Examples of functions with a natural domain $\Bbb R\setminus 2\Bbb N$ 
I would like to find a few examples of a function $f$ with a natural domain $\Bbb R\setminus 2\Bbb N$ which can be defined in a single expression.


Thoughts:
Since $\lfloor x\rfloor=\lceil x\rceil\iff x\in\Bbb Z$, I took
$$f(x)=\begin{cases}\frac1{\left|\left\lfloor\frac{x}2\right\rfloor\right|-\left\lceil\frac{x}2\right\rceil},&x\ne 0\\0,&x=0\end{cases}\space$$
Similarly, it can be defined as:
$$f(x)=\begin{cases}\ln\left(\left|\left|\left\lfloor\frac{x}2\right\rfloor\right|-\left\lceil\frac{x}2\right\rceil\right|\right),&x\ne 0\\0,&x=0\end{cases}$$
Another attempt is:
$$f(x)=\begin{cases}\tan\left(\left\lfloor\left(\frac32\right)^{\operatorname{sgn}(x)}\right\rfloor\frac{x+1}2\pi\right),&x\ne0\\0,&x=0\end{cases}$$
but $0\notin\Bbb N$, so I had to define $f(0)$ separately (however, $\tan\left(\operatorname{sgn}(x)\left\lfloor\left(\frac32\right)^{\operatorname{sgn}(x)}\right\rfloor\frac{x+1}2\pi\right)$ works if one takes $\operatorname{sgn}(0)=0$).
Furthermore, it is constant on $(-\infty,0]$, which isn't quite interesting.
How can I fix this problem if I want $f$ to be composed of $\tan(x)$? Are there any other examples of such functions defined by one expression?
Thank you in advance!
 A: What about $$f(x)=\sum_{k=0}^{+\infty} \frac{1}{(x-2k)^2} \quad ?$$
A: In the meantime, I found a function that isn't defined piece-wisely and I figured out what to do with the problematic $x=0$ in my previous attempts where I assumed $\boxed{\operatorname{sgn}(0)=1}$ as in some literature.
Let's consider a bijection $f:(a,b)\to\Bbb R$ defined by: $$f:=\tan\left(\frac\pi{b-a}\left(x-\frac{a+b}2\right)\right)$$
with a prime period $P=b-a$ and shifted by $x_0=\frac{a+b}2$ on the $x$-axis.
Since we don't want even natural numbers (I don't include $0$ in $\Bbb N$), let $a=2k, b=2(k+1),k\in\Bbb N$.
My idea was to make the desired function constant on $(-\infty,2)$ and use the above pattern on $(2,+\infty)$.
Let's take $a=2,b=4$ and proceed as follows:
$$f(x)=\tan\left(\left\lfloor2^{\operatorname{sgn(x-2)}}\right\rfloor\frac\pi4(x-3)\right)$$
this works because $\left\lfloor 2^{\operatorname{sgn}(x-2)}\right\rfloor=\begin{cases}0,&x<2,\\2,&x\ge 2\end{cases}$ if one takes $\operatorname{sgn}(0)=1$.
