Problem 261

Are there infinitely many $n$ such that there exists some $t\geq 2$ and distinct integers $a_1,\ldots,a_t\geq 1$ such that

\frac{n}{2^n}=\sum_{1\leq k\leq t}\frac{a_k}{2^{a_k}}?

Is this true for all $n$ ? Is there a rational $x$ such that

x = \sum_{k=1}^\infty \frac{a_k}{2^{a_k}}

has at least $2^{\aleph_0}$ solutions?

Source: erdosproblems.com/261

Number theory

Wiki page

Status

Open

No claim settles this problem.

Open. The site labels the problem OPEN (page last edited 1 December 2025; accessed 2026-10-07). Its remarks credit Borwein and Loring [BoLo90] with an identity giving infinitely many $n$ , which answers the first question yes, and Tengely, Ulas and Zygadło [TUZ20] with a check of the second question for every $n\le10000$ ; both are refereed results with accepted partial claim pages, Borwein and Loring 1990 and Tengely, Ulas and Zygadło 2020, and Borwein and Loring's reduction of the second question to a termination conjecture is a conditional claim page (Borwein Loring Conditional, 1990). No claim settles or pends on the second or the third question, so the derived standing is open, claim none. The three questions are the parts infinitely_many, all_n and continuum of the frontmatter.

BoLo90: Borwein, Peter and Loring, Terry A., Some questions of Erdős and Graham on numbers of the form $\sum g_n/2^{g_n}$ . Math. Comp. 54 (1990), no. 189, 377-394.
Er88c: Erdős, P., On the irrationality of certain series: problems and results. New advances in transcendence theory (Durham, 1986) (1988), 102-109.
TUZ20: Tengely, Szabolcs and Ulas, Maciej and Zygadło, Jakub, On a Diophantine equation of Erdős and Graham. J. Number Theory 217 (2020), 445-459.

Problem 261

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