I have a problem with these two questions:
- Let $P_E(n)$ be the number of partitions of $n$ with an even number of parts, $P_O(n)$ the number of partitions of n with an odd number of parts, and $P_D(n)$ the number of partitions of $n$ with distinct odd parts. Show combinatorially that $P_E(n)- P_O(n) = (-1)^n P_D(n)$
- Show the following equalities.
$\Pi_{i \ge 1}(1-xq^i)^{-1} = \sum_{k \ge 0} {x^k q^k}/{(1-q)(1-q^2)...(1-q^k)}$
<p>$\Pi_{i \ge 1}(1+xq^i) = \sum_{k \ge 0} {x^k q^\binom{k+1}{2}}/{(1-q)(1-q^2)...(1-q^k)}$</p>
What I know is:
Penthagonal Number Theorem(proved it by a bijective proof)
Number of partitions with odd parts, distinct parts
Generating function for $P(n)$ and for the partitions with distinct parts
I don't know much about generating functions and partitions, but I happened to take algebraic combinatorics. Those look quite easy but I couldn't go any further since my basic knowledge is rather shallow.
For #2, I tried to use the identity:
$\sum_{k \ge 0} {1}/{(1-q)(1-q^2)...(1-q^k)} = \sum_{n \ge 0}P_{\le k}(n)q^n$
I just plugged it in the equality and changed the order of the summation, but the result was something not I've wanted. For the second part of the generating function, I thought about using some $q$-binomial coefficient but was rather tough for me now.
I think #1 maybe I can do it only by having some hint about the required bijection.
Thanks!
$\prod_{i \geq 1}\frac{1}{1-qz^i} = \sum_{k \geq 0} \sum_{n \geq 0} p(n; k)z^n q^k$ where $p(n; k)$ is the number of integer partitions of $n$ into exactly $k$ parts.
I'm not sure how to show the first (or second) equations in (2), though.
Any help? Should I create my own question for this?
– Tyler Durden Oct 25 '16 at 02:31