Still Recurring Theme

We know that $a_{n+3}+a_{n+2}=a_{n+1}+a_n$ so that

$\sum_{k=1}^{2016}a_k=(a_1+a_2)+(a_3+a_4)+...+(a_{2015}+a_{2016})=1008(a_1+a_2)=1008$

and $\sum_{k=2}^{2015}a_k$ $=1007(a_2+a_3)=2014$ . Subtraction yields $a_1+a_{2016}=-1006$ and $a_{2016}=\boxed{-1007}$

$\begin{aligned}a_{n+3}&=\color{#20A900}{-a_{n+2}+a_{n+1}+a_n}\\&=2a_{n+1}-a_{n+1}\\&\color{#20A900}{=-2a_n+a_{n-1}+2a_{n-2}}\\&=3a_{n-1}-2a_{n-3}\\&\color{#20A900}{=-3a_{n-2}+a_{n-3}+3a_{n-4}}\\&\cdots\\&\cdots\\&\color{#20A900}{=-\dfrac{n+1}{2}(a_3)+a_2+\dfrac{n+1}{2}(a_1)}, \color{#D61F06}{n\rightarrow \text{odd}}\\&=(\dfrac n2 +1)a_3-\dfrac n2 a_1,\color{#D61F06}{n\rightarrow\text{even}}\end{aligned}$ In our case since $n=2013$ is odd putting we get: $a_{2016}=-1007a_3+a_2+1007a_1$ $\huge = \boxed{-1007}$

We have the recursion : $a_n = -a_{n-1} + a_{n-2} + a_{n-3}$

Now, we write the characteristic equation of this recursion by replacing $a_n$ with $x^n$

We get: $x^n = -x^{n-1} + x^{n-2} + x^{n-3}$

Or, $x^3 = -x^2 + x + 1$

Or, $x^3 + x^2 - x - 1 = 0$

$\Rightarrow (x-1)(x+1)^2 = 0$

So, the roots of the characteristic equation are $1, -1, -1$

So, $a_n = u \cdot (1)^n + (v+wn) \cdot (-1)^n$

Now, we have, $a_1=1, a_2=0, a_3=2$

So, we form three equations, putting $n=1,2,3$ :

$a_1 = u-v-w=1$

$a_2 = u+v+2w=0$

$a_3 = u-v-3w=2$

Solving these three equations, we get : $u = \frac{3}{4}, v = \frac{1}{4}, w = \frac{-1}{2}$

Therefore, $a_n = \frac{3}{4} + (-1)^n \cdot (\frac{1}{4} - \frac{n}{2})$

i.e. $a_n = \frac{3}{4} + (-1)^n \cdot \frac{1-2n}{4}$

Putting $n=2016$ , we get $a_{2016} = \boxed {-1007}$

This question can also be solved by the method of generating functions, but that method is a bit lengthy....

An explicit form of the sequence is $a_{2n+1} = n + 1$ and $a_{2n} = -n + 1$ .

This can be proven by induction. $a_{4} = -1$ and $a_{5} = 3$ , so this is already ok.

Now suppose $a_{2n +1} = n + 1$ , for some n, then we should prove that $a_{2n+2} = -n$ .

Proof : $a_{2n+2} = - a_{2n +1} +a_{2n} + a_{2n-1} = -n - 1 - n + 1 + n = -n.$

Now suppose $a_{2n} = -n + 1$ , for some n, then we should prove that $a_{2n+1} = n+1$ .

Proof : $a_{2n+1} = - a_{2n} +a_{2n-1} + a_{2n-2} = n - 1 +n - n + 2= n + 1.$

So $a_{2016} = a_{2.1008} = -1008 + 1 = -1007.$

Checking out few terms we have the series as $1,0,2,-1,3,-2,4....$ . Hence $a_{2016}$ is clearly equal to $\boxed{\boxed{-1007}}$

It is easy to recognize the pattern is that when n is odd a(n) is simply (n+1)/2,then a(n+1) is -n/2+1/2.Above conclusion can be proven by mathematical induction.

a2016 =1-1+2-3+4-5+...+2014-2015=-1007 so the answer is -1007