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Statement

Setting (p. 638). For an odd prime $p$,

$$R_p=\Big\{0,1,\ldots,\tfrac12(p-1)\Big\},\qquad r_p=\#R_p=\tfrac12(p+1),\qquad \theta_p=\frac{\log r_p}{\log p}.$$

By Kummer's theorem, $p\nmid\binom{2n}{n}$ exactly when every base-$p$ digit of $n$ lies in $R_p$.

Lemma 1 (p. 638), quoted: "For each odd prime $p$ and all real numbers $x\geq2$, the number of integers $1\leq n\leq x$ with $p\nmid\binom{2n}{n}$ is at most $px^{\theta_p}$."

For $p=2$ the paper notes (p. 638) that $\binom{2n}{n}$ is always even.

Heuristic that follows (p. 639, unlabeled, not a theorem). Reading Lemma 1 as a probability, at most $px^{\theta_p-1}$ when $x$ is an integer, that $p\nmid\binom{2n}{n}$ for $n$ random in $[1,x]$, with $\theta_p-1$ greater than $-0.37$, $-0.32$ and $-0.29$ for $p=3,5,7$, and assuming the events for different primes independent, the paper expects at least $x^{0.02}$ integers $n\le x$ with $\binom{2n}{n}$ coprime to $105$. The same heuristic for the four primes $3,5,7,11$ suggests only finitely many $n$ with $\binom{2n}{n}$ coprime to $1155$, such as $n=3160$; the paper reports that no larger example is known, though the search has gone up to $10^{10^4}$, citing Mauldin and Ulam.

Context the paper gives (p. 638). The numbers $n=1,10,756$ have $\binom{2n}{n}$ coprime to $105$; whether there are infinitely many is Graham's problem, with a prize as reported in the paper's references [2, 4]. For $m=pq$ a product of two odd primes, $\binom{2n}{n}$ is coprime to $m$ for infinitely many $n$, a result of Erdős, Graham, Ruzsa and Straus (1975).

Source. Carl Pomerance, Divisors of the middle binomial coefficient, Amer. Math. Monthly 122 (2015), no. 7, 636--644, doi:10.4169/amer.math.monthly.122.7.636: the setting, Lemma 1 and its proof in Section 4 (pp. 638--639), the heuristic on p. 639. The edition read is identified on the source card.

Read depth. Claims checked: the setting, the statement and the proof were read clause by clause on the printed pages. Nothing here is independently reviewed.

Proof pointer

Pages 638--639. With $D=\lfloor1+\log x/\log p\rfloor$, every $n\le x$ has at most $D$ base-$p$ digits, and restricting each to $R_p$ leaves at most $r_p^D<pr_p^{\log x/\log p}=px^{\theta_p}$ choices.

Dependencies

Kummer's theorem (Section 3, p. 637).

Bears on

• Problem 376: the problem asks whether infinitely many $n$ have $\binom{2n}{n}$ coprime to $105$. Lemma 1 bounds, for each of $p=3,5,7$ separately, how many $n\le x$ escape divisibility by $p$; it gives no lower bound and says nothing about the three primes jointly. The independence heuristic built on it predicts infinitely many such $n$ but is not a proof, and the paper does not settle the problem.