Let $\mathbf{T}$ be a fully symmetric tensor of order $3$ and size $N$. Its components can be represented as $T_{ijk}$ for all $1\leq i,j,k\leq N$. By symmetric I mean that if I permute any indices the value stays the same: \begin{equation} T_{ijk}=T_{ikj}=\dots=T_{jki} \end{equation} Now, I would like to rewrite the following sum, making use of the fact that $\mathbf{T}$ is symmetric: \begin{equation} \boxed{\sum_{i}^N\sum_{j}^N\sum_k^{N}T_{ijk}} \end{equation}
That is, I would like to decompose the sum into:
\begin{equation} \sum_{ijk}^NT_{ijk}=\alpha\sum_{i\leq j\leq k} T_{ijk}+\beta\sum_{i=j\leq k} T_{iik}+\gamma\sum_{i=j=k} T_{iii}+\text{ something else? } \end{equation}
I would like to find the values of $\alpha,\beta\gamma$ but I am unsure of the values.
In the case of \textbf{symmetric} matrices I could have written:
\begin{align*} \sum_{i}\sum_j M_{ij}&=\sum_{i<j}M_{ij}+\sum_i M_{ii}+\sum_{i>j}M_{ij}\\ &=2\sum_{i<j}M_{ij}+\sum_i M_{ii} \end{align*}
But with tensors I have trouble decomposing this sum.
The bigger scope is trying to compute the following integral:
\begin{equation} I=\int\left(\prod_{i\leq j\leq k}\mathrm{d}T_{ijk}\right)\exp\left\{-\frac{1}{2}\sum_{ijk}T_{ijk}^2\right\} \end{equation}
Note the sum over all $ijk$ in the exponential, which is why I tried to decompose the sum into different parts.
$$\sum_{i,j,k}T_{ijk}=6\sum_{i<j<k}T_{ijk}+3\sum_{i<j}T_{iij}+3\sum_{i<j}T_{ijj}+\sum_iT_{iii}$$ The $6$ accounts for all ways to permute $(i,j,k)$. For $(i,i,j)$, there are only $3$ ways, since switching the first two $i$'s has no effect, so you divide by $2$.
Another valid way to write this is $$\sum_{i,j,k}T_{ijk}=6\sum_{i<j<k}T_{ijk}+3\sum_{i\neq j}T_{iij}+\sum_iT_{iii}$$