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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): $\nu(m)$ is defined on p. 428, the conjecture is on p. 429.

Read depth. Claims checked: the conjecture and the known case it extends were read clause by clause on the printed page.

Statement

Setting (p. 428). $\nu(m)$ denotes the number of distinct prime factors of $m$.

Known case (p. 429). The paper recalls as well known, and as following easily from the prime number theorem, that

$$\limsup_{n\to\infty}\nu(n)\frac{\log\log n}{\log n}=1.$$

Conjecture (p. 429, unnumbered). The authors write that one could conjecture that for every $k$

$$\limsup_{n\to\infty}\sum_{i=0}^{k}\nu(n+i)\frac{\log\log n}{\log n}=1,$$

adding that this, if true, will be difficult. The sum has the $k+1$ terms $i=0,\ldots,k$. It is posed, not proved.

In the other direction the paper says it cannot even prove that

$$\limsup_{n\to\infty}\Bigl(\max_{1\le m\le n}\bigl(\nu(m)+\nu(m+1)\bigr) -\max_{1\le m\le n}\nu(m)\Bigr)=\infty .$$

Bears on

• Problem 890: the problem's second question is this conjecture, with the sum written over $0\le i<k$; as $k$ ranges over all values the two indexings give the same family of statements. The paper poses it and proves nothing towards it beyond the case of a single term.