Answer
$a_{n}= \displaystyle \frac{n[1+(-1)^{n}]+(n-1)[1+(-1)^{n+1}]}{4}, \quad n=1,2,3...$
Work Step by Step
For even indices, we want
$a_{2}=1, a_{4}=2, a_{6}=3,\displaystyle \qquad a_{n}=\frac{n}{2}$, when n is even.
When n is odd,
$a_{1}=0=\displaystyle \frac{1-1}{2}$
$a_{3}=1=\displaystyle \frac{3-1}{2}$
$a_{5}=2=\displaystyle \frac{5-1}{2},...$ so for odd n, $a_{n}=\displaystyle \frac{n-1}{2}$
Using: $\quad (-1)^{n}=\left\{\begin{array}{ll}
+1, & \text{ when n is even}\\
-1, & \text{ when n is odd}
\end{array}\right.$
we see that
$1+(-1)^{n}=\left\{\begin{array}{ll}
2, & \text{ when n is even}\\
0, & \text{ when n is odd}
\end{array}\right.$
$n[1+(-1)^{n}]=\left\{\begin{array}{ll}
2n, & \text{ when n is even}\\
0, & \text{ when n is odd}
\end{array}\right.$
$\displaystyle \frac{n[1+(-1)^{n}]}{4}=\left\{\begin{array}{ll}
\frac{n}{2}, & \text{ when n is even}\\
0, & \text{ when n is odd}
\end{array}\right.$
$\displaystyle \frac{n[1+(-1)^{n}]}{4}+\frac{n-1}{2}\cdot\frac{[1+(-1)^{n+1}]}{2}=\left\{\begin{array}{ll}
\frac{n}{2}+0, & \text{ when n is even}\\
0+\frac{n-1}{2}, & \text{ when n is odd}
\end{array}\right.$
and we have what we wanted, both for even and for odd n's.
$a_{n}= \displaystyle \frac{n[1+(-1)^{n}]}{4}+\frac{n-1}{2}\cdot\frac{[1+(-1)^{n+1}]}{2}=, \quad n=1,2,3...$
