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Research notes on every problem, and a library of the papers behind them. Built from the open erdos repository.

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Claim. Let nkn_k be the least integer nn such that each of n+1,…,n+kn+1,\ldots,n+k has a prime factor greater than kk. Display (6) of Erdős's paper, printed p. 273 and paged at inequality (6), states that nk<klog⁡k/log⁡log⁡kn_k<k^{\log k/\log\log k} for k>k0k>k_0. The printed proof is four lines: the number of kk-smooth integers below nkn_k is at least nk/kn_k/k, since each block of kk consecutive integers below nkn_k contains one, and comparing this with de Bruijn's asymptotic U(kα,k)=(cα+o(1))kαU(k^\alpha,k)=(c_\alpha+o(1))k^\alpha for the count of kk-smooth integers up to kαk^\alpha bounds the exponent of nkn_k.

Some m≤nm\le n works for kk in Problem 962 exactly when nk≤nn_k\le n, so k(n)=max⁡{k:nk≤n}k(n)=\max\{k:n_k\le n\}. With log⁡k=clog⁡nlog⁡log⁡n\log k=c\sqrt{\log n\log\log n} the exponent (log⁡k)2/log⁡log⁡k(\log k)^2/\log\log k of (6) equals (2c2+o(1))log⁡n(2c^2+o(1))\log n, which is at most log⁡n\log n for large nn when c<1/2c<1/\sqrt2; then nk≤nn_k\le n and k(n)≥kk(n)\ge k. Hence log⁡k(n)≥(1/2−o(1))log⁡nlog⁡log⁡n\log k(n)\ge(1/\sqrt2-o(1))\sqrt{\log n\log\log n}. The site prints the weaker form log⁡k(n)≫log⁡nlog⁡log⁡n\log k(n)\gg\sqrt{\log n\log\log n}.

Covers. The lower bound on k(n)k(n) only. The estimate of k(n)k(n) and the displayed question log⁡k(n)≤(log⁡n)1/2+o(1)\log k(n)\le(\log n)^{1/2+o(1)}, the inverse of Erdős's conjecture (7), stay open.

Depends on. No page of this wiki; the proof rests on de Bruijn's smooth-number asymptotic (Indag. Math. 13 (1951), 50--60), which the paper cites.

Acceptance. Refereed: P. Erdős, Problems and results on consecutive integers, Publ. Math. Debrecen 23 (1976), no. 3--4, 271--282. The site labels the problem OPEN, so its commentary crediting the argument is not reviewed evidence. The Crossref record gives the volume year without a day, so the day in the page name is a placeholder.