Wiki
Wiki

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

Updated


Statement

Let EN={∑n=1Ntn/n:t1,…,tN∈{0,1}}E_N=\{\sum_{n=1}^Nt_n/n:t_1,\ldots,t_N\in\{0,1\}\} (display (1), p. 1), the set of values of sums of distinct unit fractions with denominators at most NN, the empty sum included; so ∣EN∣|E_N| is the S(N)S(N) of Problem 320. Write ln⁡k\ln_k for the kk-th iterate of the natural logarithm.

Theorem 1 (p. 2). Let kk and NN be positive integers. In each case below whose condition holds,

ln⁡(∣EN∣) ≥ 2ln⁡2 Nln⁡N×{1if ln⁡2N≥1,ln⁡3Nif ln⁡3N≥1,(1−3/2ln⁡kN)∏j=3kln⁡jNif k≥4 and ln⁡kN≥3/2.\ln(|E_N|)\ \ge\ 2\ln2\,\frac{N}{\ln N}\times \begin{cases} 1 & \text{if }\ln_2N\ge1,\\[2pt] \ln_3N & \text{if }\ln_3N\ge1,\\[2pt] \Bigl(1-\dfrac{3/2}{\ln_kN}\Bigr)\displaystyle\prod_{j=3}^{k}\ln_jN & \text{if }k\ge4\text{ and }\ln_kN\ge3/2. \end{cases}

The abstract writes the third case as ln⁡(∣EN∣)ln⁡2≥(2−3ln⁡kN)Nln⁡N∏j=3kln⁡jN\frac{\ln(|E_N|)}{\ln2}\ge\bigl(2-\frac3{\ln_kN}\bigr)\frac{N}{\ln N}\prod_{j=3}^k\ln_jN, the same bound. The proof (p. 9) gives the constant 1.41.4 in place of 3/23/2.

Source. S. Bettin, L. Grenié, G. Molteni and C. Sanna, A lower bound for the number of Egyptian fractions, arXiv:2509.10030v1 (12 September 2025; the only arXiv version listed on 2026-09-18), 12 pages; Theorem 1 on p. 2 (page image), the definition (1) on p. 1, the proof in Section 2, pp. 3--9 (text layer). Published in Mathematics of Computation, DOI 10.1090/mcom/4190, online 22 January 2026 (Crossref record read; the record gives no volume or pages yet); the published text was not compared, and the locators here are v1 locators.

Read depth. Claims checked: Theorem 1, the definition of ENE_N and Lemmas 1 and 2 were read clause by clause. The proof was read for structure (below) and is not verified here.

Proof pointer and sketch (Section 2)

Let U\mathcal U be the set of N≥1N\ge1 such that ∑n=1N−1wn/n≠1/N\sum_{n=1}^{N-1}w_n/n\ne1/N for all w1,…,wN−1∈{−1,0,1}w_1,\ldots,w_{N-1}\in\{-1,0,1\}, and U(x)=U∩[1,x]\mathcal U(x)=\mathcal U\cap[1,x]. Lemma 1: N∈UN\in\mathcal U if and only if ∣EN∣=2∣EN−1∣|E_N|=2|E_{N-1}| (the union EN=EN−1∪(EN−1+1/N)E_N=E_{N-1}\cup(E_{N-1}+1/N) is disjoint exactly then). Lemma 2: ∣EN∣≥2∣U(N)∣|E_N|\ge2^{|\mathcal U(N)|}. Lemma 4: if m∈Um\in\mathcal U and pp is a prime compatible with mm (in particular p>gm=lcm(1,…,m)∑j≤m1/jp>g_m=\mathrm{lcm}(1,\ldots,m)\sum_{j\le m}1/j), then mpk∈Ump^k\in\mathcal U for every k≥1k\ge1; Lemma 5 lists U(100)\mathcal U(100) explicitly. Lemma 7 turns this into the integral recursion ∣U(x)∣≥xln⁡x∫1y∣U(v)∣v−2 dv|\mathcal U(x)|\ge\frac{x}{\ln x}\int_1^y|\mathcal U(v)|v^{-2}\,dv for y≥1y\ge1, x≥18⋅3yx\ge18\cdot3^y, which Lemmas 8--10 iterate through the functions G(z)=ln⁡xx∣U(x)∣G(z)=\frac{\ln x}{x}|\mathcal U(x)|, x=exp⁡(exp⁡z)x=\exp(\exp z), and TkT_k; the proof ends (p. 9) with ln⁡∣EN∣ln⁡2≥2(1−1.4ln⁡kN)Nln⁡N∏j=3kln⁡jN\frac{\ln|E_N|}{\ln2}\ge2(1-\frac{1.4}{\ln_kN})\frac{N}{\ln N}\prod_{j=3}^k\ln_jN for k≥4k\ge4, obtained as a lower bound for ∣U(N)∣|\mathcal U(N)| and then Lemma 2. The first two cases come from Lemma 2 with Lemma 6's bounds ∣U(x)∣≥2x/ln⁡x|\mathcal U(x)|\ge2x/\ln x (x≥13x\ge13) and the case k=1k=1 of Lemma 8.

