How many degrees of freedom does the covariant derivative of the Riemann tensor have? It is a commonly cited result that the Riemman tensor $R_{abcd}$ has $\frac{1}{12}n^2(n^2-1)$ degrees of freedom in $n$ dimensions, which follows from the identities of the Riemann tensor. This gives $20$ degrees of freedom in $3+1$ dimensions. How many does the covariant derivative $R_{abcd;e}$ have?
 A: The covariant derivative of the Riemann tensor should add a factor of $n$ to the degrees of freedom you had before, giving $\frac{1}{12}n^3(n^2-1)$. However, the second Bianchi identity restricts the amount of components:
$$R_{ab[cd;e]} = 0.$$
It is then enough to count how many tensors of rank 5 with the last three indices antisymmetrized there are and subtract it from $\frac{1}{12}n^3(n^2-1)$.
There are $\frac{n!}{r!(n-r)!}$ independent components in a rank $r$ antisymmetric tensor. Thus, $R_{ab[cd;e]} = 0$ gives us
$$\frac{n(n-1)}{2} \times \frac{n!}{3!(n-3)!} = \frac{1}{12}n^2(n-1)^2(n-2)$$
independent components, where the factor $\frac{n(n-1)}{2} $ comes from the first two indices, which are antisymmetric. So if I didn't miss anything, the final result should be:
$$\frac{1}{12}n^3(n^2-1) - \frac{1}{12}n^2(n-1)^2(n-2) = \frac{1}{12}n^2(n-1)\Big(n(n+1) - (n-1)(n-2)\Big)$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\frac{1}{12}n^2(n-1)(4n-2) = \frac{1}{6}n^2(n-1)(2n-1).$$
In 4D we get 56 independent components.
