Find the coefficient of $x^{100}$ in $\frac{x^4}{\prod_{i=1}^{4}(1-x^{i})}$ Find the coefficient  of $x^{100}$ in $$\frac{x^4}{\prod_{i=1}^{4}(1-x^{i})}$$
This problem came from finding  total number of 4-partition of an integer 100.
One uses mathematica and it maybe easy, but I want to find more Mathematical proof.
f[m_, n_] := 
  Coefficient[Series[x^m/Product[(1 - x^i), {i, 1, m}], {x, 0, n}], x,
    n];
f[4, 100]

$7153$ is obtained. More general:
FullSimplify[
 SeriesCoefficient[x^4/Product[(1 - x^i), {i, 1, 4}], {x, 0, n}], 
 Assumptions -> Element[n, PositiveIntegers]]


$$
\frac{1}{288} \left(-32 U_n\left(-\frac{1}{2}\right)+(n+1) \left(2 n
   (n+2)+9 (-1)^n-13\right)+36 \cos \left(\frac{\pi  n}{2}\right)\right)$$

Partial fractions:
$$\begin{split}
\frac{288x^4}{(1-x)(1-x^2)(1-x^3)(1-x^4)}
&=\frac{288x^4}{(1-x)^4(1+x)^2(1+x+x^2)(1+x^2)}\\
&=\frac{12}{(1-x)^4} - \frac{12}{(1-x)^3} + \frac{13}{(1-x)^2} \\
&\quad + \frac{9}{(1+x)^2} - \frac{32}{1+x+x^2} + \frac{36}{1+x^2} \\
&=\frac{12}{(1-x)^4} - \frac{12}{(1-x)^3} + \frac{13}{(1-x)^2} \\
&\quad + \frac{9}{(1+x)^2} - \frac{32(1-x)}{1-x^3} + \frac{36(1-x^2)}{1-x^4}
\end{split}$$
Taking coefficients:
$$\begin{split}
[x^n]\frac{288x^4}{(1-x)(1-x^2)(1-x^3)(1-x^4)} &=
12\binom{n+3}{4}-12\binom{n+2}{3}+13\binom{n+1}{2} \\
&\quad + 9(-1)^n\binom{n+1}{2}  -32([3\mid n]-[3\mid(n-1)]) + 36([4\mid n]-[4\mid(n-2)])
\end{split}$$
