This is more a "recreational" problem. By another question I came to the question for a closed-form-formula for this sequence $\small 1 , 2 , 3 , 6 , 9 , 18 , 27, \ldots $ which is just the mixture of the sequences $\small 3^k $ and $\small 2 \cdot 3^k $ .
I tried to find the "Binet"-type expression for it (like for instance for the Fibonacci-sequence) but do not find the initial "key". What is the way to such a formula?
On
A natural approach employs geometric generating functions and partial fraction decomposition.
$$\rm\frac{1}{1-3\ x^2}\ =\ 1 + 3\ x^2 + 9\ x^4+\ \cdots$$
$$\rm\ \ \ \frac{2\ x}{1-3\ x^2}\ =\ 2\ x + 6\ x^3 + 18\ x^5+\ \cdots$$
So your sought intermingled sequence $\rm\:f_n\:$ has generating function being their sum
$$\rm \frac{1+2\ x}{1-3\ x^2}\ =\ \frac{1/2 + 1/\alpha}{1-\alpha\ x}\ +\ \frac{1/2-1/\alpha}{1+\alpha\ x},\quad\ \alpha = \sqrt{3}$$
Thus, comparing coefficients $\rm\ \ f_n\ =\ (1/2 + 1/\alpha)\ \alpha^n\ +\ (1/2 - 1/\alpha)\ (-\alpha)^n $
Substituting the values $a_0=1$ and $a_1=2$ into the ansatz $a_k=c_1\sqrt3^k+c_2(-\sqrt3)^k$ yields
$$ \begin{eqnarray} c_1+c_2&=&1\;,\\ c_1-c_2&=&2/\sqrt3\;, \end{eqnarray}$$
which gives $c_1=(1+2/\sqrt3)/2$ and $c_2=(1-2/\sqrt3)/2$ and thus
$$\begin{eqnarray} a_k&=&\frac{(1+2/\sqrt3)\sqrt3^k+(1-2/\sqrt3)(-\sqrt3)^k}2\\ &=&\frac{\sqrt3^k+(-\sqrt3)^k}2+\sqrt3^{k-1}+(-\sqrt3)^{k-1}\;.\\ \end{eqnarray} $$
The factors $\sqrt3$ and $-\sqrt3$ can either be guessed or derived from the recurrence relation $a_{k+2}=3a_k$, which leads to the characteristic equation $\lambda^2=3$.