ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passverified 2026-08-03 (gpt-5.6-sol-codex-subscription)

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 ratio $ι (D_{n}) / ι (n!)$ computed for small $n$ as a quotient of two counts, with no probability space claimed

Example

For $n \in N$ consider the real number

$r_{n} : = \frac{ι (D_{n})}{ι (n!)},$

the quotient of the count of derangements of an $n$ -element set (The derangement number $D_{n}$ : the number of bijections of an $n$ -element set with no fixed point) by the count of all its bijections, $∣ Bij (n) ∣ = n!$ (The factorial $n!$ and the falling factorial $n^{\underline{k}}$ , defined by recursion in $N$ ). The quotient is legitimate because $n! \neq 0$ . Dividing the derangement formula by $ι (n!)$ gives

$r_{n} = \sum_{i < n + 1} \frac{(- 1)^{i}}{ι (i!)},$

so $r_{n}$ is the truncated alternating sum itself. Its first values, obtained from $D_{0} = 1$ , $D_{1} = 0$ , $D_{2} = 1$ and the first recurrence $ι (D_{n}) = ι (n) ι (D_{n - 1}) + (- 1)^{n}$ ( $ι (D_{n}) = ι (n) ι (D_{n - 1}) + (- 1)^{n}$ for $n \geq 1$ , and $D_{n} = (n - 1) (D_{n - 1} + D_{n - 2})$ for $n \geq 2$ ), are

$n$	$D_{n}$	$n!$	$r_{n}$
$0$	$1$	$1$	$1$
$1$	$0$	$1$	$0$
$2$	$1$	$2$	$1 / 2$
$3$	$2$	$6$	$1 / 3$
$4$	$9$	$24$	$3 / 8$
$5$	$44$	$120$	$11 / 30$
$6$	$265$	$720$	$53 / 144$

This is a ratio of two counts and nothing else. Nothing among this page's declared prerequisites defines a probability space, a measure or an expectation, so $r_{n}$ is not called a probability here and no statement about random behaviour is made. What is asserted is exactly that the numerator counts the fixed-point-free bijections, that the denominator counts all of them, and that the quotient is the displayed alternating sum.

Facts & Assumptions

Given: The derangement numbers $D_{n}$ , the factorials $n!$ , and the canonical natural $ι$ (The canonical natural $ι (n) = n \cdot 1_{F}$ of a field).

[L1]

The derangement formula: $ι (D_{n}) = ι (n!) \sum_{i < n + 1} (- 1)^{i} / ι (i!)$ ( $ι (D_{n}) = ι (n!) \sum_{i < n + 1} (- 1)^{i} / ι (i!)$ , with the term at $i = 0$ equal to $1$ and $D_{0} = 1$ ).

[L2]

$∣ Bij (n) ∣ = n!$ and $n! \neq 0$ , hence $ι (n!) \neq 0$ (The derangement number $D_{n}$ : the number of bijections of an $n$ -element set with no fixed point, The factorial $n!$ and the falling factorial $n^{\underline{k}}$ , defined by recursion in $N$ , Laws of finite sums and products in $N$ , and $ι (\sum_{k < n} a_{k}) = \sum_{k < n} ι (a_{k})$ , clause 7).

[L3]

The first recurrence: $ι (D_{n}) = ι (n) ι (D_{n - 1}) + (- 1)^{n}$ for $n \geq 1$ ; and $D_{0} = 1$ , $D_{1} = 0$ , $D_{2} = 1$ ( $ι (D_{n}) = ι (n) ι (D_{n - 1}) + (- 1)^{n}$ for $n \geq 1$ , and $D_{n} = (n - 1) (D_{n - 1} + D_{n - 2})$ for $n \geq 2$ , The derangement number $D_{n}$ : the number of bijections of an $n$ -element set with no fixed point).

[L4]

$R$ is an ordered field, so division by a nonzero element is available and the displayed arithmetic is legitimate (Ordered field, Field); $(- 1)^{p + 1} = - (- 1)^{p}$ and $(- 1)^{0} = 1$ (Integer powers $a^{m}$ ); and real finite sums obey the recursion clause (Finite sums and finite products, by recursion, Laws of finite sums and finite products).

[L5]

Factorials: $0! = 1$ , $1! = 1$ , $2! = 2$ , $3! = 6$ , $4! = 24$ , $5! = 120$ , $6! = 720$ (The factorial $n!$ and the falling factorial $n^{\underline{k}}$ , defined by recursion in $N$ ).

Verification

technique · direct

1.1

The quotient form. Dividing the identity of [L1] by the nonzero $ι (n!)$ gives $r_{n} = \sum_{i < n + 1} (- 1)^{i} / ι (i!)$ for every $n \in N$ .

L1L2L4

1.2

The derangement numbers up to $6$ . From $D_{2} = 1$ and [L3]: $ι (D_{3}) = 3 \cdot 1 - 1 = 2$ , $ι (D_{4}) = 4 \cdot 2 + 1 = 9$ , $ι (D_{5}) = 5 \cdot 9 - 1 = 44$ and $ι (D_{6}) = 6 \cdot 44 + 1 = 265$ ; since $ι$ is injective these are the natural numbers $D_{3} = 2$ , $D_{4} = 9$ , $D_{5} = 44$ , $D_{6} = 265$ .

L2L3L4

2.1

The tabulated ratios. Dividing the base values in [L3] and the values from step 1.2 by the factorials of [L5] gives $r_{0} = 1 / 1 = 1$ , $r_{1} = 0 / 1 = 0$ , $r_{2} = 1 / 2$ , $r_{3} = 2 / 6 = 1 / 3$ , $r_{4} = 9 / 24 = 3 / 8$ , $r_{5} = 44 / 120 = 11 / 30$ and $r_{6} = 265 / 720 = 53 / 144$ .

step 1.2L3L4L5

2.2

A cross-check at $n = 4$ through step 1.1: $\sum_{i < 5} (- 1)^{i} / ι (i!) = 1 - 1 + 1 / 2 - 1 / 6 + 1 / 24 = 9 / 24 = 3 / 8$ , which is $r_{4}$ .

step 1.1L4L5

3.1

So $r_{n}$ is the truncated alternating sum, and its values through $n = 6$ are as tabulated.

step 2.1step 2.2∎

Remarks

The alternation is visible in the table. $r_{1} = 0$ is below $r_{2} = 1 / 2$ , which is above $r_{3} = 1 / 3$ , which is below $r_{4} = 3 / 8$ ; each successive value differs from the previous one by the single term $(- 1)^{n} / ι (n!)$ , whose sign alternates and whose size decreases.
No limit is claimed. The quotient is computed at each $n$ from two counts, and nothing here asserts convergence or names a limiting value; the exponential function that would be needed to state such a limit is not among this page's declared prerequisites.
Why the division is legitimate at every $n$ , including $n = 0$ . The denominator is $ι (n!)$ and $n!$ is never $0$ , its recursion starting at $0! = 1$ and multiplying by nonzero successors.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

56 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

Derangement (Wikipedia) (standard reference, not scraped)
Rencontres numbers (Wikipedia) (standard reference, not scraped)
Derangements (OpenText at the University of Lethbridge) (standard reference, not scraped)

The ratio ι(Dn)/ι(n!)ι(Dn​)/ι(n!) computed for small nn as a quotient of two counts, with no probability space claimed