<s>[INST] Nth value of Function Given x and y we define a function as follow : 
f(1)=x
f(2)=y
f(i)=f(i-1) + f(i+1) for i>2

Now given x and y, how to calculate f(n)
Example : If x=2 and y=3 and n=3 then answer is 1
as f(2) = f(1) + f(3), 3 = 2 + f(3), f(3) = 1.
Constraints are : x,y,n all can go upto 10^9.
 [/INST] The general way to solve $$F(i+1)= F(i) - F(i - 1)$$
Is to first to see that if $F(i + 1)$ can be written as the sum of 2 geometric series, so that $$F(n) = ar^n + bs^n$$
So $$ar^{n+1} + bs^{n+1} = ar^{n} + bs^n - ar^{n-1} - bs^{n-1}$$
 $$ar^{n-1}(1 - r + r^2) = -bs^{n-1}(1 - s + s^2)$$
Since $r\ne s$, it follows that both $r$ and $s$ must satisfy the quadratic equation $$1-x+x^2 = 0$$
So $$r = \frac{1 + \sqrt{3}i}{2}, s = \frac{1 - \sqrt{3}i}{2}$$
Now we find $a$ and $b$, using $F(1)$ and $F(2)$
$$F(1) = x = a\left(\frac{1 + \sqrt{3}i}{2}\right)^1 + b\left(\frac{1 - \sqrt{3}i}{2}\right)^1$$
$$F(2) = y = a\left(\frac{1 + \sqrt{3}i}{2}\right)^2 + b\left(\frac{1 - \sqrt{3}i}{2}\right)^2$$
And then you find the closed form of your recurrence series.
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