Dimension of the space of algebraic Riemann curvature tensors
Solution 1:
Its symmetries imply that an algebraic curvature tensor on the $n$-dimensional vector space $V$ can be viewed as an element of the space $S^{2}(\wedge^{2}V^{\ast})$ of symmetric bilinear forms on the second exterior power $\wedge^{2}V^{\ast}$, so the algebraic curvature tensors constitute a subspace of $S^{2}(\wedge^{2}V^{\ast})$.
An exterior four-form on $V$ can also be viewed as an element of $S^{2}(\wedge^{2}V)$, so $\wedge^{4}V^{\ast}$ is a subspace of $S^{2}(\wedge^{2}V^{\ast})$. Using the algebraic Bianchi identity, it is not hard to check that this subspace is the orthogonal complement of the subspace of curvature tensors. (Remembering this claim makes it easy to compute the dimension when one cannot remember it.)
Hence the dimension of the space of curvature tensors is \begin{align} \dim S^{2}(\wedge^{2}V^{\ast})& - \dim \wedge^{4}V^{\ast} = \frac{1}{2}\binom{n}{2}\left(\binom{n}{2} -1\right) - \binom{n}{4}\\ & = \frac{n(n-1)(n(n-1) - 2)}{8} - \frac{n(n-1)(n-2)(n-3)}{24}\\ & = \frac{n(n-1)\left(3n^{2} - 3n - 6 - n^{2} + 5n + 6\right)}{24}\\ & = \frac{n^{2}(n^{2}-1)}{12}. \end{align}
Solution 2:
I saw that somewhere...
and that somewhere happens to be Weinberg, Gravitation and cosmology, around p.142-143, Description of curvature in N dimension. The calculation is made as you describe because Weinberg assumes in the following order (p.141):
$R_{abcd}=R_{cdab} $
$R_{abcd}= -R_{bacd}=-R_{abdc} $
$R_{abcd}+ R_{bcad}+ R_{cabd}=0 $
The first and third are indeed redundant
$3. \Rightarrow 1.$ : (so one can actually discard 1.) Consider $R_{abcd}-R_{cdab} \stackrel{?}{=}0$.
Replace both term with 3. ($R_{abcd}= -R_{bcad}- R_{cabd} $ same for the other) and use antisymmetry to cancel two terms. The equality holds for all $a,b,c,d$ so write the obtained equation again, substituting $"a":=b, "b":=c, "c":=d , "d":=a$. Cancel some terms.
Conversely, if two of the four indexes $a,b,c,d$ coincide, then $1. \Rightarrow 3.$ (assuming 2.) by a direct check.
So 3. only adds extra independent constraints if the four indexes are different.
ps: - i see no miscalculation. - of course the extra ${n\choose 4}$ equations 3. are independent since for $\{a_1,b_1,c_1,d_1 \}\neq \{a_2,b_2,c_2,d_2 \} $ no permutation can map a list of elements of the first set to a list of elements of the second set