# 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

• If $n$ is a sum of an even number of odd parts, then $n$ must be even. So I think your two types of partitions need a different definition. For example $3=2+1$ is a partition of $3$ into an even number of distinct parts, but there is no partition of $3$ into an even number of odd parts. – coffeemath Jul 19 '13 at 18:45
• By the way your proof using generating functions is the standard one for showing # partitions into odd parts = # partitions into distinct parts. However that result does not imply anything about the numbers of partitions into even number of odd parts compared to even number of distinct parts. In fact these last counts do not come easily from generating functions, and do not seem related anyway. – coffeemath Jul 19 '13 at 23:58
• @coffeemath: That verifies my initial hunch ,that this proof is not enough. Any hints how can I prove the claim ,then ? – JAN Jul 20 '13 at 3:54
• Well I think my remark in the first comment above shows your equality is not right as stated. In my opinion it would be best if you could include in your question a few examples of the two types of partitions for some values of $n$ which turn out to be the same in number. That way someone might see what is supposed to be equal to what. – coffeemath Jul 20 '13 at 7:54
• @coffeemath: You're correct , fixed . It was supposed to be "distinct parts ,number of ODD parts is even" . – JAN Jul 20 '13 at 9:02

## 1 Answer

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.