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Does anyone know of an efficient way to find all self conjugates of a partition n? I know that I could guess and check using Ferrers Diagrams and following some basic rules (same number of rows as number as magnitude of first number in partition) and what not but does anyone have a better method?

Also, could you give some examples, like n = 15 and n = 16? Thanks in advance.

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The number of self-conjugate partitions of an integer $n$ is equal to the number of partitions of $n$ into distinct odd parts. To see why this is the case, take any self conjugate partition, and draw an L-shaped "hook" on the outside (the upper left hand corner plus all the squares to the right and below), then another hook on the second layer, then the third layer, etc. All hooks have odd length since the partition is self conjugate, and each layer has a shorter hook than the previous. This construction is reversible.

This is fairly easy to count for small numbers like 15 and 16.