Consider the inequality $x1+x2+x3+x4 ≤n$ where $x1, x2, x3, x4, n ≥ 0$ are all integers. Suppose also that $x2 ≥ 2$, $x3$ is a multiple of 4, and $1 ≤ x4 ≤ 3$. Let cn be the number of solutions of the inequality subject to these restrictions. Find the generating function for the sequence {$Cn : n ≥ 0$} and use it to find a closed formula for $Cn$.
So far I have $x2 = (x^2 + x^3 + x^4 + x^5 ... )$
$x3 = (1 + x^4 + x^8 + x^12 + ...)$
$x4 = (x + x^2 + x^3)$
$x^2/( 1- x) + 1/($1 - x^4) + (1/(1 - x) - 1 - x^4/(1-x))
I don't know how to get the coefficient.
Presumably $x_1,x_2,\dots$ are all non-negative integers, else there are clearly infinitely many solutions.
$x_1$ has no restriction beyond being a non-negative integer, so we will use $(1+x+x^2+x^3+\dots)=\frac{1}{1-x}$ in the generating function.
$x_2$ being restricted to integers greater than or equal to two implies we will use a $(x^2+x^3+x^4+\dots)=\frac{x^2}{1-x}$ term in the generating function.
$x_3$ being restricted to non-negative multiples of four implies we will use $(1+x^4+x^8+x^{12}+\dots)=\frac{1}{1-x^4}$ in the generating function.
$x_4$ being restricted to the values of $1,2$ or $3$ implies we will use $(x+x^2+x^3)$ in the generating function.
That we wished for them to add up to at most $n$, let us instead use an overflow variable $x_5=n-x_1-x_2-x_3-x_4$ so that $x_1+x_2+\dots+x_5=n$ exactly, subject to the conditions that $x_5$ is again a non-negative integer. We use another term of $(1+x+x^2+\dots)=\frac{1}{1-x}$ as a result.
Putting all of this together by multiplying the respective pieces, the generating function for the number of solutions will be:
$\sum\limits_{n=0}^\infty c_n x^n = C(x) = \frac{1}{1-x}\cdot \frac{x^2}{1-x}\cdot \frac{1}{1-x^4}\cdot (x+x^2+x^3)\cdot \frac{1}{1-x}=\frac{x^3+x^4+x^5}{(1-x)^3(1-x^4)}$
where the coefficient of $x^n$ is going to be the value of $c_n$. wolfram
I do not personally see this simplifying much more nicely, but here are the first terms of the sequence $0,0,1,4,10,19,32,50,74,104,141,\dots$
As @achillehui points out in the comments below, with the help of software, one can take this further and find it decomposes as $\frac{3}{4(1-x)^4}-\frac{15}{8(1-x)^3} + \frac{19}{16(1-x)^2}+\frac{3}{32(1-x)} - \frac{1}{32(1+x)}-\frac{1-x}{8(1+x^2)}$, yielding the simplification:
$$\begin{align} c_n &= \frac34\binom{n+3}{3} - \frac{15}{8}\binom{n+2}{2} + \frac{19}{16}(n+1) + \frac{3}{32} - \frac{1}{32}(-1)^n -\frac18\Re[(1-i)(-i)^n]\\ &= \left\lfloor\frac{(n^2-2)(2n-3)}{16}\right\rfloor \end{align}$$