<s>[INST] Partition Proving 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 $A$.
For example, for $n=6$, the partitions of the first type are
$$6,~~~5+1,~~~4+2,~~~3+2+1,~~~2+2+2,$$
and the partitions of the second type are
$$5+1,~~~4+1+1,~~~3+3,~~~3+1+1+1,~~~1+1+1+1+1+1,$$
and there are $5$ of each type.

On first thought, my mind is blank.  I simply do not know how to approach this problem.  Solutions are greatly appreciated.  Thanks in advance!
 [/INST] Let $f(n)$ be the partitions of $n$ with at most one occurrence of each odd number, and $F(z)=\sum_{n} f(n)z^n$. 
Then $$\begin{align}F(z) &= (1+z)(1+z^2+z^4+\cdots)(1+z^3)(1+z^4+z^8+\cdots)\cdots \\&=\frac{(1+z)(1+z^3)(1+z^5)\cdots}{(1-z^2)(1-z^4)(1-z^6)\cdots}
\end{align}$$
Now pair off the terms $\frac{1+z^{2k+1}}{1-z^{4k+2}}=\frac{1}{1-z^{2k+1}}$, and you get:
$$\prod_{j\notin A}\frac{1}{1-z^j}$$
And this expands to give you the generating function for counting partitions of $n$ with no instance from $A$. 
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