Problem Statement
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 $\epsilon>0$)?
What if $1_A\ast 1_A(n) >\epsilon \log n$ (for all large $n$, for arbitrary fixed $\epsilon>0$)?
Categories:
Number Theory Additive Basis
Progress
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].
Source: erdosproblems.com/868 | Last verified: January 19, 2026