Relation between binomial coefficient and partitions with largest part

Question

Denote by $p(n,k)$ the number of partitions of a natural number $n$ such that the largest part in them is $k$ (can be multiple occurrences or single, doesn't matter). Is there a relation between $p(n,k)$ and $\binom{n}{k}$? Intuition says $p(n,k) \leq \binom{n}{k}$

Answer 1

We have $$\sum_{n\geq 0} p(n) x^n = \left(1+x+x^2+\ldots\right)\left(1+x^2+x^4+\ldots\right)\cdots = \prod_{m\geq 1}\frac{1}{1-x^m} $$

$$\begin{eqnarray*}\sum_{n\geq 0}p(n,k)x^n &=& \left(1+x+x^2+\ldots\right)\left(1+x^2+x^4+\ldots\right)\cdots\left(x^k+x^{2k}+x^{3k}+\ldots\right)\\&=&x^k\prod_{m=1}^{k}\frac{1}{1-x^m}.\end{eqnarray*} $$ On the other hand, if we transpose the Ferrer diagram of a partition whose largest part has size $k$, we get a partition with exactly $k$ components$^{(*)}$, hence $$p(n,k)=[y^k x^n]\prod_{m\geq 1}\left(1+y x^m+y x^{2m}+y x^{3m}+\ldots\right)=[y^k x^n]\prod_{m\geq 1}\frac{1+x^m(y-1)}{1-x^m}. $$ On the other hand the coefficient of $x^{n-k}$ in $\prod_{m=1}^{k}\frac{1}{1-x^m}=f_k(x)$ can be found by standard means. $f_k(x)$ is a meromorphic function with poles along $S^1$; the most relevant of them is the pole of order $k$ at $x=1$. Since by stars and bars $$ \frac{1}{(1-x)^k} = \sum_{h\geq 0}\binom{k-1+h}{k-1}x^{h} $$ it is not surprising that $p(n,k)$ is close to $\binom{n-1}{k-1}$, but the other poles of $f_k(x)$ (at $-1,\omega,\omega^2,i,-i,\ldots$) tend to provide a significative perturbation if $k$ (and so $n$) is large.

$(*)$ Stars and bars also gives that the number of $(a_1,a_2,\ldots,a_k)$ in $\left(\mathbb{N}+\right)^k$ such that $a_1+a_2+\ldots+a_k=n$ equals $\binom{n-1}{k-1}$.