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Statement

Printed p. 124, with $f(k)$ "the least integer so that the product of $f(k)$ consecutive integers, each greater than $k$ always contains a prime greater than $k$": Theorem 1. "There is a constant $c_1>1$ so that

$$f(k)\le c_1\frac{k}{\log k}. \tag{1}$$

In other words the sequence $u+1,u+2,\ldots,u+t$, $t=[c_1\frac{k}{\log k}]$, $u\ge k$ has at least one prime $>k$."

The constant $c_1$ is not specified in the paper: the proof takes $c_1>6$ for $u>k^{3/2}$ and $c_1$ sufficiently large for $u\le k^{3/2}$, where the gap constant of the Hoheisel--Ingham bound enters. Erdős's 1976 survey (Publ. Math. Debrecen 23, printed p. 271) restates the result as $f(k)<3k/\log k$, citing this paper; the site's Problem 961 page prints that form and attributes it to this paper.

Source. P. Erdős, On consecutive integers, Nieuw Arch. Wisk. (3) 3 (1955), 124--128; the five-page scan (printed pp. 124--128 = PDF pp. 1--5); Theorem 1 on printed p. 124 (PDF p. 1), read on the page image.

Read depth. Claims checked: the definition of $f(k)$ and the statement were read clause by clause on the page image. The proof (pp. 125--126) was read on the page images for the sketch below; it is not verified.

Proof pointer

The paper first records (p. 125) two consequences of the Hoheisel--Ingham theorem, $\pi(x+x^\theta)-\pi(x)\sim x^\theta/\log x$ for $5/8\le\theta\le1$, hence $p_{n+1}-p_n=O(p_n^{5/8})$, and deduces that for $u\le k^{3/2}$ one of $u+1,\ldots,u+t$ is a prime when $c_1$ is large. The range $u>k^{3/2}$ is handled on p. 126 by a binomial-coefficient argument. Take $t<k$, as the Sylvester--Schur theorem allows. If every prime factor of $\binom{u+t}{t}$ were at most $k$, the lemma that a prime power exactly dividing $\binom{u+t}{t}$ is at most $u+t$ would give $(u/t)^t<\binom{u+t}{t}\le(u+t)^{\pi(k)}$, and with $u>k^{3/2}$ and $\pi(k)<3k/(2\log k)$ this becomes $u^{t/3}<u^{2k/\log k}$, a contradiction for $c_1>6$.

Dependencies

The Hoheisel--Ingham prime-counting theorem (cited to Ingham, Quart. J. Math. 8 (1937), 255--266); Legendre's formula; the Sylvester--Schur theorem; the bound $\pi(k)<3k/(2\log k)$.

Bears on

• Problem 961: the second classical upper bound for $f(k)$, superseded in order by the Jutila--Ramachandra--Shorey bound reported in Erdős's 1976 survey.
• Problem 683: Theorem 1 refines the Sylvester--Schur theorem behind the problem's classical bound $P(\binom nk)>k$ ($n\ge2k$): a prime greater than $k$ already divides the product of any $[c_1k/\log k]$ consecutive factors of $n(n-1)\cdots(n-k+1)$ that exceed $k$. The problem's claim page Erdős 1955 derives $P(\binom nk)\gg\min(n-k+1,k\log k)$ for $k\le n/2$ from it; the theorem gives no bound of the form $k^{1+c}$.