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Statement

Printed p. 128. For integers $n\ge0$ and $k$, consider the block $n+1,n+2,\ldots,n+k$ (the paper's display (9), here with only $n\ge0$ assumed, "and not $n\ge k$").

Theorem 3. "Amongst the integers (9) there are at least $(\frac12+o(1))\frac{k}{\log k}$ which do not divide the product of the others."

The paper adds that the theorem is best possible: for $n=0$ and $k>5$ the block contains exactly $\pi(k)-\pi(k/2)=\frac12\frac{k}{\log k}+o(k/\log k)$ members that do not divide the product of the others.

Source. P. Erdős, On consecutive integers, Nieuw Arch. Wisk. (3) 3 (1955), 124--128; Theorem 3 with its proof and the best-possible remark on printed p. 128.

Read depth. Claims checked: the statement and the remark were read clause by clause on the page image. The proof was read for the sketch below; it is not verified.

Proof pointer

For $n\ge k$ it follows from Theorem 2: a prime greater than $k$ divides at most one member of a block of $k$ consecutive integers, so a member with such a prime factor does not divide the product of the others. For $n<k$, each prime $p$ with $n+k/2<p<n+k$ divides only one member of the block, and there are $\frac12k/\log k+o(k/\log k)$ such primes.

Dependencies

Theorem 2; the prime number theorem.

Bears on

No problem page of this corpus.