Let $(b_n)_n$ be the sequence defined by $b_1=b_2=1$ and the inductive formula:
$$\begin{cases} b_{2n} & = b_n \\ b_{2n+1} & = b_n + b_{n+1} \end{cases}$$
I am to find a simple expression of this sequence.
I've tried defining $V_n = \begin{pmatrix} b_n \\ b_{n+1}\end{pmatrix}$ and therefore getting:
$$\begin{cases} V_{2n} = \begin{pmatrix} b_n \\ b_n+b_{n+1} \end{pmatrix} &= \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} V_n = M_0 V_n\\ V_{2n+1} = \begin{pmatrix} b_n+b_{n+1} \\ b_{n+1} \end{pmatrix} &= \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} V_n = M_1 V_n \end{cases}$$
Thus, if $n=n_0+...+n_p2^p$, then $V_n = M_{n_{p-1}}...M_{n_0} V_1$.
I don't think this helps very much other than formalizing the problem. These matrices don't commute and their product has no simple expression short of calculation $b_n$ explicitly.
I don't really know where to go from here.
EDIT: after a bit of research, it's a well-known sequence (thanks to Hw Chu). It doesn't seem to have a simple expression, but I did find out:
$$b_n = \sum_{k=0}^{n-1} ({k \choose n-k-1} \mod 2)$$
This is what I am now trying to prove.