I recently came acrossed the following notation:
$\sum_{\substack{
j\leq j'=1}}^{2^i-1} a_{j,j'}$
Frankly, I don't know how to interpret the index portion correctly. Say that i=2 (as is the case I am trying to work with). The top is then equal to 3. Would that mean we would compute: $a_{1,1}+a_{1,2}+a_{2,2}+a_{2,3}+a_{3,3}$?
edit: I came across the notation in this paper, specifically in equation (4).
edit 2: I reached out to some of the authors of the paper, one of which got back to me and explained what was intended by the notation. @ConMan provided an answer that is equivalent to the authors so I have marked their reply as correct.
Best Answer
The notation is a bit fuzzy, since it's combining two different standard notations - on the one hand, you can write $\sum_{j=1}^{2^i-1}$ to mean "sum from $j = 1$ to $j = 2^i-1$", and on the other you can write $\sum_{j \leq j'}$ the mean "sum over values of $j$ (probably natural numbers) such that $j \leq j'$".
I would personally read this as being the sum over all values of $j'$ from $1$ to $2^i - 1$, and all values of $j$ that are less than or equal to $j'$, i.e. it's equivalent to the double sum notation
$$\sum_{j'=1}^{2^i-1} \sum_{j \leq j'} a_{j,j'}$$
and in the case of $i = 2$, that would mean it's equivalent to $a_{1,1} + a_{1,2} + a_{2,2} + a_{1,3} + a_{2,3} + a_{3,3}$.