If you neither know the metric nor the holonomy group, how do you recognize a curvature tensor is Riemannian?

I assume a curvature, by definition, satisfies Bianchi identities. I know it is Riemannian if there exists a symmetric non degenerate tensor $g_{ab}$ such that these satisfy the condition $g_{ea}R^e{}_{bcd}+g_{eb}R^e{}_{acd}=0,$ but its solutions are not unique for the metric (a homogeneous equation)

In $d$ dimensions, these $\frac{n(n+1)}{2}\frac{n(n-1)}{2}$ conditions bring us down from the $\frac{n^2(n+1)(n-1)}{3}$ components of the curvature to $\frac{n^2(n+1)(n-1)}{3.2.2}$ of the Riemann Tensor. Yet it seems to me these equations don't tell me what the metric should be

If you neither know the metric nor the holonomy group, how do you recognize a curvature tensor is Riemannian?

I assume a curvature, by definition, satisfies Bianchi identities. I know it is Riemannian if there exists a symmetric non degenerate tensor $g_{ab}$ such that these satisfy the condition $$g_{ea}R^e{}_{bcd}+g_{eb}R^e{}_{acd}=0 \, ,$$ but its solutions are not unique for the metric (a homogeneous equation).

In $n$ dimensions, these $\frac{n(n+1)}{2}\frac{n(n-1)}{2}$ conditions bring us down from the $\frac{n^2(n+1)(n-1)}{3}$ components of the curvature to $\frac{n^2(n+1)(n-1)}{3\cdot 2 \cdot 2}$ of the Riemann Tensor. Yet it seems to me these equations don't tell me what the metric should be.