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Let $1\leq a_1<\cdots <a_k\leq n$ be integers such that all sums of the shape $\sum_{u\leq i\leq v}a_i$ are distinct. Let $f(n)$ be the maximal such $k$.

How does $f(n)$ grow? Is $f(n)=o(n)$?
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Asked by Erdős and Harzheim. In [Er77c] Erdős asks about an infinite such set of integers, and whether such a set must have density $0$. He notes that a simple averaging process implies $a_k \gg k\log k$ for infinitely many $k$, and so the lower density is $0$. He also asks whether $\sum\frac{1}{a_k}$ must converge.

Weisenberg in the comments observes that any set which satisfies [874] also has this property, which implies $f(n)\geq (2+o(1))n^{1/2}$.

If $g(n)$ is the maximal $k$ such that there are $1\leq a_1,\ldots,a_k\leq n$ with all consecutive sums distinct (i.e. we drop the monotonicity assumption in the definition of $f$) then Hegyvári [He86] has proved that\[\left(\frac{1}{3}+o(1)\right) n\leq g(n)\leq \left(\frac{2}{3}+o(1)\right)n.\]The upper bound of Coppersmith and Phillips in [867] implies\[g(n) \leq \left(\frac{2}{3}-\frac{1}{512}+o(1)\right)n.\]A similar question can be asked if we replace strict monotonicity with weak monotonicity (i.e. we allow $a_i=a_j$).

Erdős and Harzheim also ask what is the least $m$ which is not a sum of the given form? Can it be much larger than $n$? Erdős and Harzheim can show that $\sum_{x<a_i<x^2}\frac{1}{a_i}\ll 1$. Is it true that $\sum_i \frac{1}{a_i}\ll 1$?

See also [34], [356], [670], and [867]. The multiplicative analogue is [421].

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This page was last edited 12 January 2026.

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Additional thanks to: Sarosh Adenwalla, Dogmachine, and Desmond Weisenberg

When referring to this problem, please use the original sources of Erdős. If you wish to acknowledge this website, the recommended citation format is:

T. F. Bloom, Erdős Problem #357, https://www.erdosproblems.com/357, accessed 2026-01-16