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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 2.8 (p. 6). If $a_1<\cdots<a_k$ is a solution of (1), then

$$a_k\le2n+2k\log_2k.$$

Combined with the bound $n\le2^{k+1}-k-2$ of Theorem 2.1 it gives Corollary 2.9, a bound on $a_k$ in terms of $k$ alone.

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 2.8 and its proof on p. 6.

Read depth. Claims checked: the statement was read clause by clause on the page image, and the bound was checked here against every solution listed in Theorem 2.5. The proof was read but not checked step by step. Nothing here is independently reviewed.

Proof pointer

Page 6. For $k\le8$ the paper checks the bound on the complete list of Theorem 2.5. For $k\ge8$, if $a_k>2n+2k\log_2k$, then since $x^{k-1}/2^x$ decreases for $x>(k-1)/\ln2$, Corollary 2.4 ($a_k^{k-1}/2^{a_k}\ge2^{-a_1}$) and $a_1\le n+3$ give $\bigl((2n+2k\log_2k)/k^2\bigr)^{k-1}>2^{n-3+2\log_2k}$; this fails at $n=1$, and raising $n$ by one doubles the right side while multiplying the left side by less than $e^{1/2}<2$.

Dependencies

Theorem 2.1 ($a_1\le n+3$), Corollary 2.4 of the same paper, and Theorem 2.5 for $k\le8$.

Bears on

• Problem 261: for a given $n$ and number of terms $k$, the theorem bounds every term of a representation of $n/2^n$, so whether one exists is a finite search. It gives no bound on $k$ in terms of $n$, so it does not reduce the problem's second question to a finite computation for any $n$, and it does not touch the other two questions.