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Statement

Conjecture 3.7 (p. 11). If the equation $n/2^n=\sum_{i=1}^{k}a_i/2^{a_i}$ has a solution $(n,k,a_1,\ldots,a_k)$ with $a_1<a_2<\cdots<a_k$, then

$$k+n\le a_k\le2(k+n).$$

In particular $a_k\le4(2^k-1)$.

The paper bases the conjecture on its numerical data (Figure 2, p. 11, plots $a_k(n)/2(k+n)$ for the greedy solutions with $n\le5000$). The "in particular" follows from the upper bound and $n\le2^{k+1}-k-2$ (Theorem 2.1).

Remark 3.8 (p. 12). The lower bound cannot be raised: for $n=2^{k+1}-k-2$ the solution $a_i=n+i$ has $a_k=n+k$. The upper bound holds under the extra hypothesis $n\ge2^k-k$. Both of the paper's points are recorded here; the lower bound itself already follows from $a_1\ge n+1$ (Theorem 2.1) and $a_1<\cdots<a_k$, so the conjecture's content is the upper bound for $n<2^k-k$. The bound was also checked here against every solution listed in Theorem 2.5.

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. Conjecture 3.7 and Figure 2 on p. 11, Remark 3.8 on p. 12.

Read depth. Claims checked: the conjecture and Remark 3.8 were read clause by clause on the page images; the argument of the remark was read but not checked step by step. Nothing here is independently reviewed.

Proof pointer

For Remark 3.8 (p. 12): bounding $a_i\ge n+i$ for $i<k$ gives $(n+k+1-2^k)/2^{n+k-1}\le a_k/2^{a_k}$; if $a_k>2(k+n)$ and $2^k-k\le n\le2^{k+1}-k-2$, monotonicity of $x/2^x$ turns this into $2^{n+k-1}\le2^k-1$, a contradiction.

Dependencies

Theorem 2.1.

Bears on

• Problem 261: the conjecture and the remark bound the terms of a representation of $n/2^n$ that is already given; they say nothing on whether one exists for a given $n$, nor on the problem's other questions.