Solving Recurrence $T(n) = T(n − 3) + 1/2$; I have to solve the following recurrence. 
$$\begin{gather}
T(n) = T(n − 3) + 1/2\\
T(0) = T(1) = T(2) = 1. 
\end{gather}$$
I tried solving it using the forward iteration. 
$$\begin{align}
T(3) &= 1 + 1/2\\
T(4) &= 1 + 1/2\\
T(5) &= 1 + 1/2\\
T(6) &= 1 + 1/2 + 1/2 = 2\\
T(7) &= 1 + 1/2 + 1/2 = 2\\
T(8) &= 1 + 1/2 + 1/2 = 2\\
T(9) &= 2 + 1/2
\end{align}$$
I couldnt find any sequence here. can anyone help!

The generating function is $$g(x)=\sum_{n\ge 0}T(n)x^n = \frac{2-x^3}{2(1+x+x^2)(1-x)^2}$$, which has the partial fraction representation
$$g = \frac{2}{3(1-x)} + \frac{1}{6(1-x)^2}+\frac{x+1}{6(1+x+x^2)}$$. The first
term contributes $$\frac{2}{3}(1+x+x^2+x^3+\ldots)$$, equivalent to $T(n)=2/3$ the second term contributes $$\frac{1}{6}(1+2x+3x^2+4x^3+\ldots)$$ equivalent to $T(n) = (n+1)/6$, and the third term contributes $$\frac{1}{6}(1-x^2+x^3-x^5+x^6-\ldots)$$ equivalent to $T(n) = 1/6, 0, -1/6$ depending on $n\mod 3$ being 0 or 1 or 2.
$$T(n) = \frac{2}{3}+\frac{n+1}{6}+\left\{\begin{array}{ll} 1/6,& n \mod 3=0\\
0,& n \mod 3=1 \\
-1/6,&n \mod 3 =2\end{array}\right.$$
