How do you derive the row and column of an element in a tensor unfolding? Tensor Unfolding defined with example:

I'm trying to understand tensors and their unfoldings into matrices.
I'm hitting a roadblock at the notation used for rows and columns mapped to the tensor elements respectively. What exactly is $i_n$ as a row index ? What is $n$ there? I am not able to get my head around the notation and how they translate to the elements in the given example. 
I've tried reverse engineering from the example as well in vain. Can anyone take a couple of elements from the tensor in the given example and explain how they are mapped to any one of the unfoldings, please?
 A: I hate multidimensional arrays, nonetheless, I'll try to oblige.
Here's what the formula says, translated into an algorithm rather than a hideous formula.
First, if $T$ has indices $i_1,\ldots,i_N$, then we choose an index $n$ with respect to which we are unfolding. Translating the formula it essentially says to cyclically permute the indices so that the $n$th index is now the first index. I.e. create a new tensor $S$ with indices $j_1,\ldots, j_N$ where $$j_{\alpha}=i_{\left(\alpha+n-1\mod{N}\right)}.$$
Then $T_{(n)}=S_{(1)}$, so let's focus on how to expand $T_{(1)}$.
The expansion index becomes our row index, so our matrix will have rows $R_i=\operatorname{Vectorization} T|_{i_1=i}$. I.e. for the $i$th row, we restrict to the entries of the tensor with $1$st index equal to $i$, and then vectorize that remaining $N-1$ dimensional tensor by iterating through its entries with the innermost loop being over the rightmost index and the outermost loop being over the leftmost index.
I.e., in your example, for $T_{(1)}$ to get the first row, we restrict to the upper block of numbers and we look at the remaining indices, they are $i_2$ and $i_3$ in that order. Thus we first let $i_2=i_3=1$, then first iterate over $i_3$, getting $1,11,21$. Then we increment $i_2$, and again iterate over $i_3$ getting $2,12,22$, and again increment $i_2$ and again iterate over $i_3$ to get $3,13,23$. Hopefully you're getting the idea.
I'll try to explain how the cyclic permutation of the indices works in the other examples. For $T_{(2)}$, our primary index is 2, so for the first row, we fix $i_2=1$, and our remaining indices are $i_3$ and $i_1$ in that order. Thus we first iterate over $i_1$ getting $1,5$, then increment $i_3$ and iterate again getting $11,15$, and increment $i_3$ again and iterate over $i_1$ getting $21,25$ to finish the row. Then we move on to the next row.
Finally for $T_{(3)}$ the primary index is 3, and the remaining indices are $i_1$ and $i_2$ in that order, so the inner loop is over $i_2$ and the outer loop is over $i_1$. So the first row vectorizes as $1,2,3,4,5,6,7,8$.
