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Statement

Inequality (7) (p. 90). The paper states that "it is not difficult to prove", by methods similar to those used earlier, that

$$\sum_{p\mid\binom{2n}{n},\ p\le n}\frac1p>c\log\log n ,$$

the sum running over primes. The print fixes neither the constant $c$ nor the range of $n$, and gives no proof.

Remark on the constant (p. 90): "There is no doubt that (7) holds for any $c>1-\epsilon$", and this would follow from the boundedness of $f(n)=\sum_{p\nmid\binom{2n}{n},\,p\le n}1/p$. Read literally the phrase asks for every $c>1-\epsilon$, which no constant above $1$ can meet, since $\sum_{p\le n}1/p=\log\log n+O(1)$; the corpus reads it as $c=1-\epsilon$ for every $\epsilon>0$, which is what a bounded $f$ would give.

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; (7) and the remark on p. 90. The edition is identified on the source card.

Read depth. Claims checked: the statement and the remark were read clause by clause on the page image. The paper gives no proof.

Proof pointer

None in the paper, which asserts (7) as provable by methods similar to those of Theorems 2 and 3.

Dependencies

None stated.

Bears on

• Problem 377: the problem asks whether $f(n)$ is bounded; the paper notes that a bounded $f$ would give (7) "for any $c>1-\epsilon$", read above as $c=1-\epsilon$. (7) itself does not bound $f$.
• Problem 726: the paper states the problem's conjecture (starred sum) "in this connection" right after (7). By the digit criterion (1) of p. 84, each prime $p\le n$ with $n\equiv r\pmod p$, $p/2<r<p$, divides $\binom{2n}{n}$, so the problem's sum is part of the sum in (7); this is an observation of this page, and (7) gives no bound on the problem's sum.