A sequence of numbers prime to $4n$ with $n$ being odd Let $n$ be odd. If I consider the sequence of $n$ numbers in the form $4k-1$ with $k$ running from $1$ to $n$ and take those with greatest common divisor with $4n$ being $1$ ( means those being prime to $4n$ ) I get exactly $\phi(n)$ of them.
I miss now a proof/argument why this true.
I tested the case $n=15$.

The function $f_4: \mathbb{Z}_{n}\rightarrow \mathbb{Z}_{n},\;f_4(k) \equiv 4k-1 \pmod n$ is a bijection because $\gcd(4,n)=1$. The same applies if you substitute $4$ with any number $m$ with $\gcd(n,m)=1,\,$ and the additive constant $1$ can be replaced be any number. 
Check this e.g. for $n=15, m=7, f(k) = 7k+4$ 
