Convert for loop into math sigma notation

Have some code (written in C), need the equivalent notation for mathematics. Is it possible using if-statements and sigma notation for this, and if so, how?

int res = 0;
for (int i=0; i < (N/2)+1; i=i+1)
if (N % pow(2, i) == 0) res = i;

I have gotten as far as:

$$\sum _{i=0}^{{N\over2}+1}\:(N \mod 2^i )$$ (missing the == 0 and res = i)

Ideas or a solution are welcome.

• You need to look at what the code does. I suppose you're trying to write down an expression for the value of the variable res when the code terminates. You're claiming that it's a sum of modulus expressions, but that's very unlikely to be correct because the code doesn't add a bunch of things to res. The most natural expression for the final value of res isn't going to be a sum of terms. – David Richerby Jun 14 '18 at 15:31
• Im not claiming anything, I know what the code does. res is not a sum of terms, you are right about that, but it is a value that is updated inside the for-loop (in the code). Wether or not a simple solution for math exists I don't know. I ask because maybe someone knows. – Natural Number Guy Jun 14 '18 at 15:51
• Let's not quibble about language. Your question states that the answer has something to do with the sum you quote; I explained why it doesn't. I'm trying to encourage you to work out the answer yourself by reasoning about what the code does, rather than by applying "rules" such as "it's a for loop so it must be some kind of summation. – David Richerby Jun 14 '18 at 15:54

Let me define $$f(x) = \begin{cases} 0 & x \ne 0 \\ 1 & x = 0\end{cases}$$ Now it is $$\sum _{i=0}^{{N\over2}+1}\:f(N \text { mod } 2^i)$$ The code calculates how many times number is divisible by two (which by the way could be simplified to checking number of trailing zeros in binary representation, say hacky method x & -x).