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Let $\bar p_k(n)$ be the number of partitions of $n$ with largest part at most $k$ (equivalently, into at most $k$ parts). Is there an elementary formula for the asymptotic behavior of $\bar p_k(n)$ as $n \rightarrow \infty$ that still involves $k$ in a nontrivial way?

I would also like to know of any references that discuss the numbers $\bar p_k(n)$. Browsing Stanley v.1 isn't much help, though it's possible that I am unaware of a section in the later parts or in v.2.

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up vote 3 down vote accepted

The number of compositions of $n$ into $k$ non-negative parts is ${n+k-1 \choose n}$. If $n \gg k^2$ then these parts are usually all different, and if so we could divide the number of compositions by $k!$ to get partitions, so the number of partitions of $n$ with at most $k$ parts is greater than but asymptotic to $$\dfrac{n+k-1 \choose n}{k!} = \dfrac{(n+k-1)!}{n! \; (k-1)! \; k!} $$

For large $n$ and small $k$, this is clearly much the same as Gerry Myerson's answer. For example, the number of partitions of 10,000 with largest part at most 10 (equivalently, into at most 10 parts), this Java applet gives 778,400,276,435,728,381,405,745 as the true value, i.e. about $7.78\times 10^{23}$ while Gerry's expression gives about $7.59\times 10^{23}$ and mine gives about $7.63\times 10^{23}$.

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The number of partitions of $n$ with largest part at most $k$ is asymptotic, as $n\to\infty$, to $n^{k-1}/((k-1)!k!)$. This is a special case of a formula of Sylvester. Sylvester's theorem appears, with proof, as Theorem 4.2.1 in J L Ramirez Alfonsin, The Diophantine Frobenius Problem. My guess is that the special case goes back earlier, maybe even to Euler.

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