$U^{ijkm}$ has the same symmetry properties as the Riemann tensor $R^{ijkm}$.
Let me remind the symmetries of the Riemann tensor, even most of you know it. As I know most common symmetry properties are:
$$R^{ijkm}=-R^{jikm}$$
$$R^{ijkm}=-R^{ijmk}$$
$$R^{ijkm}=R^{kmij}$$
$$R^{ijkm}+R^{imjk}+R^{ikmj}=0$$
There is an equation for the derivative of $U^{ijkm}$ wrt to $x^j$:
$${U^{ijkm}}_{,j}=2\Gamma^a_{aj}U^{ijkm}-\Gamma^i_{aj}U^{ajkm}-\Gamma^j_{aj}U^{iakm}-\Gamma^k_{aj}U^{ijam}-\Gamma^m_{aj}U^{ijka}$$
And the interesting part is using the symmetry of $U^{ijkm}$ to obtain from the last equation $$ {U^{ijkm}}_{,j}=\Gamma^a_{aj}U^{ijkm}-\Gamma^k_{aj}U^{ijam}-\Gamma^m_{aj}U^{ijka} $$
I have tried the first symmetry property. But the result was different. What hint should be used to derive the proper result?