# Asymptotics of partitions in at most n parts, bounded by r

For every positive integers $n,r,w$ define $$p_w(n,r)=\#\{ (i_1,...,i_r) | \, 0\leq i_1 \leq \dots \leq i_r\leq n, \, i_1+\dots+i_r=w\}$$ as the number of partitions of $w$ in at most $r$ piece bounded by $r$. Equivalently, this is the number of Young tableaux for $w$ that can be inscribed into an $r\times n$ rectangle.

Suppose now that $r=2s$ for a certain positive integer $s$. Then, for every fixed $s$ and $n$, I think that the the biggest of the numbers $p_w(n,2s)$ is $p_{ns}(n,2s)$, even if I have not proven this.

What I would like to know, then, would be the limit

$$\lim_{n\to +\infty} \frac{\sum_{w=0}^{2ns}p_w(n,2s)}{p_{ns}(n,2s)}$$

i.e. the ratio between the total numbers of partitions and the biggest number of partitions.

Even better would be to know the asymptotic behaviour of $$\sum_{w=0}^{2ns}p_w(n,2s) \qquad p_{ns}(n,2s)$$ as $n\to +\infty$.

What I have tried:

I know that the generating function for the numbers $p_w(n,2s)$ is the Gaussian polynomial

$$G_{n+2s,2s}(t)=\frac{(1-t^{n+2s})\dots(1-t^{n+1})}{(1-t^{2s})\dots(1-t)}$$

and from this, at least in the case that $n$ is divided by all numbers $1,\dots,2s$ it can be seen that

$$G_{n+2s,2s}(t)=\prod_{k=1}^{2s} P_k(t)$$

where $P_k(t)=\frac{t^{n+k}-1}{t^k-1}=\sum_{h=0}^{\frac{n}{k}}t^{hk}$. So that

$$\sum_{w=0}^{2s}p_w(n,2s) = G_{n+2s,2s}(1) = \prod_{k=1}^{2s}(\frac{n}{k}+1)\sim \frac{n^{2s}}{(2s)!}$$

But now to know $p_{ns}(n,2s)$ I should know something about the $ns$th-derivative of $G_{n+2s,2s}(t)$.

• @Danielle: the generating function for the numbers pw(n,2s), where the first factor in the denominator has a typo btw, yields a closed form in terms of the q-Pochhammer function. – Wouter M. Jan 17 '14 at 23:05
• @Danielle: Mathematica gets QPochhammer[t^n,t,1+q]/(1-t^n)/QPochhammer[t,t,q] with q from 2 to 2n in steps of 2. Its derivatives look awfully messy though. – Wouter M. Jan 17 '14 at 23:15
• @Wouter: yes, I had seen something related to the q-Pochammer function as well, but I did not manage to obtain a result from it. Thanks for the segnalation of the typo, it should be fixed now. – Daniele A Jan 18 '14 at 15:24