Is there a generating function for the number of symmetric Ferrers diagrams contained in a square ? I think we need to sum up the number self-conjugate partitions of n starting from $n=1$ to $n=k^2$ with highest part $k$ but I am wondering if there is a direct formula for this. Thanks
2026-03-30 12:14:46.1774872886
self-conjugate partitions with restrictions
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If you mark the sides of the square according to where the diagram jumps, both sides will be marked the same, and can be marked however we please, which means these partitions are in bijection with the possible sets of marks, which is $2^k$ where $k$ is the side length.