So we know that given a Riemannian manifold $(M,g)$ we can calculate its Riemannian curvature $R$. This covariant $4$-tensor field then satisfies some important symmetries \begin{gather*} R_{ijkl} = - R_{ijlk},\quad R_{ijkl} = R_{jikl},\quad R_{ijkl} = R_{klij}, \\ R_{ijkl} + R_{iklj} + R_{iljk} = 0. \end{gather*}

The main question of the global existence of a metric for a prescribed curvature was already explained by Robert Bryant (thank you to @Deane Yang for the summary). However, this answer only describes why such a curvature exists. Hence, I'd like to know whether there is a concrete example of a curvature that is not generated by a metric.