Erdős–Turán additive basis conjecture

Choose nonnegative integers so every sufficiently large integer can be made by adding two choices. Must some totals have arbitrarily many ways? A small example uses 0, 1, 3, 4: total 4 has 0+4 and 1+3, counted once each; total 6 can use 3+3. Reversing a pair adds no new way. This finite set illustrates the counting rule, not coverage of all large integers.[1]

Can a set cover every large integer by two-term sums while keeping the number of ways uniformly small? Take $A\subseteq\mathbb N_0$ and count $r_A(n)=\#\{(a,b)\in A^2:a\leq b,\ a+b=n\}$. Repeated elements are allowed. An order-two asymptotic basis has $r_A(n)\geq1$ for all sufficiently large $n$. The conjecture requires $\limsup_{n\to\infty}r_A(n)=\infty$, not many representations of every large integer. For example, the pairs $0+4$ and $1+3$ count separately, whereas reversing a pair does not.[1]

Taking every nonnegative integer works, and large sums then have many representations. The missing result must cover every set that eventually represents all integers, including much sparser choices. No fixed ceiling on the number of ways is known to be possible in this nonnegative setting.[1]

Unique-representation constructions in ℤ do not answer the nonnegative target.[1] [2]

Choosing numbers to cover all large totals competes with keeping their pair sums distinct. The conjecture says that complete coverage eventually forces collisions of arbitrarily high multiplicity. It connects two different measures of a set: which totals it reaches and how many ways it reaches them.[1] [2]

Representation functions distinguish exhaustive coverage from control of additive collisions. Sidon sets optimize the latter by avoiding collisions; bases require the former. Erdős–Turán would rule out every bounded-multiplicity compromise for an eventual nonnegative two-sum cover. The contrasting unique-representation bases over all integers show that this tension depends on the domain, rather than on addition alone.[1] [2]