DISPROVED
This has been solved in the negative.
If $A$ is an additive basis of order $2$, and $1_A\ast 1_A(n)\to \infty$ as $n\to \infty$, then must $A$ contain a minimal additive basis of order $2$? (i.e. such that deleting any element creates infinitely many $n\not\in A+A$)
What if $1_A\ast 1_A(n) >\epsilon \log n$ (for all large $n$, for arbitrary fixed $\epsilon>0$)?
A question of Erdős and Nathanson
[ErNa79], who proved that this is true if $1_A\ast 1_A(n) > (\log \frac{4}{3})^{-1}\log n$ for all large $n$.
Härtter
[Ha56] and Nathanson
[Na74] proved that there exist additive bases which do not contain any minimal additive bases.
Erdős and Nathanson
[ErNa89] proved that, for any $t$, there exists $A$ such that $1_A\ast 1_A(n)\geq t$ for all large $n$ and yet $A$ does not contain a minimal asymptotic basis of order $2$.
See also
[870].
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This page was last edited 06 December 2025.
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T. F. Bloom, Erdős Problem #868, https://www.erdosproblems.com/868, accessed 2026-01-16
