Unexplicit sum evaluation (Putnam) 
For positive integers $n,$ let the numbers $c(n)$ be determined by the rules $c(1)=1,c(2n)=c(n),$ and $c(2n+1)=(-1)^nc(n).$ Find the value of $$S = \sum_{n=1}^{2013}c(n)c(n+2).$$

Let $S_k$ represent the $k$th partial sum.
$$S_1 = c(1)c(3)$$
Since, $c(2n+1) = (-1)^nc(n)$ it follows: $c(3) = (-1)^1c(1) = -1$
$a_2 = c(2)c(4)$. $c(2*1) = 1$ and $c(2*2) = 1$ hence, $a_2 = 1$
$$\implies S_2 = a_1 + a_2 = 0$$
$$a_3 = c(3)c(5) = (-1)(1) = -1$$
$$\implies S_3 = -1$$
But the $1, -1, 1, -1, ....$ pattern doesnt work because of $S_2 = 0$. 
$$S = -1 + \sum_{n=2}^{2013} c(n)c(n+2) = -1 + c(2013)c(2015) + \sum_{n=2}^{2012} c(n)c(n+2)$$
But apart from that I don't really have an idea.

Split into even and odd terms. 
$$\begin{align} S &= \sum_{n=1}^{1006} c(2 n) c(2 n+2) +  \sum_{n=1}^{1006} c(2 n+1) c(2 n+3) + c(1) c(3) \\ &= \sum_{n=1}^{1006} c(n) c(n+1) +  \sum_{n=1}^{1006} (-1)^n c(n) (-1)^{n+1} c(n+1) - 1 \\ &= \sum_{n=1}^{1006} c(n) c(n+1) -  \sum_{n=1}^{1006} c(n) c(n+1) - 1 \\ &= -1\end{align} $$
