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Statement

Estimate (6) (p. 89). The paper observes that the proofs of Theorems 2 and 3 show the following: for $0<\alpha,\beta<1$ with $\beta-\alpha>\eta>0$, for almost all $n$,

$$\sum_{p\nmid\binom{2n}{n},\ n^\alpha<p<n^\beta}\frac1p =\sum_{1/\beta\le k\le1/\alpha}\frac1{2^k}\log\Bigl(1+\frac1k\Bigr)+o(1),$$

uniformly in $\eta$, the sum on the left running over primes.

Theorem 4 (p. 89). For $\alpha<1$,

$$\Bigl|\Bigl\{m:\ 1\le m\le n^\alpha,\ m\nmid\binom{2n}{n}\Bigr\}\Bigr| =c(\alpha)n^\alpha+o(n^\alpha),$$

"where $c(\alpha)\to1$ as $\alpha\to0$. (In fact, $c(\alpha)$ can be explicitly calculated.)" (p. 89). The paper does not give $c(\alpha)$.

Notes. The printed statement carries no quantifier on $n$. It is derived "by the sieve method" from (6), which holds for almost all $n$, so the corpus reads Theorem 4 as an asymptotic for almost all $n$; the print does not say so. The limit $c(\alpha)\to1$ as $\alpha\to0$ is printed as quoted, but this page observes that (6) points the other way: the primes below $n^\alpha$ not dividing $\binom{2n}{n}$ have reciprocal sum about $\sum_{k\ge1/\alpha}2^{-k}\log(1+1/k)$, which tends to $0$ with $\alpha$, so a vanishing proportion of the $m\le n^\alpha$ would have such a prime factor. The page does not settle which limit the authors intended.

Source. P. Erdős, R. L. Graham, I. Z. Ruzsa and E. G. Straus, On the prime factors of $\binom{2n}{n}$, Math. Comp. 29 (1975), no. 129, 83--92; estimate (6) and Theorem 4 on p. 89. The edition is identified on the source card.

Read depth. Claims checked: (6) and Theorem 4 were read clause by clause on the page image. The paper gives no proof beyond the sentence deriving Theorem 4 from (6); the observation on the limit of $c(\alpha)$ is this page's and is not a checked computation of $c(\alpha)$.

Proof pointer

Page 89, one sentence: Theorem 4 follows from (6) "by the sieve method". No details are given.

Dependencies

Theorem 2 and Theorem 3, whose proofs give (6).

Bears on

No problem in the corpus asks for this count. The least integer not dividing $\binom{2n}{n}$, the subject of Problem 731, is treated separately in (8); the paper does not apply Theorem 4 to it.