# Thread:

1. ## Number Theory

An unrestricted partition (order does not count, equality of sizes is ok) partition is called self-conjugate if it is identical with its conjugate. E. g 8 = 4 + 2 + 1 + 1.

Show that the number of self conjugate unrestricted partitions of n is equal to the number of partitions of n into distinct odd parts

(optional) express this result as an identity of generating functions.

2. Originally Posted by Waikato
An unrestricted partition (order does not count, equality of sizes is ok) partition is called self-conjugate if it is identical with its conjugate. E. g 8 = 4 + 2 + 1 + 1.

Show that the number of self conjugate unrestricted partitions of n is equal to the number of partitions of n into distinct odd parts

(optional) express this result as an identity of generating functions.
A partition $\lambda$ is self-conjugate if $\lambda=\lambda'$ in terms of Ferrers diagram.

The bijection can be shown graphically. See wiki.

The generating function for this is $\prod_{\text{k=odd}}(1+x^k)=(1+x)(1+x^3)(1+x^5) \cdots$.

3. ## partition

Hello,

Have you shown that the number of self conjugate unrestricted partitions of n is equal to the number of partitions of n into distinct odd parts?

4. Originally Posted by Waikato
Hello,

Have you shown that the number of self conjugate unrestricted partitions of n is equal to the number of partitions of n into distinct odd parts?

Yes. If your definition of an "unrestricted" partition is simply a partition without any further constraint being given.

See the claim in the above link:

Claim: The number of self-conjugate partitions is the same as the number of partitions with distinct odd parts.

For instance, (5, 5, 4, 3, 2) |- n, where n=19, is a self-conjugate partition. It converts into the 9+7+3=19. Read the link I provided. It will give you some explanations graphically.

The link I provided only shows the sketch of the proof. I encourage you to make a full proof on your own by using the sketch of the proof in the link.
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