Velammal 1995 is binomial coefficient squarefree

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main_theorem: Velammal's proof of the Erdős conjecture that the binomial coefficient of 2n choose n is not squarefree for any n greater than 4.

theorem_2: Velammal's digit criterion: if at least two base-P digits of n are at least (P+1)/2, for a prime P, then P^2 divides the binomial coefficient of 2n choose n.

theorem_p24: Velammal's explicit form of Sárközy's theorem: for every n at least 2^8000 the binomial coefficient of 2n choose n is not squarefree.

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Velammal, G., Is the binomial coefficient {$\binom {2n}n$} squarefree?. Hardy-Ramanujan J. 18 (1995), 23--45. DOI 10.46298/hrj.1995.132. No notice is printed in the scan (its first page, p. 23, carries only the header "Hardy-Ramanujan Journal Vol.18 (1995) 23-45"); the article's page shows only "Hal authorisation v1", a deposit authorization and not a reuse grant (https://hrj.episciences.org/132, read 2026-10-02); the journal's home page offers the collection free of charge and names no license (https://hrj.episciences.org/, read 2026-10-02), and its publishing-policies page states Diamond Open Access under "Creative Commons - Attribution - CC BY 4.0" for published articles without stating that the policy covers the digitized back volumes (https://hrj.episciences.org/page/publishing-policies, read 2026-10-02); the term is unstated.

The paper proves Erdős's conjecture that the central binomial coefficient $\binom{2n}{n}$ is not squarefree for any $n>4$. Sárközy had shown this for all sufficiently large $n$, using Jutila's estimates for sums $\sum_{p\le x}e(2\pi i\theta/p)$ to estimate $\sum_{p\le x}\log p\,e^{2\pi i\theta/p}$; Velammal instead applies Vaughan's identity and the theory of exponent pairs, which gives better estimates, and computes the constants explicitly. The unnumbered Theorem (p. 24) states that $\binom{2n}{n}$ is never squarefree for $n\ge2^{8000}$. Writing $\binom{2n}{n}=(s(n))^2q(n)$ with $q(n)$ squarefree, the proof bounds $\log s(n)$ below by the sum of $\log p$ over primes $p\in(\sqrt n,\sqrt{2n}\,]$ with $\{n/p\}\ge1/2$, each of which divides $\binom{2n}{n}$ to at least the second power, and shows that sum positive. The range $4<n<2^{8000}$ is handled by direct methods: Theorem 2 (p. 43) gives $P^2\mid\binom{2n}{n}$ when at least two base-$P$ digits of $n$ are at least $(P+1)/2$; the paper's $P=2$ step leaves only $n=2^j$, $2<j\le8000$, and a computer check finds a prime $P<100$ meeting the hypothesis of Theorem 2 for each such $j$ except $j=4$, where $3^2\mid\binom{32}{16}$ is recorded directly. A postscript (p. 45) records that J. W. Sander, J. Number Theory 46 (1994), 372--384, points out that G. Velammal, A. Granville and O. Ramaré proved the conjecture independently of each other.

Source: https://hrj.episciences.org/132.

Bears on. #175: the main theorem (abstract, p. 23; proof completed p. 43) is the problem's statement, that $\binom{2n}{n}$ is not squarefree for any $n\ge5$, proved for every such $n$; the Theorem (p. 24) proves it for $n\ge2^{8000}$ and Theorem 2 (p. 43), with the computation reported after it, is the paper's means for the range $4<n<2^{8000}$.

Results.

• Main theorem (abstract, p. 23; proof completed p. 43): $\binom{2n}{n}$ is not squarefree for any $n>4$.
• Theorem (p. 24): for $n\ge2^{8000}$, $\binom{2n}{n}$ is never squarefree.
• Theorem 2 (p. 43): if at least two base-$P$ digits of $n$ are at least $(P+1)/2$, for a prime $P$, then $P^2\mid\binom{2n}{n}$.

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