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TheoremStatement: Literature-sourcedProof: AI-generatedSession-authored (Fable 5 assisted)precheck 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

n0k=0nS(n,k)ukxnn!=exp ⁣(u(ex1)).

In particular,

n0Bnxnn!=exp(ex1).

Proof

technique · direct
1.1

A set partition is a labelled set of nonempty labelled sets. Marking each block by a factor of u replaces the basic block EGF ex1 by u(ex1). Applying the labelled-set rule therefore gives n0k=0nS(n,k)ukxnn!=exp ⁣(u(ex1)).

given
2.1

Setting u=1 sums over all block counts and therefore replaces k=0nS(n,k) by Bn. This yields n0Bnxn/n!=exp(ex1).

step 1.1given
3.1

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

step 1.1step 2.1

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