$Z$-transform of floor($\frac{n}{5}$)

Question

I found problem about floor array that I can't solve. Find Z transform of f(n) = floor(n/5). I tried writing this array and for n from 1 to infinity, I got n/5 = 1/5 + 2/5 + 3/5 + 4/5 + 5/5 + 6/5... Now, taking floor for these numbers floor(n/5) = 0 + 0 + 0 + 0 + 0 + 1 + 1.. +2 +2... +3... This means that for n>0 my array can be rewritten as floor(n/5) = 5n. I was wondering if this is right and if Z{floor(n/5)} = Z{5n}?

Answer 1

For $x[n]=\lfloor n/N\rfloor$ we have $N$ identical (but shifted) summations

$$X(z) = \sum_{k=0}^\infty k z^{-Nk} + \cdots + \sum_{k=0}^\infty k z^{-Nk-N+1}$$

$$= \frac{z^{-N}z^{-0}}{(1-z^{-N})^2} + \cdots + \frac{z^{-N}z^{-(N-1)}}{(1-z^{-N})^2} = \frac{z^{-N}}{(1-z^{-N})(1-z^{-1})}$$

which does not line up with your guess.