\begin{align} f(0) & = 1 \\ f(1) & = 1 \\ f(2) & = 2 \\ f(2n) & = f(n)+f(n+1), \;\;\;n\gt1 \\ f(2n+1) & = f(n-1)+f(n), \;\;\;n\ge1 \\ \end{align}
I am trying to figure out the generating function $G(x)$ for these given parameters, but am not sure how to proceed. I know for the fibonacci series, you just define a function $G(x) = f_0 + f_1x + ... $ and manipulate that so that each term has the appropriate summations, but for this, it doesnt seem so simple. Any help would be appreciated.
For $n\gt 1$ we have $$\begin{cases}f(2n)=f(n)+f(n+1)\\f(2(n+1)+1)=f(n)+f(n+1)\end{cases}\Rightarrow \boxed {f(2n)=f(2n+3)}$$ It follows $$G(x)=(1+x+2x^2+2x^3+3x^5)+\sum_{n\gt 1}a_{2n}(x^{2n}+x^{2n+3})$$ where I have calculated the coefficients $2,3$ of $x^3,x^5$ respectively.
We need the coefficients $a_4,a_6,a_8,...,a_{2n},…$ for which it holds the property $$a_{2n+2}-a_{2n}\in \{1,2\}$$ If someone wants to finish go ahead. I have no time now.