Defining a repeating series with a simple expression. I'm looking for a simple expression with which I can define a repeating series.
$$f(0)=0$$
$$f(1)=1$$
$$f(2)=2$$
$$f(3)=3$$
$$f(4)=f(n-4)$$
So, the series goes $0, 1, 2, 3, 0, 1, 2, 3...$ and on and on. I feel like this should have something to do with exponents because in Discrete Math we figured out how to define a series that goes $1, -1, 1, -1...$ but I'm not sure how to even begin defining this.
I don't want anything recursive if I can possibly avoid it. Not sure if that will be a problem.
Thanks a bunch!
EDIT: Spoke to a professor about how to solve this and he says there is a way to do it involving only +/-/*/^/÷, but it involves complex numbers.
 A: You could just write
$$
f(n) = n \pmod{4}
$$
( a loose use of notation but perfectly clear)
or
$$
f(n) = n -  4 \lfloor n/4 \rfloor .
$$
I think the first would be kinder to your reader.
A: If you want something along the lines of $(-1)^n$
(which produces $1,-1,1,-1,\ldots$),
first consider
$$ f_1(n) = \tfrac12\left(1 - (-1)^n\right),$$
which implies that $f_1(n) = 0,1,0,1,\ldots$ for $n = 0,1,2,3,\ldots.$
Example 1
We can slow down the rate of alternation of $f_1$
by making the exponent grow slower:
$$ f_2(n) = \tfrac12\left(1 - (-1)^{\lfloor n/2\rfloor}\right),$$
where $\lfloor n/2\rfloor$ is the greatest integer less than or equal
to $n/2.$ This function follows the pattern
$f_2(n) = 0, 0, 1, 1, 0, 0, 1, 1,\ldots$ for $n=0,1,2,3,4,5,6,7,\ldots.$
Put this together in the form $f(n) = f_1(n) + 2f_2(n)$ and you have the function you asked for. That is, let
$$
f(n) = \tfrac32 - \tfrac12(-1)^n - (-1)^{\lfloor n/2\rfloor}.
$$
Example 2
If you don't like the greatest-integer function 
(used in the expression $\lfloor n/2\rfloor$), we can work around this
by fiddling with sinusoidal functions.
Let $g_1(n) = \cos\left(\frac12 n\pi - \frac14\pi\right),$ so that 
$\newcommand{s}{\frac{\sqrt2}{2}}
 g_1(n) = \s,\s,-\s,-\s,\s,\s,-\s,-\s,\ldots$ 
for $n=0,1,2,3,4,5,6,7,\ldots.$
Then let
\begin{align}
f(n) &= f_1(n) + 1 - \sqrt2g_1(n) \\
&= \tfrac32 - \tfrac12(-1)^n
      - \sqrt2\cos\left(\tfrac12 n\pi - \tfrac14\pi\right).
\end{align}
Example 3
If you're willing to use complex numbers, notice that
$\cos z = \tfrac12\left(e^{iz} + e^{-iz}\right).$
Using the function notation $\exp(z) = e^z$ to make the first equation below more readable, we get
\begin{align}
\cos\left(\tfrac12 n\pi - \tfrac14\pi\right) &=
\tfrac12 \left(\exp\left(i\left(\tfrac12 n\pi - \tfrac14\pi\right)\right)
      + \exp\left(-i\left(\tfrac12 n\pi - \tfrac14\pi\right)\right)\right)\\
&= \tfrac12 \left(e^{in\pi/2} e^{-i\pi/4} + e^{-in\pi/2} e^{i\pi/4}\right)\\
&= \tfrac12 e^{i\pi/4} \left(e^{in\pi/2} e^{-i\pi/2} + e^{-in\pi/2}\right).
\end{align}
Then, using the facts that $e^{i\pi/2}=i,$ $e^{-i\pi/2}=-i,$
and $e^{i\pi/4}=\tfrac{\sqrt2}{2}(1+i),$
\begin{align}
\cos\left(\tfrac12 n\pi + \tfrac14\pi\right)
&= \tfrac12\left(\tfrac{\sqrt2}{2}(1+i)\right)\left(i^n(-i)+(-i)^n\right)\\
&= \tfrac{\sqrt2}{4}(1+i) \left((-1)^n - i\right) i^n.
\end{align}
It follows that
$$
f(n) = \tfrac32 - \tfrac12(-1)^n - \tfrac12(1+i)\left((-1)^n - i\right)i^n.
$$
And indeed the right-hand side is also equal to
$\tfrac32-\tfrac12(-1)^n-\tfrac12(1+i)(-i)^n-\tfrac12(1-i)i^n,$
an expression that was given in a comment under another answer.

