Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I posted this question on MathStackexchange (http://math.stackexchange.com/questions/639878/asymptotics-of-partitions-in-at-most-n-parts-bounded-by-r) some time ago, but it did not receive any answer, so that I thought of reposting it here.

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 $G_{n+2s,2s}(t)$.

share|improve this question

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.