Prove : $p$(n│even number of ODD parts)=$p$(n│distinct parts ,number of ODD parts is even ) I'm trying to prove the following Integer Partition claim : 
$p$(n│even number of ODD parts) = $p$(n│distinct parts ,number of ODD parts is even) .
So I tried to prove a stronger claim : 
$p$(n│number of odd parts) = $p$(n│number of distinct parts) :
Using generating functions : 
DISTINCT PARTS = 
$$(1+x)\cdot(1+x^2 )\cdot(1+x^3 )\cdot(1+x^4 )\cdot\cdot\cdot\cdot$$
=
$$\frac {1-x^2} {1-x} \cdot \frac {1-x^4} {1-x^2} \cdot \frac {1-x^6} {1-x^3} \cdot \frac {1-x^8} {1-x^4}\cdot\cdot\cdot\cdot$$
$$ \frac {1} {1-x} \cdot \frac {1} {1-x^3} \cdot \frac {1} {1-x^5} \cdot \frac {1} {1-x^7} \cdot\cdot\cdot\cdot
$$
= ODD PARTS
But I'm not sure regarding the proof , does it really prove that even number of odd parts equals distinct parts ,where number of ODD parts is even ? 
Thanks 
 A: Given the rewritten descriptions and comments, I think this equality is not really so much a case where some new clever manipulation of generating functions or other argument is necessary, but is really an exercise to see if one can unravel the definitions. It really comes down to the already proved statement that the number of partitions into odd summands equals the number of partitions into distinct summands.
A look at either side shows the requirement of an even number of odd parts, so for either side to be nonzero, $n$ must be even. [In a proof one would say "suppose $n$ is odd, then both sides are zero because..."]
Now if $n$ is even, and we look at the right side, which is partitions of $n$ into distinct parts, it follows even without saying so that the number of odd parts will automatically be even. So the right side (in the $n$ even case) boils down simply to the number of partitions of $n$ into distinct parts.
Now look at the left side. We again can see that the number of odd parts must automatically be even, since $n$ itself is even. So here again the extra term "even number of" may be dropped, and the left side is simply the number of partitions of the (even) number $n$ into odd parts. [At the same time this makes it clear the left side should not allow even summands at all, otherwise the count on the left will exceed the right side.]
So the overall statement holds, but is really just a re-work of the already mentioned (and shown in the OP) statement about odd parts and distinct parts.
