Is it possible to change the step interval on a sum? Suppose we wanted to create a sum that says, from n=0 to n=16, n will add 4 to itself and add the function like a regular sigma form. Quite similar to:
i = 0;
sum = 0;
while (i <= 16) { 
    i += 4; 
    sum += 4*i^2/(log(i)); 
};

I considered a form like this:
$$\sum_{n=0}^{16} \left(\frac 1 {0^{n \bmod 4}}\right) \frac {4n^2}{\log n}$$
Where you ignore the undefined terms. This isn't a very convenient or mathematically pure form, though. It would be much more convenient to create something like:
$$\sum_{n=0, \, 4}^{16} \frac {4n^2}{\log n}$$
Where $4$ is the interval.
 A: You could use $$\sum_{n=0}^{4} \frac{4(4n)^2}{\log{(4n)}}$$
That is, substitute 4n for n and continue using the step size 1.  This way as n increases by 1, wherever it's being use in the formula is being increased by 4.
You can similarly do this for any progression you could write a nice formula for.
Alternatively, you can define an index set, like $I=\{0,4,8,12,16\}$ and then write $$\sum_{n\in I} \frac{4n^2}{\log{n}}$$
A: You are allowed to write any condition you want under the summation, as long as it is clear.  So for example
$$\sum_{n \in\{0,4,8,12,16\}} \cdots$$ (this is a variation on matt's answer)
or $$\sum_{\textstyle{{0\le n\le16}\atop{n \text{ a multiple of 4}}}}\cdots$$
or $$\sum_{\textstyle{{n=4k}\atop{k\in\{0\ldots 4\}}}} \cdots$$
You could probably write $$\sum_{n =0,4,8,12,16} \cdots$$ and people would understand.
You do not have to adhere to one particular specific form as if you were filing a tax return.
A: Write $i$ as $4k$.  Then you have $$ \sum_{k=0}^4\frac{4(4k)^2}{\log(4k)} $$
