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Source. P. Erdős and J. L. Selfridge, Some problems on the prime factors of consecutive integers, Illinois J. Math. 11 (1967), 428--430 (source card): the last paragraph of p. 430, citing A. Schinzel, Problem 31, Elem. Math. 14 (1959), 82--83.

Read depth. Claims checked: the paragraph was read clause by clause on the printed page. The paper reports Schinzel's result without proof, and Schinzel's note was not read here.

Statement

Let $p_1=2<p_2<\cdots$ be the consecutive primes.

Schinzel's result, as reported (p. 430, unlabeled). Schinzel deduces from Pólya's theorem that, with the possible exception of a finite number of cases, among any $p_1\cdots p_{k-1}p_{k+1}$ consecutive integers there is always one with more than $k$ prime factors.

Question (p. 430). The authors say that it seems possible that $p_1\cdots p_k$ is the right value, and that even for $k=2$ they cannot improve Schinzel's value.

The paragraph says "prime factors" without saying whether multiplicity is counted; it follows the paper's discussion of $\nu$, the number of distinct prime factors.

Proof pointer

None in this paper; it attributes the deduction to Schinzel's Problem 31 and to Pólya's theorem on gaps between integers composed of a fixed finite set of primes, stated on the same page in the derivation of inequality (1).

Dependencies

A. Schinzel, Problem 31, Elem. Math. 14 (1959), 82--83; G. Pólya, Zur arithmetischen Untersuchung der Polynome, Math. Z. 1 (1918), 143--148 (source card).

Bears on

• Problem 891: the problem asks whether, for $k\ge2$ and all large $n$, the interval $[n,n+p_1\cdots p_k)$ contains an integer with more than $k$ prime factors, which is the authors' question here. Schinzel's result, as reported, gives this with the longer length $p_1\cdots p_{k-1}p_{k+1}$ and possibly finitely many exceptions; the authors say that even for $k=2$ they cannot improve Schinzel's value.