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Statement

Setting as on Theorem 2.1: equation (1) is $n/2^n=\sum_{i=1}^{k}a_i/2^{a_i}$ with $k>1$ and $a_1<\cdots<a_k$.

Theorem 3.5 (p. 8), quoted: "For each $2\le n\le10^4$ the Diophantine equation (1) has a solution in variables $k,a_1,\ldots,a_k$ satisfying $a_i=n+1$ [sic] and $a_i\ge n+i$ for $i=2,\ldots k$."

The first condition is read as $a_1=n+1$, the reading of the paper's abstract ("a solution in integers $n+1=a_1<a_2<\ldots<a_k$", p. 1); the second already follows from the first and $a_1<\cdots<a_k$. The case $n=1$ is not covered by the theorem but is solved in Theorem 2.5, for instance $1/2=3/2^3+6/2^6+8/2^8$. The paper reports (p. 8) that Borwein and Loring had proved solvability for each $n\le10^3$, and that the question for every $n$ is essentially Borwein and Loring's Conjecture 1, which it does not answer.

The computation behind Table 2 and Figures 1--3 (pp. 10--12) records the number of terms $k(n)$ and the largest term $a_k(n)$ that the greedy algorithm returns; it is irregular, for example $k(5588)=460536$.

Source. Sz. Tengely, M. Ulas and J. Zygadło, On a Diophantine equation of Erdős and Graham, J. Number Theory 217 (2020), 445--459, doi:10.1016/j.jnt.2020.05.006, read in arXiv:2008.01501v1 as identified on the source card; labels and pages are that preprint's. Theorem 3.5 on p. 8, the algorithm on pp. 8--9, Table 2 on p. 10.

Read depth. Claims checked: the statement was read clause by clause on the page image. The computation was not rerun. Nothing here is independently reviewed.

Proof pointer

Pages 8--9, by computer. The greedy strategy appends at each step the smallest $j$ with $j/2^j$ not exceeding what remains of $n/2^n$. The paper implements a variant of Borwein and Loring's Algorithm 2: for rational $0<x<2$ it starts from $k_0=\min\{k\ge1:k/2^k<x\}$ and $x_{k_0}=x\cdot2^{k_0-1}$, and iterates $x_{i+1}=2x_i-i$ when this is nonnegative and $x_{i+1}=2x_i$ otherwise; the run terminates when some $x_i=0$, and the indices $j$ with $x_{j+1}\ne2x_j$ are the terms of the representation. For $x=n/2^n$ the first term chosen is $n+1$.

Dependencies

None in this paper; the algorithm modifies Borwein and Loring's Algorithm 2 (Borwein and Loring 1990).

Bears on

• Problem 261: with the case $n=1$ from Theorem 2.5, every $n\le10^4$ has the property of the problem's second question, $n/2^n$ being a sum of at least two distinct terms $a/2^a$. The theorem says nothing about $n>10^4$ and leaves the second question open; it does not address the first or the third.