Yes. All the components where the first two indices are the same, or the last two indices are the same, are zero.
Sometimes it is useful to think of this tensor as a $6\times6$ symmetric matrix where the “indices” are $01$, $02$, $03$, $12$, $13$, and $23$.
However, don’t conclude from this that there are $6+5+4+3+2+1=21$ independent components. There are actually only $20$ because of the algebraic Bianchi identity.
Note that without any of these relations between components, there would be $4^4=256$ components! So the Riemann curvature tensor is about 13 times less complicated than it might appear.