Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-27
How statement and proof provenance work

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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

The signless first-kind Stirling numbers satisfy their recurrence and expand the rising factorial

Statement

For every n≥1 and every k≥1,

c(n,k)=c(n−1,k−1)+(n−1)c(n−1,k).

Moreover, for every n∈N,

xn‾=∑k=0nc(n,k)xk.

Facts & Assumptions

Proof

technique · direct
1.1given

To build a permutation of [n] with exactly k cycles, start from a permutation of [n−1]. Either n forms a new one-cycle, which contributes c(n−1,k−1) possibilities, or else n is inserted into one of the n−1 cyclic slots of a permutation with k cycles, which contributes (n−1)c(n−1,k). These two constructions are disjoint and exhaustive, so the recurrence follows.

2.1step 1.1algebra

Let Pn(x):=∑k=0nc(n,k)xk. Step 1.1 gives Pn(x)=xPn−1(x)+(n−1)Pn−1(x)=(x+n−1)Pn−1(x), and also P0(x)=1.

3.1step 2.1given∎

The rising factorial satisfies the same recursion: x0‾=1 and xn‾=(x+n−1)xn−1‾ by The rising factorial. Therefore Pn(x)=xn‾ for all n by induction on n.

Depends on

Used by

Dependency tree · two levels

7 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