I am interested in knowing the exact form of the anti-commutation of two generators of $su(N)$ lie algebra.
Let us denote $T^a$ to be the generator of $su(N)$ lie algebra in the defining representation. Since the number of generators is $n^2-1$, the index takes value in $a=1, ..., n-1$. The normalization of $T^a$ is $$Tr(T^aT^b)=\frac{1}{2} \delta^{ab}$$ The anti commutation of two such generators is $$\{T^a, T^b\}=T^a T^b+T^bT^a=\frac{1}{N}\delta^{ab}I+d^{abc}T^c$$ where $d^{abc}$ is a totally symmetric tensor in all the three indices. In https://pdfs.semanticscholar.org/1101/914fc76a36d4fb0ab0022f8c4ec6295d8d1f.pdf, it was shown that $$d^{abc}d^{abh}=\frac{N^2-4}{N}\delta^{ch}$$ where the repeated indices are to be summed over. In the above, we contract over two indices from each $d$-tensor.
My questions are:
1) Is there a simple expression for $$d^{abc}d^{agh}$$ where we only contract one index for each $d$-tensor? (in terms of $N$)
2) Is there a simple expression for $d^{abc}$ itself? (in terms of $N$)
There is no simple expression for the $d^{abc}$ tensors for the simple reason that they are basis-dependent. To wit, if I'm given a set of generators $T^a$ that generate $SU(N)$, and a unitary matrix $A$, then the matrices
$$(T')^a \equiv A^a{}_b T^b$$
also generate $SU(N)$ and obey
$$\text{tr}\; T'^a T'^b = \frac{1}{2} \delta^{ab}.$$
However, unless $A$ is the identity matrix, the $d$-coefficients after the change of basis read
$$d'^{abc} = A^a{}_{\alpha} \, A^{b}{}_\beta \, d^{\alpha \beta \gamma} \,(A^{-1})^c{}_\gamma \neq d^{abc}.$$
Some identities, like the one you showed, are clearly basis-independent, however the numerical values of the coefficients $d^{abc}$ are not universal.