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Statement

Theorem 2 (p. 43). Let $P$ be a prime and write $n$ in base $P$, with digits $a_1$ (the units digit), $a_2,\ldots$, each between $0$ and $P-1$. If at least two of the digits $a_i$ are at least $\frac{P+1}{2}$, then $P^2$ divides $\binom{2n}{n}$.

The print writes the expansion as $n=a_{r+1}P^r+\cdots a_2P+a$ with "$0\leq a_r<P$", bounding only $a_r$, and gives the hypothesis as stated above; the indexing above follows its proof, where $a_i$ is the coefficient of $P^{i-1}$.

Proof pointer

P. 43. If $a_i\ge\frac{P+1}{2}$ then $\{n/P^i\}\ge\frac12$, and by (1) and (2) on p. 24 each such $i$ adds one to the exponent of $P$ in $\binom{2n}{n}$.

Use in the paper

P. 43. Taking $P=2$, the paper states that $2^2$ divides $\binom{2n}{n}$ except when $n$ is a power of 2. As printed, the hypothesis of Theorem 2 needs a digit at least $3/2$, which no binary digit reaches; the step for $P=2$ rests on (1) and (2) directly, under which $\{n/2^i\}\ge1/2$ exactly when the binary digit of $n$ at $2^{i-1}$ is 1. For $n=2^j$, $2<j\le8000$, the paper reports a computer check of the last few base-$P$ digits of $2^j$: for each such $j$ except $j=4$ some prime $P<100$ meets the hypothesis of Theorem 2, and for $j=4$ it records $3^2\mid\binom{32}{16}$, though $2^4$ does not meet the hypothesis.

Read depth

Claims checked: the statement, its proof and the computation reported after it were read on the page image of p. 43. The computer check was not rerun. Nothing here is independently reviewed.

Dependencies

None in the corpus. Within the paper: formulas (1) and (2) (p. 24), the expression of the exponent of a prime in $\binom{2n}{n}$ through $[2n/p^i]-2[n/p^i]$.

Source. G. Velammal, Is the binomial coefficient $\binom{2n}{n}$ squarefree?, Hardy-Ramanujan J. 18 (1995), 23--45, DOI 10.46298/hrj.1995.132; the edition read is named on the source card.

Bears on

• Problem 175: with the computation reported on p. 43 it is the paper's means of settling the range $4<n<2^{8000}$ left by the Theorem on p. 24.