ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-01

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Factoring a central binomial coefficient to detect interval primes

Example

At $n = 10$ ,

$(\binom{20}{10}) = 184756 = 2^{2} \cdot 11 \cdot 13 \cdot 17 \cdot 19.$

Thus every prime in $(10, 20]$ appears, exactly as the valuation lemma predicts.

Facts & Assumptions

Given: The value $n = 10$ .

[L1]

The central binomial coefficient satisfies $\frac{4^{10}}{21} \leq (\binom{20}{10}) \leq 4^{10}$ (Central binomial coefficient bounds).

[L2]

Every prime $p$ with $10 < p \leq 20$ divides $(\binom{20}{10})$ exactly once (Prime valuations in the central binomial coefficient).

Verification

technique · direct

1.1L2givenalgebra

A direct factorization gives $(\binom{20}{10}) = 184756 = 2^{2} \cdot 11 \cdot 13 \cdot 17 \cdot 19.$ The primes in $(10, 20]$ are precisely $11, 13, 17, 19$ , and each appears with valuation $1$ .

2.1L1step 1.1algebra

The bound in [L1] reads $\frac{4^{10}}{21} \leq 184756 \leq 4^{10},$ so this concrete value sits comfortably inside the abstract central-binomial window used in Chebyshev's proof.

3.1L2step 1.1step 2.1∎

This example shows exactly how the factorization of $(\binom{2 n}{n})$ detects the interval primes: the large primes appear once, while only small primes contribute higher powers.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

11 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

Victor Shoup, A Computational Introduction to Number Theory and Algebra, Version 2 (standard reference, not scraped)