Show that the recursively defined function z is neither surjective or injective

My question is:

The function z from $\displaystyle \mathbb{N}$ to $\displaystyle \mathbb{N}$ is defined recursively by:

z(1)=1 and

$\displaystyle z(n+1) = \begin{cases} 1/2(z(n)+3), & \text{if }z(n)\text{ is odd} \\ z(n)+5, & \text{if }z(n)\text{ is even} \end{cases}$

for all $\displaystyle n\geq 1$.

Show that z is neither an injection or a surjection.

What I've done so far is find the first couple of values of z(n) up to n=7, where I got z(2)=2, z(3)=7, z(4)=5, z(5)=4, z(6)=9, and z(7)=6. At this point, I'm not sure about what to do, so any help would be greatly appreciated!

Re: Show that the recursively defined function z is neither surjective or injective

z(9)=z(3), so it's not injective.

From 9 onwards, the sequence is periodic, because z(3+k+6n)=z(3+k), so it's not surjective.

Re: Show that the recursively defined function z is neither surjective or injective

I'm a bit confused about k, could you explain how you got z(3+k+6n) and z(3+k)? Thank you