Generating Functions and Linear Diophantine Inequalities The following exercise is from Analytic Combinatorics by Philippe Flajolet and Robert Sedgewick, page 46. 
A $k$-composition of $n$ is an ordered $k$-tuple of non-negative integers whose sum is $n$.

Consider the class $\mathcal{F}$ of compositions of integers into four summands $(x_1, x_2, x_3, x_4)$ such that
  $$ x_1 \ge 0, \, x_2 \ge 2 x_1, \, x_3 \ge 2 x_2, x_4 \ge 2 x_3,$$ 
  where the $x_j$ are in $\mathbb{Z}_{\ge 0}$.The ordinary generating function is
  $$F(z) = \dfrac{1}{(1 − z)(1 − z^3)(1 − z^7)(1 − z^{15})}.$$
Generalize to $r \ge 4$ summands and a similar system of inequalities. Work out elementarily the OGFs corresponding to the following systems of inequalities:
  $$\{x_1+x_2 \le x_3\},\,\, \{x_1 + x_2 \ge x_3\},\,\, \{x_1+x_2 \le x_3 + x_4\}, \,\,\{x_1 \ge x_2, x_2 \ge x_3,x_3 \le x_4\}.$$
  More generally, the OGF of compositions into a fixed number of summands (in $\mathbb{Z} \ge 0$), con- strained to satisfy a linear system of equations and inequalities with coefficients in $\mathbb{Z}$, is rational; its denominator is a product of factors of the form $(1 − z^j)$.

I am looking for help developing a systematic approach to determine the generating functions of these types of inequalities. I have tried working out the first two, and my proposed OGF's are respectively
$$\dfrac{1}{(1-z)(1-z^2)^2}$$
and
$$\dfrac{1 + z + z^2}{(1-z)(1-z^2)^2}$$
However, I did not use the so-called symbolic method, instead I looked at a few elementary cases and then reasoned about what the coefficients should be in a combinatorial manner. If you are experienced with generating functions, I would truly appreciate any insight you could shed on these problems. 
Even if you approach the problem differently than Sedgewick, but can still give combinatorial interpretations of each of the generating functions corresponding to the above restricted partitions, I would accept your answer. All I really want is help building intuition on this subject.
Thank you.
 A: I am not exactly sure what it means to do the symbolic method, but I'll explain the method I would use for computing generating functions of this type by example.
Say we wanted the one for $\{x_1+x_2\leq x_3\}$.  We can introduce $\xi\geq0$ and write $x_3=x_1+x_2+\xi$, so we have as the generating function
\begin{align*}
\sum_{x_1+x_2\leq x_3} z^{x_1+x_2+x_3}
&= \sum_{x_1}\sum_{x_2}\sum_{\xi} z^{2x_1+2x_2+\xi}\\
&=\sum_{x_1}z^{2x_1}\sum_{x_2}z^{2x_2}\sum_{\xi}z^\xi\\
&=\frac{1}{(1-z)(1-z^2)^2}.
\end{align*}
Combinatorially, we could have noticed that this introduced variable shows what we are wanting to do is count partitions of $n$ in the form $n=x_1+x_2+x_3=2x_1+2x_2+\xi$, which is to say as a sum of a non-negative integer and two even non-negative integers.  Hence we multiply $1/(1-z)$ (a generating function for the number of ways of partitioning a number as a single non-negative integer) and $1/(1-z^2)$ twice (a generating function for the number of ways of partitioning a number as a single even non-negative integer).
And for $\{x_1+x_2\geq x_3\}$, we similarly introduce $\xi\geq0$ and write $x_1+x_2=x_3+\xi$ to compute
\begin{align*}
\sum_{x_1+x_2\geq x_3} z^{x_1+x_2+x_3}
&= \sum_{x_3}\sum_{\xi}\sum_{x_1+x_2=\xi+x_3} z^{2x_3+\xi}\\
&= \sum_{x_3}\left(z^{2x_3}\sum_{\xi}\sum_{x_1+x_2=x_3+\xi}z^\xi\right)\\
&= \sum_{x_3}z^{2x_3}\sum_\xi(x_3+\xi+1)z^\xi\\
&= \sum_{x_3}z^{2x_3}\left(\sum_\xi x_3z^\xi + \sum_\xi (\xi+1)z^\xi\right)\\
&= \sum_{x_3}z^{2x_3}\left(\frac{x_3}{1-z}+\frac{1}{(1-z)^2}\right)\\
&= \frac{1}{1-z}\cdot\frac{z}{(1-z^2)^2} + \frac{1}{(1-z)^2}\cdot\frac{1}{1-z^2}\\
&= \frac{1+z+z^2}{(1-z)(1-z^2)^2}.
\end{align*}
Unfortunately, I'm having a hard time producing a combinatorial description for this one.
A: For a set $A\subseteq {\mathbb N}^4$, we put 
$G(A)=\sum_{(x_1,x_2,x_3,x_4)\in A} z^{x_1+x_2+x_3+x_4}$. Then
$$
\begin{array}{lcl}
G(\lbrace x_1+x_2\leq x_3+x_4 \rbrace)
&=& \frac{1}{2}.\bigg(G({\mathbb Z}^4)+G(\lbrace x_1+x_2= x_3+x_4 \rbrace)
\bigg) \ \text{(symmetry argument)} \\
&=& \frac{1}{2}.\bigg(\frac{1}{(1-z)^4}+\sum_{t}\sum_{x_1+x_2=t}z^{x_1+x_2}
\sum_{x_3+x_4=t}z^{x_3+x_4}
\bigg)\\
&=& \frac{1}{2}.\Bigg(\frac{1}{(1-z)^4}+\sum_{t}\bigg(\sum_{x_1+x_2=t}z^{x_1+x_2}
\bigg)^2
\Bigg)\\
&=& \frac{1}{2}.\Bigg(\frac{1}{(1-z)^4}+\sum_{t}(t+1)^2z^{2t}
\Bigg)\\
&=& \frac{1}{2}.\Bigg(\frac{1}{(1-z)^4}+\sum_{t}\big(2\frac{t^2+3t+2}{2}-(t+1)\big)z^{2t}
\Bigg)\\
&=& \frac{1}{2}.\Bigg(\frac{1}{(1-z)^4}+\sum_{t}\big(2\binom{t+2}{2}-\binom{t+1}{1}\big)z^{2t}
\Bigg)\\
&=& \frac{1}{2}.\Bigg(\frac{1}{(1-z)^4}+\frac{2}{(1-z^2)^3}-\frac{1}{(1-z^2)^2}
\Bigg)\\
&=& \frac{1 - 2z + 2z^2 - 4z^3 + 5z^4 - 2z^5}{(1-z)^4(1-z^2)^3}\\
\end{array}
$$
For the last one, we can use the same ideas as in Kyle Miller’s answer :
putting $t_4=x_4-x_3,t_2=x_2-x_3,t_1=x_1-x_2$, we have
$$
\lbrace x_1 \ge x_2, x_2 \ge x_3,x_3 \le x_4 \rbrace=
\lbrace (x_3+t_1+t_2,x_3+t_2,x_3+t_4) | (t_1,t_2,t_3,t_4)\in {\mathbb N}^4 \rbrace
$$
and hence
$$
\begin{array}{lcl}
G(\lbrace x_1 \ge x_2, x_2 \ge x_3,x_3 \le x_4 \rbrace)
&=& \sum_{t_1,t_2,x_3,t_4}z^{t_1+2t_2+4x_3+t_4} \\
&=& \frac{1}{(1-z)^2(1-z^2)(1-z^4)}\\
\end{array}
$$
