$ A = x + y + z$, number of solutions in $Z$ if $x, y, z$ are bounded in intervals For the equation $x + y = A$, it's easy, when you notice that when iterating over all possible $x$, the number of solutions for $y$ is $0$ at the beginning, then increases by $1$, then stays constant, then decreases by $1$, and at the end $0$. This can be calculated in $O(1)$.
$l_1 < a < u_1$, $l_2 < b < u_2$, $l_3 < c < u_3$.
I'm looking for a general answer for all $u_i$, $l_i$, $A$ ($i = 1, 2, 3$).
I assume this problem is just as easy, but it's hard to find the formula, since i have to think in $3$ dimensions instead of $2$.
example: $2 < x < 5$, $1 < y < 5$, $3 < z < 7$, $A = 11$, the answer is 5.

The number of non-negative integer solutions to $x+y+z=11$ where $2<x<5, 1<y<5, 3<z<7$ is equal to the number of integer solutions to $a+b+c=2$ where $0\leq a\leq 1$, $0\leq b\leq 2$, $0\leq c\leq 2$ by setting $a=x-3$, $b=y-2$, $c=z-4$.
In this case it is easy to count the number of solutions: $\binom{2+3-1}{3-1}-1=\binom{4}{2}-1=5$, since there is only one 'bad' option: $a=2,b=0,c=0$.
