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Statement

Theorem 1 (printed p. 617). If the integers $1\le a_1<\cdots<a_n$ have pairwise distinct subset sums, that is, the $2^n$ sums $\sum_{i=1}^n\varepsilon_ia_i$ with each $\varepsilon_i\in\{0,1\}$ all differ, then their reciprocals sum to less than $2$:

$$\sum_{i=1}^n\frac1{a_i}<2.$$

The paper introduces the theorem through the divisors of an integer: $n$ has property P if the $2^k$ sums $\sum\varepsilon_id_i$ over its divisors $d_i$ are distinct, and Theorem 1 gives $\sigma(n)/n<2$ for such $n$; "We conjectured this and the simple and ingenious proof is due to C. Ryavec" (p. 617).

Refinement (printed p. 619, after the proof). The paper says that the same argument, under the same distinct-subset-sum hypothesis, gives the sharper bound

$$\sum_{i=1}^n\frac1{a_i}\le2-\frac1{2^{n-1}},$$

with equality only for the powers of two, $a_i=2^{i-1}$ for $i=1,\ldots,n$. (The set $\{1,2,4,\ldots,2^{n-1}\}$ attains the bound, its reciprocal sum being $2-2^{1-n}$; the converse direction is what is printed.) The page then recalls Erdős's conjecture that distinct subset sums force $a_n>2^{n-C}$, with the offer of a prize (the site's Problem 1).

Source. S. J. Benkoski and P. Erdős, On weird and pseudoperfect numbers, Math. Comp. 28 (1974), no. 126, 617–623, DOI 10.1090/S0025-5718-1974-0347726-9 (Crossref record read); the copy read is a seven-page scan, printed p. $n$ on PDF p. $n-616$. Theorem 1 on printed p. 617 (PDF p. 1), the proof on pp. 617–619 (PDF pp. 1–3), the refinement on p. 619 (PDF p. 3), read on the page images.

Read depth. Claims checked: Theorem 1 and the refinement were read clause by clause on the page images. The one-page proof was read for its structure (below); it is not reconstructed or independently reviewed here.

Proof pointer

Pages 617–619 (Ryavec's argument). For $0<x<1$ the distinctness of the $2^n$ subset sums gives $\prod_{i=1}^n(1+x^{a_i})<\sum_{k\ge0}x^k=1/(1-x)$, hence $\sum_i\log(1+x^{a_i})<-\log(1-x)$; dividing by $x$ and integrating over $(0,1)$ (display (1)), then substituting $y=x^{a_i}$ in each term, $\sum_i\frac1{a_i}\int_0^1\frac{\log(1+y)}y\,dy<-\int_0^1\frac{\log(1-x)}x\,dx$, that is $\sum_i\frac1{a_i}\cdot\frac{\pi^2}{12}<\frac{\pi^2}6$, so $\sum1/a_i<2$. The refinement is stated as following from "the same argument" with no further detail.

Dependencies

None beyond the two classical integrals.

Bears on

• Problem 350: the status-defining source. The problem's dissociated set (all subset sums distinct) is the theorem's hypothesis (subsets of $A$ correspond to the vectors $(\varepsilon_i)$; a dissociated set contains no $0$, so its elements are $1\le a_1<\cdots<a_n$), and the conclusion is the problem's $\sum_{n\in A}1/n<2$. The refinement is the site's "$\sum1/n\le2-2^{1-|A|}$, with equality if and only if $A=\{1,2,\ldots,2^k\}$", where the source writes the extremal set as $a_i=2^{i-1}$, the powers of two up to $2^{n-1}$.
• Problem 469 (not linked from this page; the source card carries its row): the property P context in which the theorem is stated.