Balls and Urns Modification Question:
How many ways are there to distribute $b$ balls in $u$ urns such that none of the urns are empty and all of the urns have a distinct number of balls? I do not consider permutations of the balls in the urns to generate distinct cases, nor does permuting the urns (ie, all the balls and urns are identical).
Thoughts:
The restrictions above are similar to those defining the Stirling numbers of the second kind - except I'm also requiring that the urns have a unique number of balls in them.
I tried counting the number of cases where the final balls-and-urns state has repetitions (with the intention of subtracting this from the Stirling numbers) but I haven't had any luck.
Follow-up Question:
Now suppose that I have some constant $C\in\mathbb{N}$. How many ways are there to distribute $b$ balls in $u$ urns, where $b+u=C$, such that each urn has a unique and non-zero number of balls? This is basically the same problem as above, but now summing over several cases for $b$ and $u$ (specifically, all $b$ and $u$ such that $b+u=C$).
 A: The O.E.I.S  can provide an answer of sorts... Consider the formal power series in two indeterminates $x,y$:
$$ (1+xy)(1+xy^2)(1+xy^3)(1+xy^4) \cdots $$
The coefficient of $x^u y^b$ in the expansion will equal the number of ways to write
\begin{align*}
 b = n_1 + n_2 + \ldots + n_u && \text{where} && 0 < n_1 < n_2 < \cdots n_u.
\end{align*}
If you want to consider the urns distinct, you should multiply by $u!$ to account for order.
A: Let $p_u(b)$ denote the number of partitions of $b$ into $u$ nonzero parts, without the requirement that the parts be different. This is a well known function, counting classes of the surjections from a $b$-set to a $u$-set under permutations of both those sets. There is no closed formula for these values, but the generating series
$$
  \sum_{n\in\mathbf N}p_k(n)X^n = \frac{X^k}{(1-X)(1-X^2)\ldots(1-X^k)}
$$
that allows to compute its values easily. (The generating series more naturally counts partitions of $n$ into any number of parts each at most $k$ with at least one part of size$~k$, but by taking the conjugate partitions this translates into counting partitions into $k$ nonzero parts.) This solves the variation of your problem without the requirement that the parts (numbers of balls in each urn) be all distinct.
To accommodate for that additional requirement of distinctness, proceed as follows. Assume that $b\geq\frac{(u+1)u}2$ so that at least one solution to your problem exists (in the contrary case the answer is $0$). Now take such a solution, order the urns by (strictly) increasing number of balls, remove nothing from the first urn, one ball from the second, two balls from the third, and so forth until taking $u-1$ balls from the last urn. This removes $\binom u2$ balls in all, and leaves a solution for the problem of putting $b-\binom u2$ balls into $u$ urns without additional requirements, again distinguishing neither the balls nor the urns. Conversely given such a solution one can order the urns by weakly increasing number of balls, and then add $0,1,2,\ldots,u-1$ balls to the successive urns to find a solution to your problem. So the problems have equally many solutions, and this number is
$$f(b,u)=p_u(b-\binom u2)$$
by the definition of $p_k(n)$. From the above, the generating series by$~b$ for this is
$$
 \sum_{b\geq(u+1)u/2}f(b,u)X^b = \frac{X^{(u+1)u/2}}{(1-X)(1-X^2)\ldots(1-X^u)}.
$$
