\, {\pi\over 3}\;   J_0^a  J_0^b J_0^c +
\sum_{n\not= 0 } {i\over  n} \tilde J^a_{-n} \tilde J^b_{n}
J^c_0\;  + {\rm cyclic} + {\rm reversal}\; \right]\ ,$$ and $$\begin{aligned}
{\cal O}_{\rm reg}^{(4)}(R)&=& {\pi^2\over 2} \; {\rm Tr}_R(t^a
t^b t^c t^d)\; \Biggl[\, {\pi^2\over 6}\;   J_0^a  J_0^b  J_0^c
J^d_0 \,+ \,\sum_{n\not= 0} {i\pi \over  n} 
\tilde J^a_{-n} \tilde J^b_{n} J^c_0  J^d_0 
\nonumber \\
&& \quad + \sum_{n\not= 0 } {1 \over  n^2}\, \left( \tilde
J^a_{-n} \tilde J^b_{n} J^c_0  J^d_0 - \tilde J^a_{-n}
 J^b_{0}\tilde J^c_n  J^d_0\right)
\label{R4}  + \sum_{\stackrel{\footnotesize{m,l,n\not=
0}}{m+n+l=0}} {1\over ml}\, \tilde J^a_{m} \tilde J^b_{n}\tilde
J^c_l  J^d_0 \\
&& \quad - {1\over 2} \sum_{m,n\not= 0}{1\over mn} \tilde J^a_{-n}
\tilde J^b_{n}\tilde J^c_{-m} \tilde J^d_m
\;   +\;
{\rm cyclic} + {\rm reversal} \;   \Biggr]\ . \nonumber\end{aligned}$$ Here we have used the short-hand notation $\tilde J^a_n = J^a_n \;e^{- |n| s/2}$. Note that in deriving these formulae it helps to sum over all cyclic permutations of the currents and use the cyclic property of the trace, before performing explicitly the $\sigma_j$ integrals.

Let us pause for a moment to describe our group theoretic conventions. The $t^a$ are (traceless) generators of $g$ in the representation $R$, which we assume irreducible. The Killing form is $\delta^{ab}$, so adjoint indices 