Are the Ricci and Scalar curvatures the only "interesting" tensors coming from the Riemannian curvature tensor? In studying Riemannian geometry, one learns about the Riemann curvature tensor
$$Rm(X,Y,Z,W) = \langle \nabla_X\nabla_YZ -\nabla_Y\nabla_XZ - \nabla_{[X,Y]}Z, W\rangle$$
and its various symmetries.  From the Riemann curvature tensor, one can define the Ricci and scalar curvatures, which give us "pieces" of the curvature.
I understand that both the Ricci and scalar curvatures are important ways of measuring curvature.  My question is: are these (in any sense) the "only" interesting tensors that come from the Riemann curvature?
That is, I could imagine inventing other curvature tensors by performing various operations on $Rm$.  Is there a reason that doing so would be fruitless?  Why do we privilege the Ricci and Scalar curvatures?  Do they give us all the information that we might want?
A previous question of mine hinted at this, though my thoughts were not quite as clear.
 A: No, there's a lot more interesting stuff. One particular subject that is actively studied is identities involving curvature invariants and how formulae equating combinations of curvature invariants constrain the form of the metric. 
For example, see this article. 

If you are willing to include not strictly tensorial quantities, an obvious thing that you missed is the sectional curvature operator. But a less well-known operator, that is nevertheless very interesting, that relates to the Riemann curvature tensor is the isotropic curvature, which is defined as
$$ (X,Y,Z,W) \mapsto Rm(X,Z,X,Z) + Rm(X,W,X,W) + Rm(Y,Z,Y,Z) + Rm(Y,W,Y,W) - 2 Rm(X,Y,Z,W) $$
This operator featured prominently in the resolution of the differentiable sphere theorem by Simon Brendle and Rick Schoen, see e.g. this paper or this book. 
A: As mentioned by Harald and myself in the comments above, an important tensor that arises from the Riemann curvature tensor is the Einstein tensor. In terms of components, the Einstein tensor $ G_{\mu \nu} $ is defined from the Riemann-curvature tensor as follows:
$$
G_{\mu \nu} \stackrel{\text{def}}{=} R_{\mu \nu} - \frac{1}{2} g_{\mu \nu} R,
$$
where $ g_{\mu \nu} $ is the metric tensor, $ R_{\mu \nu} $ is the Ricci-curvature tensor and $ R $ is the Ricci scalar.
An important fact about the Einstein tensor is that it is divergence-free. Historically, this is important because when Einstein was attempting to find the correct field equations for gravity, he tried to equate the stress-energy tensor $ T_{\mu \nu} $, which describes density and flux of energy and momentum in spacetime, with a tensor $ G_{\mu \nu} $ that describes the curvature of spacetime. If $ G_{\mu \nu} $ could be found, then one would be able to determine how the distribution of matter and energy in spacetime affects its curvature, as well as how curvature of spacetime affects matter and energy.
As the Riemann-curvature tensor encodes information about spacetime curvature, it was naturally believed that $ G_{\mu \nu} $ should and would be derived from it. Einstein could narrow down his search for $ G_{\mu \nu} $ precisely because of the fact that the stress-energy tensor is divergence-free, which translates into physical terms as conservation of energy. In the paper The Four-Dimensionality of Space and the Einstein Tensor, published in 1972 in Volume 13, Issue 6, of the Journal of Mathematical Physics, David Lovelock showed that the only contravariant $ 2 $-tensors that are divergence-free on one index and that are concomitants of $ g_{\mu \nu} $, together with the first two derivatives of $ g_{\mu \nu} $, are $ g_{\mu \nu} $ itself and $ G_{\mu \nu} $ as defined above.
There are other kinds of tensors that can be derived from the Riemann-curvature tensor. For example, if you Ricci-decompose the Riemann-curvature tensor, then you obtain three kinds of tensors, one of which is the Weyl tensor as mentioned by Paul.
A: No, there is (at least) Weyl tensor. Basically Weyl tensor measure how far away the Riemannian metric from being conformally flat. Conformally flat means that the metric is conformal to flat metric. If Weyl tensor vanishes, then it is conformally flat. 
