I'm implementing a CYK algorithm in my software and I've found a pseudo-code on Wikipedia. Here's its complexity(modified version for special use, which doesn't go from j=1 to n-i+1):
However, pseudo-codes are usually indexing from 1, but in programming everything starts at 0. My goal is to make this algorithm go from i = 1, j = 0 and k = 0 and getting exactly the same sum. I kind of forgot how to do this properly and been lost in this problem for a while.
If possible, I'd like not to have anything outside of these sums. The goal is not to compute these sums, just to lower their indexes.
If you have a sum $$ \sum_{k=a}^b c_k, $$ you can make a change of index, for example, $k'=k-d$, and you get $$ \sum_{k'=a-d}^{b-d} c_{k'+d}, $$ for some $d$. You may check that this holds, for example, if $a=2$ and $b=5$: both sums are $c_2+c_3+c_4+c_5$.
Because the indices $k$ and $k'$ are just dummy summation indices (they don't have a meaning outside the sum), we can of course call $k'$ in the final expression just $k$: $$ \sum_{k=a}^b c_k = \sum_{k'=a-d}^{b-d} c_{k'+d} = \sum_{k=a-d}^{b-d} c_{k+d} $$
Let's work through your formula, $$ \sum_{i=2}^n \left(\sum_{j=1}^n\left(\sum_{k=1}^{i-1} 1\right) \right). $$ Introducing $i'=i-1$, we get (note that we have an $i$ inside the sum so that should change as well) $$ \sum_{i'=1}^{n-1} \left(\sum_{j=1}^n\left(\sum_{k=1}^{i'} 1\right) \right). $$ Introducing $j'=j-1$, we get (note that we don't have a $j$ inside the sum over $j$ so nothing changes there) $$ \sum_{i'=1}^{n-1} \left(\sum_{j'=0}^{n-1}\left(\sum_{k=1}^{i'} 1\right) \right), $$ and finally, with $k'=k-1$, we get $$ \sum_{i'=1}^{n-1} \left(\sum_{j'=0}^{n-1}\left(\sum_{k'=0}^{i'-1} 1\right) \right). $$ Finally, if we don't like those $'$s, we can rewrite this as $$ \sum_{i=1}^{n-1} \left(\sum_{j=0}^{n-1}\left(\sum_{k=0}^{i-1} 1\right) \right). $$