Okay so the sequence $a_n$ defined by $a_1=0, a_{2m}= \frac{a_{2m-1}}{2}, a_{2m+1}=\frac{1}{2}+s_{2m}$ gives me this:
$s_1 = 0, s_2 = 0, s_{2(1)+1} = \frac{1}{2}, s_4=1, s_5= \frac{3}{4}, s_6= \frac{3}{8}, s_7 = \frac{7}{8}, s_8= \frac{7}{16}, s_9 = \frac{15}{16}, s_{10}= \frac{15}{32}$, which I can see can be simplified down to
$s_n = \begin{cases} 1- \dfrac 1{2^{\frac{n-1}{2} }} & n \neq 1 \text{ odd} \\ \dfrac 12 - \dfrac 1{2^{\frac{n}{2} }} & n \text{ even} \end{cases}$
But how do I prove this via induction if I get two different things for even and odd $n$ but the $a_n$ depends on $a_{n-1}$?