Denote:
- $p_k(n)$ as the partition of integer $n$ into $k$ parts
- $q_k(n)$ as the partition of integer $n$ into $k$ different parts
It was proven by Erdos-Lehner that $p_k(n)\sim \frac{1}{k!}\binom{n-1}{k-1}$ for $k=o\left(n^{1/3}\right)$ , I want to employ the same principle for $q_k(n)$ since I saw some articles that claim the same for $q_k(n)$ (i.e. $q_k(n)\sim \frac{1}{k!}\binom{n-1}{k-1}$) and I want to prove it myself. I know that : $$\frac{1}{k!}\binom{ n-\binom{k}{2}-1}{ k-1}\leq q_{k}\left(n\right)\leq\frac{1}{k!}\binom{n-1}{k-1}$$ from the book "The Theory of Partitions" by George E. Andrews. So I want to prove that: $$\binom{ n-\binom{k}{2}-1}{k-1}\sim\binom{n-1}{k-1}$$
as $k=o\left(n^{1/3}\right)$ but I don't know exactly how to approach this, I tried to mimic the proof in the book for $q_k(n)$ (trying something else) but without success.
Example where the asymptotic analysis was mentioned AN ASYMPTOTIC FORMULA IN THE THEORY OF PARTITIONS
Here's a sketch that I hope will get you the rest of the way. (Very well asked question by the way!)
To prove that $\displaystyle\binom{n-\binom{k}{2}-1}{k-1}\sim\binom{n-1}{k-1}$, we need to prove that the quotient tends to $1$. The quotient is \begin{align*} \binom{n-\binom{k}{2}-1}{k-1}\bigg/\binom{n-1}{k-1} &= \frac{(n-\binom k2-1)(n-\binom k2-2)\cdots(n-\binom k2-(k-1))}{(n-1)(n-2)\cdots(n-(k-1))} \\ &= \prod_{j=1}^{k-1} (n-{\textstyle\binom k2}-j) \prod_{j=1}^{k-1} (n-j)^{-1} \\ &= \prod_{j=1}^{k-1} \biggl( 1-\frac{\binom k2+j}n \biggr) \biggl( 1-\frac jn \biggr)^{-1}. \end{align*} It suffices to show that the logarithm of this product tends to $0$; the logarithm is $$ \sum_{j=1}^{k-1} \biggl( \log \biggl( 1-\frac{\binom k2+j}n \biggr) - \log \biggl( 1-\frac jn \biggr) \biggr), $$ and now an approximation such as $\log(1-\delta)\sim-\delta$ should take you the rest of the way given the assumption $k=o(n^{1/3})$.