CorollaryStatement: Literature-sourcedProof: Literature-sourcedaudited 2026-08-28

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 symmetric-group cycle-index series is coefficientwise exponential

Statement

In the formal power-series ring $Q [s_{1}, s_{2}, \dots] ⟦ t ⟧$ ,

$\sum_{n \geq 0} Z (S_{n}) t^{n} = \exp (\sum_{d \geq 1} \frac{s_{d} t^{d}}{d}) .$

For each fixed $n$ , the coefficient of $t^{n}$ depends only on $s_{1}, \dots, s_{n}$ .

Facts & Assumptions

Given: the formal exponential and the symmetric-group cycle indices.

[L1]

The coefficient of $t^{n}$ in the cycle-index series of the family with one fixed structure in each degree is $Z (S_{n})$ (The cycle-index series of a graded family of S_n-actions, The cycle index of S_n is the sum over cycle types).

[L2]

Formal exponential turns finite sums into products, so for each finite truncation one may expand $\exp ⁣ (\sum_{d = 1}^{N} s_{d} t^{d} / d)$ as $\prod_{d = 1}^{N} \exp (s_{d} t^{d} / d)$ (Formal $\exp$ and $\log$ are inverse homomorphisms and formal binomial powers obey the expected addition laws).

Proof

technique · coefficient comparison

1.1L2

Fix $n \geq 0$ . Factors with $d > n$ cannot contribute to the coefficient of $t^{n}$ , so that coefficient is already the coefficient of $t^{n}$ in the finite truncation $\exp (\sum_{d = 1}^{n} \frac{s_{d} t^{d}}{d})$ . By [L2], this truncation equals $\prod_{d = 1}^{n} \exp (\frac{s_{d} t^{d}}{d}) = \prod_{d = 1}^{n} \sum_{m_{d} \geq 0} \frac{s_{d}^{m_{d}} t^{d m_{d}}}{d^{m_{d}} m_{d}!}$ .

2.1step 1.1

Expanding the finite product in step 1.1, the coefficient of $t^{n}$ is $\sum_{\begin{array}{c} m_{1}, \dots, m_{n} \geq 0 \\ \sum_{d = 1}^{n} d m_{d} = n \end{array}} \frac{1}{\prod_{d = 1}^{n} d^{m_{d}} m_{d}!} \prod_{d = 1}^{n} s_{d}^{m_{d}}$ . This depends only on $s_{1}, \dots, s_{n}$ .

3.1step 2.1L1∎

The coefficient in step 2.1 is exactly the cycle-type formula for $Z (S_{n})$ from [L1]. Since this holds for every $n$ , the whole series satisfies $\sum_{n \geq 0} Z (S_{n}) t^{n} = \exp (\sum_{d \geq 1} \frac{s_{d} t^{d}}{d})$ .

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

18 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

Ben Lynn, Polya Theory: The Cycle Index Polynomial (standard reference, not scraped)
Eric W. Weisstein, Cycle Index, Wolfram MathWorld (standard reference, not scraped)