unique integer partitions Let me motivate my general question with an explicit example:
Suppose I am looking for all unique combinations of exactly three non-negative integers that sum to five. The solutions are 005, 014, 023, 113, and 122. Which means that there are five unique combinations.
Is there a way to find the $\textit{number}$ of unique combinations of exactly $k$ non-negative integers that sum to $n$? I'd rather not generate all the unique combinations and then count. I am hoping that there is a straightforward combinatoric solution to this.
Please let me know if more clarification is needed.
Thanks!
 A: For fixed $k$ and large $n$ this is pretty doable.  You want to find solutions to
$$x_1 + x_2 + ... + x_k = n$$
where $x_1 \ge x_2 \ge ... \ge x_k$.  Letting $y_i = x_i - x_{i+1}$ and $y_k = x_k$, this is equivalent to finding solutions to
$$y_1 + 2y_2 + ... + ky_k = n$$
where $y_i \ge 0$.  If $p_k(n)$ denotes the number of ways to do this, it follows by a standard generating function trick that
$$\sum_{k \ge 0} p_k(n) x^n = \frac{1}{(1 - x)(1 - x^2)...(1 - x^k)}.$$
In principle one can find the partial fraction decomposition of the RHS, allowing us to write $p_k(n)$ as a quasi-polynomial.  
A: If you need an algorithm to calculate this number, you can use the following.
Let $a_{nk}$ be an answer to your question, then it's not hard to prove that $a_{nk} = a_{n,k-1} + a_{n - k, k}$. So you can fill in the table of all $a_{nk}$ using this formulae.
A: Nothing wrong with Qiaochu Yuan's answer, but here's an orthogonal approach; for fixed $k$, calculate the first 5 or 10 $n$ values and then look up the resulting sequence at the Online Encyclopedia of Integer Sequences. 
A: For fixed k
$p_k(n) \sim {n^{k-1} \over k!(k-1)!}.$
Maybe this limitingform will be of some use to you.
