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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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Statement

Setting (p. 109). a1<a2<⋯a_1<a_2<\cdots is any sequence of integers, and b1<b2<⋯b_1<b_2<\cdots are the integers divisible by none of the aa's.

Gap densities. Assume the density of the bb's exists, which the paper notes is certainly the case when ∑1/ai<∞\sum1/a_i<\infty. Then the density of the bib_i with bi+1−bi=tb_{i+1}-b_i=t exists (printed bi−1−bib_{i-1}-b_i [sic]). This is the paper's generalization of Lemma 1; it gives no proof and remarks only that the statement follows almost immediately from a theorem of Davenport and Erdős: if ckc_k is the density of the integers divisible by none of a1,…,aka_1,\ldots,a_k and cc the density of those divisible by no aia_i, then c=lim⁡k→∞ckc=\lim_{k\to\infty}c_k.

First-moment tail. By the same theorem, the paper says, it is easy to see that if the density of the bb's exists and is positive, then to every ε\varepsilon there is a cεc_\varepsilon with

∑bi+1≤xbi+1−bi>cε(bi+1−bi)<εx.\sum_{\substack{b_{i+1}\le x\\ b_{i+1}-b_i>c_\varepsilon}}(b_{i+1}-b_i)<\varepsilon x.

The example. No stronger result holds in general, even when ∑1/ai<∞\sum1/a_i<\infty: taking the aa's to be the integers in the intervals [2k,2k(1+1/k2)][2^k,2^k(1+1/k^2)] gives ∑1/ai<∞\sum1/a_i<\infty but

lim⁡x→∞1x∑bi<x(bi+1−bi)1+ε=∞,\lim_{x\to\infty}\frac1x\sum_{b_i<x}(b_{i+1}-b_i)^{1+\varepsilon}=\infty,

with lim⁡\lim as printed; the exponent is faint in the print and reads as 1+ε1+\varepsilon. Erdős adds that he does not know whether this can happen when ∑1/ai<∞\sum1/a_i<\infty and (ai,aj)=1(a_i,a_j)=1.

Source. P. Erdős, Some problems and results in elementary number theory, Publ. Math. Debrecen 2 (1951), 103--109, doi:10.5486/pmd.1951.2.2.04: the closing paragraphs on p. 109. The Davenport--Erdős theorem is cited there (footnote 4) from H. Davenport and P. Erdős, On sequences of integers, Acta Arithmetica 2 (1936), 147--151, and On sequences of positive integers, J. Indian Math. Soc. 15, Part A (1951), 19--24. The edition read is identified on the source card.

Read depth. Claims checked: the statements were read clause by clause on the printed page. The paper proves none of them, and nothing was checked beyond the statements.

Proof pointer

None in the paper beyond the reduction to the Davenport--Erdős theorem named above.

Dependencies

The Davenport--Erdős theorem, carded at davenport_1936_sequences_positive_integers and davenport_1951_sequences_positive_integers.

Bears on

  • Problem 489: context for the general question. The remarks concern the sequence BB of the problem for arbitrary AA, but control only the first moment; the example has ∑1/ai<∞\sum1/a_i<\infty and an unbounded (1+ε)(1+\varepsilon)-moment, yet its aa's up to xx number far more than x1/2x^{1/2} (an observation of this page, not of the paper), so it lies outside the problem's hypothesis ∣A∩[1,x]∣=o(x1/2)|A\cap[1,x]|=o(x^{1/2}). The remarks settle no case of the problem beyond the squarefree one on the page for (23) with α = 2.