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Erdős: Many old and on some new problems of mine in number theory

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problem_iii_3: Erdős and Harzheim's questions whether a sequence in which no term is a sum of consecutive terms has upper density at most 1/2, lower density 0 and logarithmic density 0, with Erdős's construction of upper density 1/2 and his bound (1) on the reciprocal sum over (x, x^2), whose printed proof has a gap.


The copy read for this card is a scan of the Congressus Numerantium 30 article, 25 pages (PDF p. n is printed p. 2+n). No notice is printed on pp. 3--4 or 26--27 of the scan; its download URL is not recorded, and its OmniPage creator and 2006 modification stamp match the Rényi archive's scans, whose site footer speaks for the site, not the paper (https://users.renyi.hu/~p_erdos/, read: "(C) 2005-2007 All rights reserved. All material on this site is for scientifics purposes only."); the publisher has no online page for Congressus Numerantium, so none was consulted, and no Crossref license is recorded; the term is unstated.

Paul Erdős, Many old and on some new problems of mine in number theory, Congressus Numerantium 30 (1981), 3–27.

Overview

This is a collection of problems and brief results, with proofs given only rarely (introduction, p. 3). Part I (§§1–6, pp. 3–14) concerns primes; Part II (§§1–2, pp. 14–19) concerns consecutive integers; Part III (§§1–16, pp. 19–27) contains miscellaneous questions. The passage relevant to E839 is III.3 (pp. 20–21).

In III.3, Erdős and Harzheim ask whether a sequence containing no term that is a sum of consecutive terms has upper density at most 1/21/2, lower density zero, and logarithmic density zero (p. 20). Erdős gives an iterative construction with upper density 1/21/2: after a finite initial segment ending at m=akm=a_k, append m4m^4, then m4+m2,m4+m2+1,…,2m4−1m^4+m^2,m^4+m^2+1,\ldots,2m^4-1 (p. 21). This is a construction, not a bound for all avoiding sequences. He conjectures that the reciprocal series might converge, then asserts the weaker bound ∑x<ak<x21/ak<c\sum_{x<a_k<x^2}1/a_k<c as III.3(1) (p. 21).

The printed proof of III.3(1) is defective. Its estimate III.3(2) assumes that all the consecutive-block sums under consideration are distinct; avoidance of individual terms does not imply this. The next displayed comparison, ∑i=uvai<(v−u)av\sum_{i=u}^{v}a_i<(v-u)a_v, also has an incorrect term count. Consequently III.3(1) cannot be treated as proved by the argument on p. 21. The paper states no theorem settling the lower-density or logarithmic-density questions.

Results.

  • Problem III.3 (pp. 20–21): the density questions, the construction of upper density 1/21/2, the conjecture ∑k1/ak<∞\sum_k1/a_k<\infty and the bound (1), whose printed proof has a gap.

Read status: claims checked for item III.3 (pp. 20–21), read clause by clause on the page images; no other item was read for this card. Nothing here is independently reviewed.

Relation to E839

Bears on. #839: III.3 (p. 20) asks the problem's two questions, as the lower-density and logarithmic-density questions, for sequences in which no term is a sum of consecutive terms, and answers neither; its construction (p. 21) decides neither, and its bound (1) (p. 21), whose printed proof has a gap, would answer the second yes if it held for every such sequence, even with a constant depending on the sequence.

For E839, write A(x)=#{n:an<x}A(x)=\#\{n:a_n<x\}. With strictly increasing positive integers, lim sup⁡an/n=∞\limsup a_n/n=\infty is equivalent to lim inf⁡A(x)/x=0\liminf A(x)/x=0; the second question is whether ∑an<x1/an=o(log⁡x)\sum_{a_n<x}1/a_n=o(\log x). These are the lower-density and logarithmic-density questions posed in III.3 (p. 20). The paper writes 1≤a1≤a2≤⋯1\le a_1\le a_2\le\cdots, allowing repeated terms; for strictly increasing sequences its condition and the problem's coincide, since a run of two or more positive terms summing to a term uses only earlier terms.

The construction in III.3 (p. 21) shows why upper density alone does not answer either question. At the end of each dense block, an/n→2a_n/n\to2 along a subsequence. At the first term m4m^4 of the next block, its index is at most m+1m+1, so an/n→∞a_n/n\to\infty along those indices. Each block contributes O(1)O(1) to the reciprocal sum and the block sizes grow by fourth powers; hence this particular sequence has reciprocal sum O(log⁡log⁡x)O(\log\log x). It therefore satisfies both conclusions asked about in E839 while having upper density 1/21/2.

If III.3(1) held uniformly for every avoiding sequence, intervals [x,x2)[x,x^2) would give ∑an<x1/an=O(log⁡log⁡x)\sum_{a_n<x}1/a_n=O(\log\log x), answering E839's logarithmic-density question. The paper does not establish that premise. For example, 4,6,7,8,94,6,7,8,9 obeys the avoidance condition, yet 4+6+7=8+94+6+7=8+9; thus the distinctness used in III.3(2) (p. 21) fails. The proposed reciprocal-sum argument is a possible route to E839, not a resolution.

No file of this source is held: no license on record permits its redistribution, and the card cites the edition it names above.