Frankly, I think any of the preceding examples is far clumsier than Ethan Bolker's formula $n -  4 \lfloor n/4 \rfloor,$ which is inspired by modular arithmetic but actually uses only ordinary integer arithmetic with the 
greatest-integer function.
That answer gets my vote.

Example 4
I have added this example after seeing that the accepted answer uses
a curly bracket with multiple cases ($n$ odd and $n$ even)
on the right-hand side of the equation.
If two cases are OK, why not four? That is,
$$
f(n) = \begin{cases}
  0 \quad & n = 4k, k \in \mathbb Z \\
  1 & n = 4k+1, k \in \mathbb Z \\
  2 & n = 4k+2, k \in \mathbb Z \\
  3 & n = 4k+3, k \in \mathbb Z \\
\end{cases}
$$
A: Just to add an alternative to the other answers: you can use generating functions to try to find an explicit solution to the recurrence relation. If $F(x)$ is the generating function for the sequence $f(n)$,
$$F(x)=\sum_{n\ge0}f(n)x^n=f(0)+f(1)x+f(2)x^2+f(3)x^3+\cdots$$
then from the recurrence given by
$$\begin{cases}f(0)=0\\f(1)=1\\f(2)=2\\f(3)=3\\f(n)=f(n-4)&\text{for }n\ge4\end{cases}$$
we have
$$\begin{align*}
f(n-4)&=f(n)\\
\sum_{n\ge4}f(n-4)x^n&=\sum_{n\ge4}f(n)x^n\\
x^4\sum_{n\ge4}f(n-4)x^{n-4}&=\sum_{n\ge0}f(n)x^n-f(0)-f(1)x-f(2)x^2-f(3)x^3\\
x^4\sum_{n\ge0}f(n)x^n&=F(x)-x-2x^2-3x^3\\
x^4F(x)&=F(x)-x-2x^2-3x^3\\
(x^4-1)F(x)&=-x-2x^2-3x^3\\
F(x)&=\frac{x+2x^2+3x^3}{1-x^4}
\end{align*}$$
Expanding into partial fractions, you have
$$F(x)=\frac32\frac1{1-x}-\frac12\frac1{1+x}-\frac1{1+x^2}-\frac x{1+x^2}$$
and expressing each term as a power series we arrive at
$$\begin{align*}
F(x)&=\frac32\sum_{n\ge0}x^n-\frac12\sum_{n\ge0}(-x)^n-\sum_{n\ge0}(-x^2)^n-x\sum_{n\ge0}(-x^2)^n\\
F(x)&=\sum_{n\ge0}\left(\frac32-\frac12(-1)^n\right)x^n-\sum_{n\ge0}(-1)^nx^{2n}-\sum_{n\ge0}(-1)^nx^{2n+1}
\end{align*}$$
each of which are valid for $|x|<1$.
Now,
$$\sum_{n\ge0}\left(\frac32-\frac12(-1)^n\right)x^n=1+2x+x^2+2x^3+\cdots$$
$$\sum_{n\ge0}(-1)^nx^{2n}=1-x^2+x^4-x^6+\cdots$$
$$\sum_{n\ge0}(-1)^nx^{2n+1}=x-x^3+x^5-x^7+\cdots$$
so it follows that
$$F(x)=\color{lightgray}{0+}\,x+2x^2+3x^3+\color{lightgray}{0x^4+}\,x^5+2x^6+3x^7+\cdots$$
Replace $n=k$ in the latter two series, then split up the first into the even and odd cases, letting $n=2k$ and $n=2k+1$, respectively, so that the generating function can be expressed as
$$F(x)=\begin{cases}
\displaystyle\sum_{k\ge0}(1-(-1)^k)x^{2k}&n=2k\text{ is even}\\[1ex]
\displaystyle\sum_{k\ge0}(2-(-1)^k)x^{2k+1}&n=2k+1\text{ is odd}
\end{cases}$$
From this it follows that a closed form for $f(n)$ can be obtained by splitting into even/odd cases:
$$f(n)_{n\ge0}=\begin{cases}1-(-1)^{n/2}&n\text{ is even}\\2-(-1)^{(n-1)/2}&n\text{ is odd}\end{cases}$$
If I'm not mistaken, there should be a way to get an "even more explicit" form in terms of complex exponentials, but perhaps this result will suffice for your needs.
A: For integer values of $x$ this: $J\left(x\right)=1-\left(\sin \left(\frac{\pi }{4}\cdot 2\cdot x\right)+\cos \left(\frac{\pi }{4}\cdot 2\cdot x\right)\right)+\frac{\left(1-\left(-1\right)^x\right)}{2}$ works.
