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Erdos 1967 problems prime factors consecutive integers

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conjecture_p429: Erdős and Selfridge's unnumbered conjecture that for every k the limsup over n of the sum of nu(n+i) for 0 <= i <= k, multiplied by log log n / log n, equals 1, the value known for a single nu(n).

inequality_1: Erdős and Selfridge's deduction from Pólya's theorem that the liminf over n of the sum of nu(n+i) for 0 <= i <= k-1 is at least k + pi(k) - 1, with their conjecture (2) that it is at most k + pi(k).

remark_p430: Erdős and Selfridge's report of Schinzel's deduction from Pólya's theorem that, with possibly finitely many exceptions, every p_1...p_{k-1}p_{k+1} consecutive integers include one with more than k prime factors, and their question whether p_1...p_k suffices.

theorem_p428: Erdős and Selfridge's unnumbered result that some n + k has at least two prime factors exceeding k, that is v_0(n) > 1, for every positive integer n other than 1, 2, 3, 4, 7, 8 and 16.


P. Erdős, J. L. Selfridge, Some problems on the prime factors of consecutive integers. Illinois Journal of Mathematics 11 (1967), 428-430. The copy read for this card is an archive scan that prints no notice; the article's Project Euclid page could not be read on 2026-10-02, its Crossref record (DOI 10.1215/ijm/1256054564) names no license, and the publisher's journal page shows the footer "© 2024 Duke University Press. All Rights Reserved." and names no license (https://www.dukeupress.edu/illinois-journal-of-mathematics, read 2026-10-02), every other right reserved.

For nn a positive integer and k≥0k\ge0, v(n;k)v(n;k) counts the primes p>kp>k dividing n+kn+k, which the paper restates as the prime factors of n+kn+k dividing no n+in+i with 0≤i<k0\le i<k, and v0(n)=max⁡k≥0v(n;k)v_0(n)=\max_{k\ge0}v(n;k) (p. 428). The authors prove v0(n)>1v_0(n)>1 for every nn except n=1,2,3,4,7,8,16n=1,2,3,4,7,8,16, by reducing v0(n)=1v_0(n)=1 to n=pαn=p^\alpha and to exponential equations in powers of 22 and 33, and say they are very far from proving v0(n)→∞v_0(n)\to\infty. They introduce vl(n)=max⁡k≥lv(n;k)v_l(n)=\max_{k\ge l}v(n;k), say it seems certain that vl(n)→∞v_l(n)\to\infty for every ll, and report that v1(n)=1v_1(n)=1 for n=n= 1-4, 6-8, 10, 12, 15, 16, 18, 22, 24, 26, 30, 36, 42, 46, 48, 60, 70, 78, 80, 96, 120, 190, 222, 330 and for no other n<2500n<2500, that 330330 is probably the largest solution, and that they cannot prove the solutions finite (p. 428).

On p. 429, V(n;k)V(n;k) counts the primes pp with pα∥n+kp^\alpha\parallel n+k and pα>kp^\alpha>k, and Vl(n)=max⁡k≥lV(n;k)V_l(n)=\max_{k\ge l}V(n;k). The paper reports V1(n)=1V_1(n)=1 for n=n= 1-4, 6-8, 12, 15, 16, 24, 30, 48, 80 and no other n<2500n<2500, probably with 8080 the largest solution, though finiteness is unproved; from factor tables, V0(n)=2V_0(n)=2 for n=94491,94492,99387,99741n=94491, 94492, 99387, 99741 and V0(n)>2V_0(n)>2 for every other nn with 94000<n<10594000<n<10^5. It puts f(n)=max⁡k≥01k+1∑i=0kv(n;i)f(n)=\max_{k\ge0}\frac1{k+1}\sum_{i=0}^kv(n;i), says f(n)→∞f(n)\to\infty seems probable but very difficult, notes that a result of Hardy and Ramanujan gives f(n)≤(1+o(1))log⁡log⁡nf(n)\le(1+o(1))\log\log n for almost all nn, and calls it not impossible that f(n)>(1−ε)log⁡log⁡nf(n)>(1-\varepsilon)\log\log n for every ε>0\varepsilon>0 and n>n0(ε)n>n_0(\varepsilon). With ν(m)\nu(m) the number of distinct prime factors of mm, it poses the limsup conjecture for ∑i=0kν(n+i)\sum_{i=0}^k\nu(n+i), and states without proof two limsup results for σ\sigma and dd over consecutive integers (p. 429, with a footnote).

On p. 430 a theorem of Pólya on gaps between integers composed of primes up to kk gives inequality (1), lim inf⁡n∑i=0k−1ν(n+i)≥k+π(k)−1\liminf_n\sum_{i=0}^{k-1}\nu(n+i)\ge k+\pi(k)-1, and the authors conjecture (2), the same liminf at most k+π(k)k+\pi(k), perhaps with equality. The last paragraph reports Schinzel's deduction from Pólya's theorem that, with possibly finitely many exceptions, among any p1⋯pk−1pk+1p_1\cdots p_{k-1}p_{k+1} consecutive integers one has more than kk prime factors, suggests that p1⋯pkp_1\cdots p_k may be the right value, and says that even for k=2k=2 the authors cannot improve Schinzel's value.

Source: https://combinatorica.hu/~p_erdos/1967-21.pdf.

Results. Labels and pages are the print's.

  • Theorem (p. 428, unnumbered; proof on p. 428): v0(n)>1v_0(n)>1 for every positive integer nn except n=1,2,3,4,7,8,16n=1,2,3,4,7,8,16.
  • Conjecture (p. 429, unnumbered; posed, not proved): for every kk, lim sup⁡n∑i=0kν(n+i)log⁡log⁡nlog⁡n=1\limsup_n\sum_{i=0}^k\nu(n+i)\frac{\log\log n}{\log n}=1.
  • Inequality (1) (p. 430; derived from Pólya's theorem): $\liminf_n\sum_{i=0}^{k-1}\nu(n+i)\ge k+\pi(k)-1$, with the conjecture (2) that the liminf is at most k+π(k)k+\pi(k).
  • Remark (p. 430, unlabeled; reported without proof): Schinzel's interval length p1⋯pk−1pk+1p_1\cdots p_{k-1}p_{k+1}, and the question whether p1⋯pkp_1\cdots p_k suffices.

Bears on.

  • #889: the problem's question whether v0(n)→∞v_0(n)\to\infty is stated as an expectation here (p. 428); the theorem gives only v0(n)≥2v_0(n)\ge2 outside the seven listed values.
  • #890: the problem's second question is the conjecture on p. 429, posed and not proved (the paper's sum runs over 0≤i≤k0\le i\le k, the problem's over 0≤i<k0\le i<k); its first question, on prime factors exceeding kk, is not the paper's (2), which counts all distinct prime factors, and the derivation of inequality (1) gives only the lower bound k−1k-1 for the problem's liminf.
  • #891: the problem is the authors' question in the remark on p. 430, where Schinzel's reported result gives the longer length p1⋯pk−1pk+1p_1\cdots p_{k-1}p_{k+1} with possibly finitely many exceptions.

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