A common term for $a_n=\begin{cases} 2a_{n-1} & \text{if } n\ \text{ is even, }\\ 2a_{n-1}+1 & \text{if } n\ \text{ is odd. } \end{cases}$

Note that in binary we have

$$\begin{align*} a_1&=1\\ a_2&=10\\ a_3&=101\\ a_4&=1010\\ a_5&=10101\;, \end{align*}$$

displaying a pattern easily shown by induction to be real. Now note that the binary expansion of $\frac23$ is $\frac23=0.\overline{10}_{\text{two}}$, so that

$$\begin{align*} 2\cdot\frac23&=1.\overline{01}_{\text{two}}\\ 2^2\cdot\frac23&=10.\overline{10}_{\text{two}}\\ 2^3\cdot\frac23&=101.\overline{01}_{\text{two}}\\ 2^4\cdot\frac23&=1010.\overline{10}_{\text{two}}\\ 2^5\cdot\frac23&=10101.\overline{01}_{\text{two}}\;, \end{align*}$$

and therefore

$$a_n=\left\lfloor 2^n\cdot\frac23\right\rfloor=\left\lfloor\frac{2^{n+1}}3\right\rfloor\;.$$

If you really want to get rid of the floor function, observe that $2^{n+1}\equiv 1\pmod3$ when $n$ is odd, and $2^{n+1}\equiv2\pmod3$ when $n$ is even, so

$$\left\lfloor\frac{2^{n+1}}3\right\rfloor=\begin{cases} \dfrac{2^{n+1}-1}3,&\text{if }n\text{ is odd}\\ \dfrac{2^{n+1}-2}3,&\text{if }n\text{ is even}\;. \end{cases}$$

Now

$$\frac12\big(1+(-1)^n\big)=\begin{cases} 0,&\text{if }n\text{ is odd}\\ 1,&\text{if }n\text{ is even}\;, \end{cases}$$

so

$$\begin{align*} \left\lfloor\frac{2^{n+1}}3\right\rfloor&=\frac13\left(2^{n+1}-1-\frac12\big(1+(-1)^n\big)\right)\\ &=\frac13\left(2^{n+1}-\frac12\left(3+(-1)^n\right)\right)\\ &=\frac16\left(2^{n+2}-3-(-1)^n\right)\;. \end{align*}$$

This answer deals with how one can solve the recursion using the standard (textbook) method. We can write your recursion on the form $$a_n - 2a_{n-1} = f(n)$$

where $f(n) = 1$ when $n$ is odd and zero otherwise. In this form it's easy to see that the homogenous solution is $a_n^{\rm h} = c\cdot2^n$ as the characteristic equation is simply $\lambda-2 = 0$. To find a particular solution note that we can write $f(n)$ as

$$f(n) = \frac{1 - (-1)^n}{2}$$

which motivates the ansatz $a_n^{\rm p} = A + B(-1)^n$ (always try a test-solution on a similar form as the right hand side). Inserting this into the recursion and solving for $A,B$ gives us the particular solution $a_n^{\rm p} = -\frac{1}{2} - \frac{(-1)^n}{6}$. Finally imposing $a_1=1$ determines $c = \frac{2}{3}$ and gives us the final solution $a_n = a_n^{\rm h} + a_n^{\rm p} = \frac{2^{n+2} - (-1)^{n} - 3}{6}$.