Say I have Lorentz tensors $A^{\mu\nu}$ and say this Lorentz tensor is symmetric under $\mu \Leftrightarrow \nu$ and there are only $p^\mu$ and $q^\mu$ as the physical Lorentz vectors involved. If so, then $A^{\mu\nu}$ can be parameterized simply in terms of these vectors as
$$A^{\mu\nu} = g^{\mu\nu}A_1 + p^\mu p^\nu A_2 + q^\mu q^\nu A_3 + (p^\mu q^\nu+q^\mu p^\nu) A_4 $$
Thus, $A^{\mu\nu}$ have four-independent components. This kind of exercise is well-known, for example, in DIS structure functions parameterization, etc. Imposing Ward Identities (if available) would reduce the number of independent components even further.
Say instead, I have Lorentz tensors with four indices. $A^{\mu\nu\rho\sigma}$ which is symmetric under $\mu \Leftrightarrow \nu$ and $\rho \Leftrightarrow \sigma$. Also, assume that there are only $p^\mu$ and $q^\mu$ as the available Lorentz vectors. The general parameterization has quite a bit of structures: $\{qqqq,qqqp,qqpp,qppp,pppp,gqq,gpp,gqp,gg\}$. Taking this full space of parameterization then imposing symmetries to reduce the number of independent components is a bit cumbersome (and certainly becomes less tractable if I consider a case with even more than four indices). How should I do the counting? Also, my intuition tells me I can just factorize this tensor into a product $A^{\mu\nu\rho\sigma} = B^{\mu\nu} C^{\rho\sigma}$, then parameterize $B^{\mu\nu}$ and $C^{\rho\sigma}$ separately into four symmetric structures, giving one total of 16 independent components. Is this correct?
Edit: one cannot use factorized product form to parameterize. For example, $$g^{\mu \sigma} g^{\nu \rho} + g^{\mu \rho} g^{\nu\sigma}$$ is an example of the tensor structure which is symmetric under $\mu \Leftrightarrow \nu$ and $\rho \Leftrightarrow \sigma$, but is not given in factorized product form.
