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. 

So I'm currently solving this problem with the AOPS intermediate counting and probability course, and if you are too, you should probably use the message board but here is an explanation for the two values:
We start by making a generating function for partitions when no odd number appears more than once.
For the even numbers that will be in the partition, we can first form the power series for each individual number. For $2,$ we have $((x^2)^0+(x^2)^1+(x^2)^2+(x^2)^3+ \dots)$ where the highest power represents the number of $2s$ there are, so in $(x^2)^3,$ there will be a $2+2+2$ in our partition. Similarly for $4$ we have $((x^4)^0+(x^4)^1+(x^4)^2+(x^4)^3+ \dots).$ And for $6$ we have $((x^6)^0+(x^6)^1+(x^6)^2+(x^6)^3+ \dots).$ And so on and so forth for all even numbers.
Multiplying the power series together we have $((x^2)^0+(x^2)^1+(x^2)^2+(x^2)^3+ \dots) \cdot ((x^4)^0+(x^4)^1+(x^4)^2+(x^4)^3+ \dots) \cdot ((x^6)^0+(x^6)^1+(x^6)^2+(x^6)^3+ \dots) \cdot \dots$
Then for the odd numbers in our partition, we are limited by the problem condition: no odd number appears more than once. Thus, $1$ can only appear $0$ times or $1$ time so $((x^1)^0+(x^1)^1$, $3$ can also only appear $0$ or $1$ times, giving us $((x^3)^0+(x^3)^1).$ This applies to all odd numbers so we have:
$$((x^1)^0+(x^1)^1) \cdot ((x^3)^0+(x^3)^1) \cdot ((x^5)^0+(x^5)^1) \dots.$$
Now all we need to do is to multiply the two functions. So our generating function for partitions of any number when no odd number appears more than once is done:
\begin{align*}
\left(((x^2)^0+(x^2)^1+(x^2)^2+(x^2)^3+ \dots) \cdot ((x^4)^0+(x^4)^1+(x^4)^2+(x^4)^3+ \dots) \cdot ((x^6)^0+(x^6)^1+(x^6)^2+(x^6)^3+ \dots) \cdot \dots \right) \cdot \left( ((x^1)^0+(x^1)^1) \cdot ((x^3)^0+(x^3)^1) \cdot ((x^5)^0+(x^5)^1) \dots \right)
&= \left( (1+x^2+x^4+x^6+ \dots)(1+x^4+x^8+x^{12}+ \dots)(1+x^6+x^{12}+x^{18}+\dots) \dots \right) \cdot \left( (1+x)(1+x^3)(1+x^5) \dots \right) \\
&= \prod_{n=1}^{ \infty } \frac{1}{1-x^{2k}} \cdot \prod_{n=0}^{ \infty } (1+x^{2k+1}). \\
\end{align*}
Now we need to make the generating function for when the partition contains no element of $A.$
To do this, we can take the generating function for partitions with no restrictions, and then remove the power series for numbers that are part of set $A.$
Thus we have $$\frac{ (1+x+x^2+ \dots) \cdot (1+x^2+x^4+ \dots) \cdot (1+x^3+x^6+ \dots) \cdot (1+x^4+x^8+ \dots) \cdot (1+x^5+x^10+ \dots) \cdot (1+x^6+x^{12}+ \dots) \dots}{(1+x^2+x^4+ \dots) \cdot (1+x^6+x^{12}+ \dots) \dots} = \prod_{n=1}^{ \infty } \frac{1}{1-x^n} \prod_{n=0}^{ \infty } (1-x^{4n+2}). $$
