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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

Throughout the paper (printed p. 28) 1≤a1<a2<⋯1\le a_1<a_2<\cdots is a sequence of integers, A(x)=∑ai≤x1A(x)=\sum_{a_i\le x}1, and a sequence has density 00 when A(x)/x→0A(x)/x\to0. Condition (1) is that the equation

ak=ai1+ai2+⋯+air,i1<i2<⋯<ir,(1)a_k=a_{i_1}+a_{i_2}+\cdots+a_{i_r},\qquad i_1<i_2<\cdots<i_r, \tag{1}

has no solution for any kk and rr: no term is a sum of distinct other terms (the display is printed with the index condition i1<⋯<iri_1<\cdots<i_r alone; the English summary on printed p. 38 writes it with ir<ki_r<k, and Theorem I's hypothesis, that no aa splits into a sum of distinct aa's, implies the same; the site's Problem 876 writes the same condition as a=b1+⋯+bra=b_1+\cdots+b_r with b1<⋯<br<ab_1<\cdots<b_r<a in AA).

Theorem I (printed p. 28). If no aa can be written as a sum of distinct other aa's, then AA has density 00.

Theorem II (printed p. 30). If (1) has no solution, then ∑i=1∞1/ai\sum_{i=1}^\infty1/a_i converges, and indeed always

∑i=1∞1ai<103.\sum_{i=1}^\infty\frac1{a_i}<103.

After the proof (printed p. 31): it would be easy to improve 103103 substantially, but the exact constant is not determined.

Theorem III (printed p. 31). If (1) has no solution, then

lim inf⁡x→∞A(x) x−(5−1)/2<∞.(13)\liminf_{x\to\infty}A(x)\,x^{-(\sqrt5-1)/2}<\infty. \tag{13}

The theorem shows that Theorem I cannot be improved for every xx but that a much sharper inequality holds for infinitely many xx.

Construction (printed pp. 32–33). There is a sequence with (1) unsolvable and A(x)>cx2/7A(x)>cx^{2/7} for every xx: put a1=1a_1=1; when a1,…,akia_1,\ldots,a_{k_i} are chosen, put Bi+1=2∑r≤kiarB_{i+1}=2\sum_{r\le k_i}a_r and

aki+l=1+lBi+1,1≤l≤[Bi+1210],(16)a_{k_i+l}=1+lB_{i+1},\qquad1\le l\le\Bigl[\frac{B_{i+1}^2}{10}\Bigr], \tag{16}

so that aki+1≤Bi+13/10+1a_{k_{i+1}}\le B_{i+1}^3/10+1. The paper then asks (printed p. 33) for the supremum β\beta of the α\alpha for which some sequence with (1) unsolvable has A(x)>cxαA(x)>cx^\alpha for every xx, and records 2/7≤β≤(5−1)/22/7\le\beta\le(\sqrt5-1)/2.

Remarks printed with the theorems: density 00 also follows when (1) has only finitely many solutions (p. 29); if (1) is excluded only for 2≤j≤r2\le j\le r summands, the upper density is at most 1/r1/r, and ak=1+kra_k=1+kr shows this cannot be sharpened (p. 29); Theorem I is best possible in the sense that for f(x)→∞f(x)\to\infty arbitrarily slowly there is a sequence with (1) unsolvable and A(x)>x/f(x)A(x)>x/f(x) for infinitely many xx (pp. 29–30).

Source. P. Erdős, Számelméleti megjegyzések, III. Néhány additív számelméleti problémáról, Mat. Lapok 13 (1962), 28–38 (Hungarian; Russian and English summaries on printed pp. 37–38); printed p. nn is PDF p. n−27n-27 of the eleven-page scan read for this page. Theorems I–III on printed pp. 28, 30 and 31 (PDF pp. 1, 3 and 4), the construction on pp. 32–33 (PDF pp. 5–6), the English summary on p. 38 (PDF p. 11), all read on the page images; the prose is rendered here in the corpus's words, and the displays keep the paper's numbering in modern notation ((13) is printed with x=∞x=\infty under the liminf and as A(x)/x(5−1)/2A(x)/x^{(\sqrt5-1)/2}).

Read depth. Claims checked: the three theorem statements, the construction (16) and the question on β\beta were read clause by clause on the page images, and the English summary (p. 38) was compared with them. The proofs were read for their structure only and are not checked or reconstructed here.

Proof pointer

Theorem I (pp. 28–29): the shifted sequences Ar={a1+⋯+ar+ak:k>r}A_r=\{a_1+\cdots+a_r+a_k:k>r\}, r≥0r\ge0, are pairwise disjoint when (1) is unsolvable, which gives (2) x≥∑i≤kAi(x)≥(k+1)Ak(x)x\ge\sum_{i\le k}A_i(x)\ge(k+1)A_k(x) and (4) A(x)≤x/(k+1)+∑i≤kai+kA(x)\le x/(k+1)+\sum_{i\le k}a_i+k for every kk. Theorem II (pp. 30–31): the indices jj are split by whether A(2j+1)−A(2j)≤2j/j2A(2^{j+1})-A(2^j)\le2^j/j^2; the first class contributes less than ∑1/j2<2\sum1/j^2<2 (display (5)), and for the second class the distinct sums a1+⋯+ar+aka_1+\cdots+a_r+a_k bound A(2jr+1)A(2^{j_r+1}) from above (displays (6)–(9)), giving (10)–(12) and the total 103103. Theorem III (pp. 31–32): if (13) failed then ak=o(k(1+5)/2)a_k=o(k^{(1+\sqrt5)/2}), and displays (14)–(15) with (2) give a contradiction. The construction's sum-freeness (pp. 32–33) is checked by comparing residues modulo Bi+1B_{i+1}, and the lower bound A(x)>cx2/7A(x)>cx^{2/7} follows from (19). An English outline of the Theorem II argument, with an unspecified absolute constant in place of 103103, is Theorem 2 of Benkoski and Erdős.

Dependencies

None; the arguments are elementary. Footnote 1 (p. 29) refers to Problem 4268 of the American Mathematical Monthly for the sharpness example.

Bears on

  • Problem 876: the density-zero theorem the site attributes to this paper; the reciprocal-sum bound 103103 that opens the reciprocal-sum question of the site's commentary (Erdős's later restatements print 103103 in 1975 and 100100 in 1977); Theorem III and the x2/7x^{2/7} construction as the 1962 bounds on how dense such a sequence can be, both superseded on the density side by Łuczak and Schoen's Theorem 3 and construction. The paper says nothing about the gaps an+1−ana_{n+1}-a_n the problem asks about beyond what its density statements imply.