Call $a_n$ the number of poles ending in a no 1-foot flag, $b_n$ if it ends in one 1-foot flag, $c_n$ if it ends in 2 1-foot flags. Clearly:
$a_0 = 1, a_1 = 0, b_0 = 0, c_0 = 0$
Also, as we can add a 2-foot flag to anything, and adding a 1-foot flag changes the number of 1-foot flags at the end:
$\begin{align*}
a_n &= a_{n - 2} + b_{n - 2} + c_{n - 2} \\
b_n &= 2 a_{n - 1} \\
c_n &= 2 b_{n - 1}
\end{align*}$
The total number of $n$ feet poles is $a_n + b_n + c_n$.
Set up generating functions $A(z) = \sum_{n \ge 0} a_n z^n$, $B(z) = \sum_{n \ge 0} b_n z^n$ and $C(z) = \sum_{n \ge 0} c_n z^n$. Multiply shifted recurrences by $z^n$, sum over $n \ge 0$ and recognize resulting sums:
$\begin{align*}
\sum_{n \ge 0} a_{n + 2} z^n
&= \sum_{n \ge 0} a_n z^n + \sum_{n \ge 0} b_n z^n + \sum_{n \ge 0} c_n z^n \\
\sum_{n \ge 0} b_{n + 1} z^n
&= \sum_{n \ge 0} a_n z^n \\
\sum_{n \ge 0} c_n z^n
&= \sum_{n \ge 0} b_n z^n
\end{align*}$
$\begin{align*}
\frac{A(z) - a_0 - a_1 z}{z^2}
&= A(z) + B(z) + C(z) \\
\frac{B(z) - b_0}{z}
&= 2 A(z) \\
\frac{C(z) -c_0}{z}
&= 2 B(z)
\end{align*}$
Solve this system to get:
$\begin{align*}
A(z)
&= \frac{1}{1 - z^2 - 2 z^3 - 4 z^4} \\
B(z)
&= \frac{z}{1 - z^2 - 2 z^3 - 4 z^4} \\
C(z)
&= \frac{z^2}{1 - z^2 - 2 z^3 - 4 z^4}
\end{align*}$
We want:
$\begin{align*}
F(z)
&= A(z) + B(z) + C(z) \\
&= \frac{1 + 2 z + 4 z^2}{1 - z^2 - 2 z^3 - 4 z^4}
\end{align*}$
We want the coeficient of $z^n$ in this. The zeroes of the denominator are quite horrible, though.
We can glean the recurrence from the denominator: It is $F_n = F_{n - 2} + 2 F_{n - 3} + 4 F_{n - 4}$, with initial values $F_0 = 1, F_1 = 2, F_2 = 3$.