Alphabeta Math
TheoremStatement: Literature-sourcedProof: Literature-sourcedSession-authored (Fable 5 assisted)audited 2026-08-28
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Durfee-square decomposition of the partition series

Statement

In Zx,

n0p(n)xn=k0xk2i=1k(1xi)2,

where the empty product at k=0 is 1.

Facts & Assumptions

Given: partitions written by Ferrers diagrams.

[L2]

Partitions with at most k parts are equinumerous with partitions whose parts are all at most k (Partitions with at most k parts are equinumerous with partitions whose parts are all at most k).

Proof

technique · decomposition
1.1

Let λ be a partition with Durfee length k. Removing its k×k Durfee square leaves two pieces: a right-hand piece α consisting of the cells to the right of the square, and a lower piece β consisting of the cells below the square. The piece α has at most k rows, while each row of β has length at most k. Conversely, given k, a partition α with at most k parts, and a partition β with all parts at most k, one reconstructs λ uniquely by adjoining α to the right side and β below the square.

construct
2.1

For fixed k, the square contributes the factor xk2. By [L2], the right-hand piece α has the same generating function as partitions with parts at most k, namely i=1k(1xi)1; the lower piece β has the same generating function for the same direct multiplicity reason. Thus partitions whose Durfee square has size k contribute xk2/i=1k(1xi)2.

step 1.1L2
3.1

Every partition has exactly one Durfee length, so summing the contributions of step 2.1 over all k0 counts every partition exactly once. Therefore the coefficient of xn on the right is p(n) for every n0, and [L1] gives the displayed identity.

step 2.1L1

Depends on

Used by

Dependency tree · two levels

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Sources