Alphabeta Math
CorollaryStatement: Literature-sourcedProof: Literature-sourcedSession-authored (Fable 5 assisted)audited 2026-08-28
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Partitions with at most k parts are equinumerous with partitions whose parts are all at most k

Statement

For every n0 and k0, the number of partitions of n with at most k parts equals the number of partitions of n whose parts are all at most k.

Facts & Assumptions

Given: integers n0 and k0.

[F1]

A partition has all parts at most k exactly when its largest part is at most k (The functions p(n), p_k(n), and the standard restricted partition families).

[L1]

For each j0, partitions with exactly j parts are equinumerous with partitions whose largest part is j (Partitions with k parts are equinumerous with partitions whose largest part is k).

Proof

technique · bijection
1.1

Let λ be a partition of n with at most k parts. If λ has exactly j parts, then jk, and [L1] sends λ by conjugation to a partition whose largest part is j. By [F1], every part of the conjugate is therefore at most k. The same reasoning in reverse sends any partition all of whose parts are at most k to one with at most k parts.

F1L1
2.1

Thus conjugation restricts to a bijection between the two displayed sets, so they have equal cardinality.

step 1.1

Depends on

Used by

Dependency tree · two levels

5 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources