!}\binom{n}{i}}
\end{align*}
and the claim (1) follows.

