How to show that the generating function $ S(z) $ of the sequence $$\begin{eqnarray} s_{n} &=& s_{n-1}+s_{\left\lfloor \frac{n}{2} \right\rfloor} \end{eqnarray}$$
satisfies the relation $$\begin{eqnarray} S(z) &=& z S(z)+S(z^{2}) + z S(z^{2})? \end{eqnarray}$$
I tried doing it like this:
$$\begin{eqnarray} s_{2k} &=& s_{2k-1}+s_{k}, \\ s_{2k+1} &=& s_{2k}+s_k \end{eqnarray}$$
$$\begin{eqnarray} \sum_k^{\infty} s_{2k}z^{2k} &=& \sum_k^{\infty} s_{2k-1}z^{2k} + \sum_k^{\infty} s_{k}z^{2k}, \\ \sum_k^{\infty} s_{2k+1}z^{2k+1} &=& \sum_k^{\infty} s_{2k}z^{2k+1} + \sum_k^{\infty} s_{k}z^{2k+1} \end{eqnarray}$$
$$\begin{eqnarray} \sum_k^{\infty} s_{2k}z^{2k} &=& z\sum_k^{\infty} s_{2k-1}z^{2k-1} + \sum_k^{\infty} s_{k}z^{2k}, \\ \sum_k^{\infty} s_{2k+1}z^{2k+1} &=& z\sum_k^{\infty} s_{2k}z^{2k} + z\sum_k^{\infty} s_{k}z^{2k} \end{eqnarray}$$
$$\begin{eqnarray} \frac{S(z)+S(-z)}{2} &=& z \frac{S(z)-S(-z)}{2}+S(z^2), \\ \frac{S(z)-S(-z)}{2} &=& z \frac{S(z)+S(-z)}{2} + zS(z^2) \end{eqnarray}$$
$$\begin{eqnarray} \frac{S(z)-S(-z)}{2} &=& z(z \frac{S(z)-S(-z)}{2}+S(z^2)) +zS(z^2) \end{eqnarray}$$
$$\begin{eqnarray} S(z)-S(-z) &=& z^2S(z)-z^2S(-z)+4zS(z^2) \end{eqnarray}$$
But this does not result in what was expected, which is $$\begin{eqnarray} S(z) &=& z S(z)+S(z^{2}) + z S(z^{2}) \end{eqnarray}$$
Where am I making a mistake? Thanks.
As @MatyMangoo wrote in a comment, just add together \begin{eqnarray} \frac{S(z)+S(-z)}{2} &=& z \frac{S(z)-S(-z)}{2}+S(z^2), \\ \frac{S(z)-S(-z)}{2} &=& z \frac{S(z)+S(-z)}{2} + zS(z^2) \end{eqnarray} and you get the expected result.