0
$\begingroup$

Let $A=\{2,6,10,14,\ldots\}$ be the set of integers that are twice an odd number.

Prove that, for every positive integer $n$, the number of partitions of $n$ in which no odd number appears more than once is equal to the number of partitions of $n$ containing no element of $A$.

Not sure where to start.Generating Functions?

$\endgroup$

2 Answers 2

Reset to default
3
$\begingroup$

Hint: Create an explicit bijection between the two sets of partitions.

A partition that satisfies both conditions will map to itself. A partition that satisfies only one of them can map to one that satisfies the other.

$\endgroup$
1
  • $\begingroup$ What exactly is this bijection? $\endgroup$
    – user118161
    Commented Jul 10, 2020 at 8:22
1
$\begingroup$

$${1+x\over1-x^2}{1\over1-x^4}{1+x^3\over1-x^6}{1\over1-x^8}\dots={1\over1-x}{1\over1-x^3}{1\over1-x^4}{1\over1-x^5}{1\over1-x^7}\dots$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .