Let $S_k$ be the set of sequences $\langle x_1,\ldots,x_n\rangle$ such that $x_i\in[k]$ for $i=1,\ldots,n$ and $x_i\le\frac{x_{i+1}}2$ for $i=1,\ldots,n-1$; $s_k=|S_k|$. Suppose that $\sigma=\langle x_1,\ldots,x_n\rangle\in S_k$. If $x_n<k$, then $\sigma\in S_{k-1}$. And $S_{k-1}\subseteq S_k$, so there are $s_{k-1}$ sequences in $S_k$ whose last term is less than $k$. If $x_n=k$, then $x_{n-1}\le\frac{x_n}2$, so $\langle x_1,\ldots,x_{n-1}\rangle\in S_{\lfloor k/2\rfloor}$. That is, every sequence in $S_k$ whose last term is $k$ is obtainted from a sequence in $S_{\lfloor k/2\rfloor}$ by appending a term $k$. Conversely, if $\langle x_1,\ldots,x_{n-1}\rangle\in S_{\lfloor k/2\rfloor}$, then $\langle x_1,\ldots,x_{n-1},k\rangle\in S_k$, so there are $s_{\lfloor k/2\rfloor}$ sequences in $S_k$ that end in $k$. Every $\sigma\in S_k$ either does or does not end in $k$, and none does both, so we’ve counted every sequence in $S_k$ once, and $s_k=s_{k-1}+s_{\lfloor k/2\rfloor}$.
We can infer the relationship
$$\frac{F(x)}{F(x^2)}=\frac{1+x}{1-x}\tag{1}$$
directly from the recurrence without determining the generating function $F(x)$ itself. Rewrite the recurrence as $s_k-s_{k-1}=s_{\lfloor k/2\rfloor}$, multiply through by $x^k$, and sum over $k\ge 0$:
$$\sum_{k\ge 0}s_kx^k-\sum_{k\ge 0}s_{k-1}x^k=\sum_{k\ge 0}s_{\lfloor k/2\rfloor}x^k\;.$$
The lefthand side is
$$\begin{align*}
\sum_{k\ge 0}s_kx^k-\sum_{k\ge 0}s_{k-1}x^k&=F(x)-x\sum_{k\ge 0}s_{k-1}x^{k-1}\\
&=F(x)-x\sum_{k\ge 0}s_kx^k\\
&=(1-x)F(x)\;,
\end{align*}$$
and the righthand side is
$$\begin{align*}
\sum_{k\ge 0}s_{\lfloor k/2\rfloor}x^k&=\sum_{k\ge 1}s_k(x^{2k}+x^{2k+1})\\
&=(1+x)\sum_{k\ge 0}s_kx^{2k}\\
&=(1+x)F(x^2)\;,
\end{align*}$$
so $(1-x)F(x)=(1+x)F(x^2)$, and $(1)$ follows immediately.
The sequence is OEIS A000123, and the generating function apparently does not have a nice form.
Best Answer
Let $2x_{n+2} = 3 \, x_{n+1} + 8 \, x_{n} + 3 \, x_{n-1}$, with $x_{0} = -1, x_{1} = 3$ and $x_{2} = 3$ for which letting $x_{n} = p^{n}$ the equation $$2 \, p^{3} - 3 \, p^{2} - 8 \, p - 3 = 0$$ is obtained. This equation can be factored to $(p-3)(p+1)(2p+1) = 0$ and leads to the roots $p \in \{ 3, -1, -1/2 \}$. From this the general form of $x_{n}$ is $$x_{n} = a_{0} \, 3^{n} + a_{1} \, (-1)^{n} + a_{2} \, \left( - \frac{1}{2} \right)^{n}.$$ Applying the initial conditions yields: \begin{align} -1 &= a_{0} + a_{1} + a_{2} \\ 3 &= 3 \, a_{0} - a_{1} - \frac{a_{2}}{2} \\ 3 &= 9 \, a_{0} + a_{1} + \frac{a_{2}}{4} \end{align} which leads to $a_{0} = \frac{1}{2}$, $a_{1} = - \frac{3}{2}$, $a_{2} = 0$. The resulting sequence is generated by \begin{align} x_{n} = \frac{3}{2} \, \left( 3^{n-1} + (-1)^{n-1} \right) \end{align}