It is well known that the Riemann curvature tensor of a metric satisfies
\begin{eqnarray}
R_{jikl}=-R_{ijkl}=R_{ijlk},(1)\\
R_{klij}=R_{ijkl},(2)\\
R_{i[jkl]}=0 \mbox{(1st Bianchi identity)}.(3)
\end{eqnarray}
The question is whether there are more (may be non-linear and/or differential) equations satisfied by $R_{ijkl}?$
(I do not mention here the 2nd Bianchi identity $R_{ij[kl;m]}=0$ since it involves covariant derivatives and hence is equation not only on the tensor $R$ but also on the original metric $g$).
One can ask this question in the following more precise form.
Question 1. Let $R_{ijkl}$ be a collection of numbers satisfying (1)-(3) with the range of indices from 1 to $n$. Does there exist a metric $g$ in a neighborhood of 0 in $\mathbb{R}^n$ such that its curvature tensor at 0 is equal to the given collection of numbers?
Question 2. Let $R_{ijkl}(x)$ be a collection of smooth (real analytic?) functions in a neighborgood of 0 in $\mathbb{R}^n$ satisfying (1)-(3). Does there exist a metric in a neighborhood of 0 such that its curvature tensor is equal to the given collection of functions $R_{ijkl}(x)$?
UPDATE: As mentioned by Peter Michor below, the answer to Q1 is positive. As mentioned by him and Vladimir Matveev, in dimensions $n>3$, $g\mapsto R(g)$ is an overdetermined system of PDE. Hence the answer to Q2 is 'No' for $n>3$, though it is not clear to me how to write explicitly any constrain on the image of the map. In $n=2$ the answer is 'Yes' according to Vladimir Matveev.
ANOTHER UPDATE: According to Robert Bryant's answer below, in $n=3$ the answer to Q2 is 'No' in general. However under some non-degeneracy assumptions it is 'Yes'.
Best Answer
The answer to Q2 for $n=3$ is actually 'no, without some nondegeneracy hypotheses'. The reason is as follows:
The curvature tensor $\mathcal{R}= R_{ijkl}\,(\mathrm{d}x^i\wedge\mathrm{d}x^j)\circ(\mathrm{d}x^k\wedge\mathrm{d}x^l)$, with all its indices lowered, is a section of the subbundle $K(M)\subset S^2\bigl(\Lambda^2(T^*M)\bigr)$ that is the kernel of the natural linear mapping $$ S^2\bigl(\Lambda^2(T^*M)\bigr)\longrightarrow \Lambda^4(T^*M). $$ You are asking, for a given section $\mathcal{R}$, whether there exists a metric $g$ such that $\mathrm{Riem}(g) = \mathcal{R}$. There is no pointwise algebraic condition on the section $\mathcal{R}$ imposed by this equation (that was what Q1 was about), but, as you note, there is the second Bianchi identity $$ \mu(\nabla^g\mathcal{R}) = 0, $$ where $\mu:K(M)\otimes T^*M\to \Lambda^2(T^*M)\otimes\Lambda^3(T^*M)$ is the natural skewsymmetrization operation. Since $\nabla^g$ depends on one derivative of the metric $g$, the above equation with a given $\mathcal{R}$ can be regarded as a first-order system of equations on $g$. When $n=3$, this is at most $3$ equations, the rank of the bundle $\Lambda^2(T^*M)\otimes\Lambda^3(T^*M)$.
Now, for a point $p\in M$ satisfying $\mathcal{R}(p)=0$, the value $\nabla^g\mathcal{R}(p)$ does not depend on $g$ (just look at the formula for $\nabla^g\mathcal{R}$ in local coordinates). Thus, if $\mathcal{R}(p)=0$, but $\mu(\nabla^{g_0}\mathcal{R})(p) \not= 0$ for some metric $g_0$, then $\mu\bigl(\nabla^{g}\mathcal{R}(p)\bigr) \not= 0$ for all metrics $g$ and hence there is no open neighborhood of $p$ on which the equations $\mu\bigl(\nabla^{g}\mathcal{R}\bigr) = 0$ have a solution $g$. In particular, the original system $\mathrm{Riem}(g) = \mathcal{R}$ has no solution in a neighborhood of such a point $p$.
To convince yourself that such examples exist when $n=3$, just note that the rank of $K(M)=S^2\bigl(\Lambda^2(T^*M)\bigr)$ in this case is $6$, so an arbitrary section that vanishes at $p$ will have $6\times 3 = 18$ independent first derivatives. Thus, the map $\mu:K(M)\otimes T^*M\to \Lambda^2(T^*M)\otimes\Lambda^3(T^*M)$ is surjective (and the rank of the target bundle is $3$), so that the generic section of $K(M)$ that vanishes at $p$ will not satisfy the second Bianchi identity at $p$ for any metric $g$.
However, suppose that $n=3$ and that $\mathcal{R}$ is a nondegenerate section of $K(M)=S^2\bigl(\Lambda^2(T^*M)\bigr)$. I proved (back in the early 1980s) that, when $\mathcal{R}$ is real-analytic, there always exist local solutions to the equation $\mathrm{Riem}(g) = \mathcal{R}$. Specifically, I showed that, in this case, in addition to the $6$ second-order equations that these equations represent on $g$ and the $3$ first-order equations on $g$ that $\mu\bigl(\nabla^{g}\mathcal{R}\bigr) = 0$ represents, there is one more first-order equation $Q_\mathcal{R}(g)=0$ on $g$ that is satisfied by any metric $g$ that satisfies $\mathrm{Riem}(g) = \mathcal{R}$. Then I proved that the combined overdetermined system of $6$ second-order equations and $4$ first-order equations for $g$ is involutive, so that an application of the Cartan-Kähler Theorem proves local solvability. Unfortunately, the involutive system is never either hyperbolic or elliptic, though it can be of real principal type.
I never published my proof, but later, Dennis DeTurck and Deane Yang studied the overdetermined system that I wrote down and published a proof of its local solvability in the smooth category. See Deturck and Yang, Local existence of smooth metrics with prescribed curvature, Nonlinear problems in geometry (Mobile, Ala., 1985), 37–43, Contemp. Math., 51, Amer. Math. Soc., Providence, RI, 1986.