CorollaryStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26

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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.

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$n + 1$ divides $(\binom{2 n}{n})$ for every $n \in N$

Statement

For every $n \in N$ the integer $n + 1$ divides $(\binom{2 n}{n})$ (Divisibility in $Z$ : $d ∣ a$ when $a = d q$ for some integer $q$ ), and the quotient is the Catalan number $C_{n}$ (The Catalan number $C_{n} : = ∣ D_{n} ∣$ ).

Facts & Assumptions

Given: a natural number $n$ .

[F1]

$(n + 1) C_{n} = (\binom{2 n}{n})$ in $N$ ( $(n + 1) C_{n} = (\binom{2 n}{n})$ ).

[F2]

$C_{n} = ∣ D_{n} ∣ \in N$ (The Catalan number $C_{n} : = ∣ D_{n} ∣$ ).

[L1]

For $d, a \in Z$ , $d$ divides $a$ when $a = d q$ for some $q \in Z$ (Divisibility in $Z$ : $d ∣ a$ when $a = d q$ for some integer $q$ ).

[L2]

The embedding of $N$ into $Z$ sending $k$ to $[(k, 0)]$ is injective and preserves addition, multiplication, and order (The naturals embed in the integers).

Proof

technique · direct

1.1

The Catalan number $C_{n}$ is a natural number, so its image in $Z$ is an integer, and the identity $(n + 1) C_{n} = (\binom{2 n}{n})$ of [F1] holds between natural numbers.

F1F2

2.1

Since the embedding preserves multiplication and addition, the same identity holds in $Z$ between the corresponding integers; taking $q : = C_{n}$ in [L1] with $d = n + 1$ and $a = (\binom{2 n}{n})$ shows that $n + 1$ divides $(\binom{2 n}{n})$ and exhibits $C_{n}$ as the quotient.

L1L2step 1.1∎

Remarks

The quotient is exhibited as a count, and that is the whole proof. No arithmetic property of $(\binom{2 n}{n})$ is used: the divisibility holds because a set of Dyck paths was counted and the count turned out to be the quotient. An argument from prime factorisations would have to be made separately for every prime dividing $n + 1$ , and would give no combinatorial meaning to the quotient.
What is not claimed. Nothing here says $n + 1$ is the largest such divisor, or that $(\binom{2 n}{n})$ has any other divisibility property. The statement is the single divisibility, for every $n$ , with $n = 0$ included: there $1$ divides $1$ .

Depends on

Used by

FALSE: the quotient C(2n, n)/(n+1) is an integer only for small n False statement

Dependency tree · two levels

24 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

D. Guichard, An Introduction to Combinatorics and Graph Theory, §3.5 Catalan Numbers (standard reference, not scraped)

n+1n+1 divides (2nn)(n2n​) for every n∈Nn∈N