Number of solutions via generating function How to find generating function for number of solutions
$x_1+x_2+x_3=n$ in set of positive integers such that $x_1 \ge x_2 \ge x_3$ and $x_1<x_2+x_3$. 

Hint: First rewrite $x_2 = x_3 + d_{2,3}$, with $d_{2,3} \geq 0$ and $x_1 = x_2 + d_{1,2} = x_3 + d_{2,3} + d_{1,2}$ with $d_{1,2} \geq 0$.  In terms of $d_{1,2} \geq 0$, $d_{2,3} \geq 0$, $x_3  \geq 1$, we have arranged for all three variables to be (so far) independent.
Now we need to implement $x_1 < x_2 + x_3$, or, what is the same thing, $x_3 + d_{2,3} + d_{1,2} < x_3 + d_{2,3} + x_3$, which is just $d_{1,2} < x_3$.   Set $x_3 = d_{1,2} + e$ with $e \geq 1$.  Now we have arranged for the problem to be in terms of the independent variables $d_{1,2} \geq 0$, $d_{2,3} \geq 0$, and $e \geq 1$.
Rolling back, we see that \begin{align*}
n &= x_1 + x_2 + x_3  \\
    &= (x_3 + d_{2,3} + d_{1,2}) + (x_3 + d_{2,3}) + x_3  \\
    &= 3 x_3 + 2 d_{2,3} + d_{1,2}  \\
    &= 3(d_{1,2} + e) + 2 d_{2,3} + d_{1,2}  \\
    &= 4d_{1,2} + 2d_{2,3} + 3e  \text{.}
\end{align*}
So the given problem is equivalent to finding a partition of $n$ into zero or more $2$s, one or more $3$s, and zero or more $4$s.
