Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-27
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The exponential formula gives the Bell-number generating function

Statement

Let u be a second formal indeterminate. In Q[u]⟦x⟧, the block-count-refined exponential generating function of set partitions is

∑n≥0∑k=0nS(n,k)ukxnn!=exp⁡ ⁣(u(ex−1)).

In particular,

∑n≥0Bnxnn!=exp⁡(ex−1).

Proof

technique · direct
1.1given

A set partition is a labelled set of nonempty labelled sets. Marking each block by a factor of u replaces the basic block EGF ex−1 by u(ex−1). Applying the labelled-set rule therefore gives ∑n≥0∑k=0nS(n,k)ukxnn!=exp⁡ ⁣(u(ex−1)).

2.1step 1.1given

Setting u=1 sums over all block counts and therefore replaces ∑k=0nS(n,k) by Bn. This yields ∑n≥0Bnxn/n!=exp⁡(ex−1).

3.1step 1.1step 2.1∎

Steps 1.1 and 2.1 prove the refined formula and its Bell-number specialization.

Depends on

Used by

Dependency tree · two levels

9 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