1
vote
1answer
207 views

An infinite product: combinatorial interpretation

It is an undergraduate exercise to show that the generating function for the sequence of unrestricted integer partitions $p(n)$ is the celebrated infinite product ...
5
votes
1answer
149 views

The number of partitions between two fixed partitions

Given two partitions M and N, with $M_i \leq N_i$ for all $1\leq i\leq \max\{l(M),l(N)\}$. Is there a formula for the generating function: $$\sum_{\lambda: M_i\leq \lambda_i\leq N_i} ...
1
vote
1answer
378 views

Recurrence relation for coefficients of product of generating functions for partition numbers

It is well known that $$Z(x,q) = \prod_{n=1}^\infty\frac{1}{(1-xq^n)} = \sum_{m=0}^\infty\sum_{k=0}^m p_{m,k}x^kq^m$$ is the generating function for the number $p_{m,k}$ of partitions of $m$ in ...
8
votes
4answers
374 views

Partitions-Sum of divisors identity

A few years ago I first read about the marvelous Euler identity: $\sum_{n\in\mathbb{N}}p(n)z^n=\prod_{k\geq1}\frac{1}{1-z^k}$, where $p(n)$ is the number of partitions of $n$ ($p(0)=1$ by ...
5
votes
2answers
776 views

What are the best known bounds on the number of partitions of $n$ into exactly $k$ distinct parts?

For example, if $n = 10$ and $k = 3$, then the legal partitions are $$10 = 7 + 2 + 1 = 6 + 3 + 1 = 5 + 4 + 1 = 5 + 3 + 2$$ so the answer is $4$. By choosing $k$ random elements of $\{1,\ldots,2n/k\}$, ...
20
votes
2answers
829 views

Partitions to different parts not exceeding $n$

Consider the polynomial $(1+x)(1+x^2)\dots (1+x^n)=1+x+\dots+x^{n(n+1)/2}$, which enumerates subj. How to prove that it's coefficients increase up to $x^{n(n+1)/4}$ (and hence decrease after this)? Or ...