Recurrence relation of the following sequence? This is the code:
for (unsigned int i = 0; i < n; ++i)
        if (i % 2 == 0)
            ++k;

And this is the output for when n and k start at 1 and 0 respectively:
\begin{array}{c|c}
n & k\\\hline
1 & 1\\
2 & 1\\
3 & 2\\
4 & 2\\
5 & 3\\
6 & 3\\
7 & 4\\
8 & 4\\
9 & 5\\
10 & 5\\
11 & 6\\
12 & 6\\
13 & 7\\
14 & 7\\
15 & 8\\
16 & 8\\
17 & 9\\
18 & 9\\
19 & 10\\
20 & 10\\
\end{array}
I need to express this as a sequence. Something like $a_n = ..$. How would I do this?
I know it would be something like $a_n = \frac{n}{2}$ when $n$ is even, and $a_n = \lceil\frac{n}{2}\rceil$ otherwise but is there a different way?
 A: The simple solution is as you mentioned
$$a_n=\left\lceil\frac n2\right\rceil$$
A different way of expressing this could be:
$$a_n=\frac n2+\frac12\!\left|\,\sin\left(\frac{\pi x}2\right)\right|$$
Derivation of above formula:
We start with the formula
$$a_n=\left\lceil\frac n2\right\rceil$$
When $n$ is even this is equal to
$$a_n=\frac n2$$
And when $n$ is odd this is equal to
$$a_n=\frac n2+\frac12$$
This means that if we find a function $f(x)$ that is equal to $0$ if $n$ is even and $\frac12$ when $n$ is odd then the following would be true.
$$a_n=\frac n2+f(x)$$
To find such a function, I'll use the periodicy of $\sin x$ meaning if $k$ is an integer then $\sin(\pi k)=0$
Another fact about the sine function is that $\sin\left(\pi\cdot\!\!\left(k + \frac12\right)\right)=\pm1$ where the $\pm1$ that for some integers it's $1$ and for others it's $-1$
Using these two facts we can deduce that
$$\sin\left(\frac{\pi k}2\right)$$
is zero for even $k$ and $\pm1$ for odd $k$, take the absolute value and half it, you have our function $f(x)$
$$f(x)=\frac12\!\left|\,\sin\left(\frac{\pi x}2\right)\right|$$

Can also be written like this:
$$a_1=1$$
$$a_2=1$$
$$a_n=1+a_{n-2}$$
