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TheoremStatement: Literature-sourcedProof: Literature-sourcedSession-authored (Fable 5 assisted)judge pass (gpt-5.6-terra)audited 2026-08-28
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Glaisher's bijection between odd-part and distinct-part partitions

Statement

For every integer n0, there is a bijection between partitions of n into odd parts and partitions of n into distinct parts.

Facts & Assumptions

Given: an integer n0.

[F1]

Partitions into odd parts and partitions into distinct parts are the two families counted by podd(n) and pdist(n) (The functions p(n), p_k(n), and the standard restricted partition families).

Proof

technique · bijection
1.1

Let λ be a partition of n into odd parts. For each odd integer m, let cm be its multiplicity in λ, and write the binary expansion

cm=j0εm,j2j,εm,j{0,1}.

Define G(λ) to contain the part 2jm once for every pair (m,j) with εm,j=1. Since

j0εm,j2jm=mcm,

the parts with odd core m have the same total mcm before and after the replacement. Summing over the odd cores shows that the total sum of the parts of G(λ) is still n. [F1, construct, algebra]

1.2

Conversely, let μ be a partition of n into distinct parts. Write each part uniquely as 2ju with u odd, and replace it by 2j copies of the odd part u. The resulting partition H(μ) has only odd parts and still sums to n.

F1construct
2.1

The parts of G(λ) are distinct. Indeed, every positive integer has a unique expression 2ju with u odd, so two equal parts in G(λ) would come from the same odd core u and the same power 2j, hence from the same binary digit.

step 1.1
3.1

The two constructions are inverse. Starting from λ, step 1.2 reconstructs from each part 2jm of G(λ) exactly the 2j copies of m encoded by the corresponding binary digit, so H(G(λ))=λ. Starting from μ, step 1.1 groups together all parts with the same odd core and reassembles exactly the original powers of two, so G(H(μ))=μ. Therefore G is a bijection.

step 1.1step 2.1step 1.2

Depends on

Used by

Dependency tree · two levels

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Sources