Seymour's second neighbourhood conjecture is a simple question about directed networks. Take any network in which no two points point at each other. Is there always a point that reaches at least as many new points in two steps as it reaches in one? The answer is known for tournaments, and for networks in which some point sends at most six arrows, but not in general. This release studies the densest case that is still open: networks whose number of points is two times the minimum number of outgoing arrows, plus three. It does not settle that case. It proves that any counterexample there would need a rigid structure, including at least four points joined to every other point. Using these facts as constraints, a large computer search rules out a counterexample with seventeen points. Every one of its three hundred and seventy-nine proof certificates is kept, and each was accepted by two separate checkers, one of them formally verified. So any counterexample must have at least eighteen points. The paper also shows that a weighted version of the classical tournament proof cannot work in general. As a result, a stronger conjecture attributed to DeVos, as printed in a 2006 survey, is false. The main limitation is this: the step from the graph question to the computer formula rests on written proofs that have not been machine-checked or reviewed by independent specialists. This is an unrefereed candidate from Evidence Press, dated September twenty-fourth, twenty twenty-six. The paper and all evidence are linked on the release page. This is an AI-generated voice summary, not additional evidence.