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Statement

Printed p. 183, item 5 of the first part: "What is the largest $k=k(n)$ for which there is an $m\le n$ so that each of the integers $m+i$, $1\le i\le k$, are divisible by at least one prime $>k$? It is not hard to prove that

$$k(n)>\exp(\log n)^{1/2-\epsilon}.$$

It seems likely that $k(n)=o(n^\epsilon)$, but I have not been able to obtain any non-trivial upper bound for $k(n)$."

The printed display has no parentheses around $(\log n)^{1/2-\epsilon}$; its natural reading, $k(n)>\exp\bigl((\log n)^{1/2-\epsilon}\bigr)$, is the one the site's Problem 962 page prints as $\log k(n)\ge(\log n)^{1/2-o(1)}$.

Source. P. Erdős, Extremal problems in number theory, Proc. Sympos. Pure Math. VIII (Theory of Numbers), Amer. Math. Soc. (1965), 181--189, DOI 10.1090/pspum/008/0174539 (Crossref record read); printed p. 183 (PDF p. 3 of the eleven-page scan read for this page), read on the page image; a site key for Problem 962.

Read depth. Claims checked: the passage was read clause by clause on the page image. The lower bound is asserted as "not hard to prove" without proof; by footnote 1 (printed p. 181) a result stated without reference refers to Erdős's Hungarian paper (Mat. Lapok 13 (1962), 228--255; erdos_1962_szamelmeleti_megjegyzesek_iv), whose problem 16 (p. 238) states the same bound, grouped as $\exp((\log n)^{1/2-\epsilon})$ and for the runs $m,m+1,\dots,m+k$, also without proof; the $o(n^\epsilon)$ statement is an expectation.

Proof pointer

None on the page. Erdős's 1976 Debrecen paper proves the stronger $n_k<k^{\log k/\log\log k}$ for the inverse function (inequality (6)).

Dependencies

None stated.

Bears on

• Problem 962: the problem's definition of $k(n)$ and its first lower bound, as the site quotes them.