Let $A$ be an additive basis of order $2$, and suppose $1_A\ast 1_A(n)\to \infty$ as $n\to \infty$. Can $A$ be partitioned into two disjoint additive bases of order $2$?

Let $A$ be an additive basis of order $2$, and suppose $1_A\ast 1_A(n)\to \infty$ as $n\to \infty$. Can $A$ be partitioned into two disjoint additive bases of order $2$?

A question of Erdős and Nathanson [ErNa88], who proved this is true if $1_A\ast 1_A(n) > c\log n$ (for all large $n$) for some constant $c>(\log\frac{4}{3})^{-1}$. Erdős and Nathanson [ErNa89] also proved that for every $t$ there exists a basis $A$ of order $2$ such that $1_A\ast 1_A(n)\geq t$ for all large $n$ and yet $A$ cannot be partitioned into two disjoint additive bases. This has been disproved by Larsen using Claude Opus 4.5 - in fact only a small modification of the argument of [ErNa89] is required.