# Partitions without twice odd numbers and where every odd number appears at most once

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 $A$.

I can't seem to find the generating function for either of these.

• Do you have a guess for either generating function? or do you know the generating function for partitions with similar restrictions? – Greg Martin Sep 27 '14 at 0:58

The generating function for partitions where no odd number appears more than once is $$\prod_{k\ge 1} \frac{1}{1-z^{2k}} \prod_{k\ge 0} (1+z^{2k+1}).$$
The number of partitions containing no element of $A$ is $$\prod_{k\ge 1} \frac{1}{1-z^k} \prod_{k\ge 0} (1-z^{4k+2}).$$ Re-write this as $$\prod_{k\ge 1} \frac{1}{1-z^{2k}} \prod_{k\ge 0} \frac{1}{1-z^{2k+1}} \prod_{k\ge 0} (1-z^{4k+2}).$$
Finally observe that $$\frac{1-z^{4k+2}}{1-z^{2k+1}} = 1 + z^{2k+1}$$ so this becomes $$\prod_{k\ge 1} \frac{1}{1-z^{2k}} \prod_{k\ge 0} (1+z^{2k+1})$$ which is the same as the first generating function.
• Can you explain how you re-wrote the generating function for "number of partitions containing no element of A" into $\prod_{k\ge 1} \frac{1}{1-z^{2k}} \prod_{k\ge 0} \frac{1}{1-z^{2k+1}} \prod_{k\ge 0} (1-z^{4k+2}).$ – qs13 Oct 16 '20 at 4:11