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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 2.5 (p. 5). For $k\in\{2,3,4,5,6,7,8\}$, with $A=(a_1,\ldots,a_k)$, the solutions $(n,A)$ of (1) are exactly the following.

Here $m,\ldots,m'$ means every integer from $m$ to $m'$. Since $1/2^1=2/2^2$, the solutions for $n=1$ and $n=2$ pair up with the same $A$.

The count on p. 6 reads $N(7)=3$, where $N(k)$ is the number of solutions of (1) with $k$ terms, but the list of the theorem has five entries for $k=7$; each of the 27 listed solutions was checked here in exact rational arithmetic and holds. The other counts on p. 6 ($N(2)=1$, $N(3)=6$, $N(4)=2$, $N(5)=4$, $N(6)=5$) agree with the list.

Corollary 2.6 (p. 5) uses the solutions with $n=1$ to give infinitely many rationals with at least three representations as an infinite sum of terms $a_i/2^{a_i}$; it is superseded by Corollary 3.6.

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 2.3 on p. 3, Corollary 2.4 on p. 4, Theorem 2.5 and Corollary 2.6 on p. 5, the counts $N(k)$ on p. 6.

Read depth. Claims checked: the list was read entry by entry on the page image, and every listed solution was verified in exact arithmetic. The completeness of the list rests on the paper's computation, which was not rerun. Nothing here is independently reviewed.

Proof pointer

Pages 3--5. Theorem 2.1 bounds $n$ by $2^{k+1}-k-2$ and fixes the first terms for large $n$. For small $n$ the paper uses Theorem 2.3 (p. 3), $2^{a_k-a_i}\le(a_{i+2}\cdots a_k)\,a_k$ for $1\le i\le k-2$, and its Corollary 2.4 (p. 4), $a_k^{k-1}/2^{a_k}\ge2^{-a_1}$, to bound $a_k$ in terms of $a_1$, and then searches; the paper reports that the case $k=8$ took more than two days of computing.

Dependencies

Theorem 2.1, with Theorem 2.3 and Corollary 2.4 of the same paper.

Bears on

• Problem 261: the list shows that $n=1,2,3,4,5,6,7,9,11,12,13,15,17,18,19,21,26,57,120,197,247,502$ each have the property of the problem's second question; in particular it supplies $n=1$, which Theorem 3.5 does not cover. It settles no part of the problem.