The paper compares its bound with Bleicher and Erdős's: their 1976 Theorems 2 and 3 give αNln⁡N∏j=3kln⁡jN≤ln⁡∣EN∣≤Nln⁡kNln⁡N∏j=3kln⁡jN\alpha\frac{N}{\ln N}\prod_{j=3}^k\ln_jN\le\ln|E_N|\le\frac{N\ln_kN}{\ln N}\prod_{j=3}^k\ln_jN with α=e−1\alpha=e^{-1} for ln⁡2kN≥1\ln_{2k}N\ge1, and their 1975 Corollaries 1--3 raise α\alpha to ln⁡2\ln2 under ln⁡kN≥k\ln_kN\ge k (p. 1); relaxing the condition to ln⁡kN≥3/2\ln_kN\ge3/2 admits larger kk and so improves the order of growth, not only the constant (p. 2). Section 3 computes ∣EN∣|E_N| exactly for N≤154N\le154 (Table 2), extending OEIS A072207's values for N≤83N\le83.

Relation to Problem 321

The set U(N)\mathcal U(N) has all its subset reciprocal sums distinct: if two distinct subsets B≠CB\ne C of U(N)\mathcal U(N) had equal reciprocal sums, one could drop their common elements and take the largest remaining element uu, say u∈Bu\in B; then 1/u1/u would equal a {−1,0,1}\{-1,0,1\}-combination of the 1/n1/n with n<un<u, contradicting u∈Uu\in\mathcal U. So, in the notation of Problem 321, R(N)≥∣U(N)∣R(N)\ge|\mathcal U(N)|, and the proof's lower bound for ∣U(N)∣|\mathcal U(N)| gives R(N)≥2(1−3/2ln⁡kN)Nln⁡N∏j=3kln⁡jNR(N)\ge2(1-\frac{3/2}{\ln_kN})\frac{N}{\ln N}\prod_{j=3}^k\ln_jN for k≥4k\ge4 and ln⁡kN≥3/2\ln_kN\ge3/2. The paper does not state this consequence; the site's Problem 321 page calls the lower bound "implicit" in the paper and the proof claim accepted there attributes it to "the dissociated set constructed in" the paper. The three-line argument above is written for that page and is not taken from a source; the accepted claim's Lean file proves the same finite statement (BGMSU_dissociated, pow_card_BGMSU_le_harmonic_subsetSums), as the problem page records.

Dependencies

Rosser's explicit bound for ψ\psi (the paper's [8], p. 228, giving ψ(m)≤1.04m\psi(m)\le1.04m in the proof of Lemma 3); Rosser and Schoenfeld's explicit bounds for π(x)\pi(x) (the paper's [9]: Th. 1 in Lemma 3, and Th. 2, Cor. 1 in Lemmas 6 and 7); Lemma 5's explicit list, checked by the authors.

Bears on

  • Problem 320: the best refereed lower bound for log⁡S(N)\log S(N), matched in order by the site-accepted upper bound of July 2026.
  • Problem 321: the lower bound for R(N)R(N) through the dissociated set U(N)\mathcal U(N), as above.