False statementConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)judge pass (gpt-5.6-terra)audited 2026-08-28

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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FALSE: $p_{k} (n)$ is counted by $(\binom{n - 1}{k - 1})$

Statement

False claim: for all positive integers $n$ and $k$ ,

$p_{k} (n) = (\binom{n - 1}{k - 1}) .$

Facts & Assumptions

Given: the case $n = 5$ and $k = 3$ .

[L1]

The recurrence $p_{k} (n) = p_{k - 1} (n - 1) + p_{k} (n - k)$ holds for $k \geq 1$ (Exact-k partition recurrence).

[L2]

Compositions of $n$ into $k$ positive parts are counted by $(\binom{n - 1}{k - 1})$ (Compositions of $n$ into $k$ positive parts are counted by $(\binom{n - 1}{k - 1})$ ).

Refutation

technique · direct

1.1

Using [L1], one gets $p_{3} (5) = p_{2} (4) + p_{3} (2) = 2 + 0 = 2$ , corresponding to the two partitions $3 + 1 + 1$ and $2 + 2 + 1$ .

L1algebra

1.2

By [L2], one has $(\binom{5 - 1}{3 - 1}) = (\binom{4}{2}) = 6$ .

L2algebra

2.1

Steps 1.1 and 1.2 give $p_{3} (5) = 2 \neq 6 = (\binom{4}{2})$ , so the displayed claim is false. The binomial coefficient counts ordered compositions, not unordered partitions.

step 1.1step 1.2∎

Depends on

Used by

Nothing in the library uses this result yet.

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

Alexander Hulpke, Combinatorics notes (standard reference, not scraped)
Darij Grinberg, Enumerative Combinatorics: class notes (standard reference, not scraped)

FALSE: pk(n)pk​(n) is counted by (n−1k−1)(k−1n−1​)