Figure-It-Out Function! - Part 1: For 4 years

We have the second formula as for any odd:- $f(n) = \dfrac{f(n) + 2 f(\dfrac{n-1}{2})}{2}$ Put $n = 2t+1$ , so that above formula is true for any natural $t$ . $2f(2t+1) = f(2t+1) + 2f(t)$ $f(2t+1) = 2f(t)$ Similarly put $n=2t$ in the first formula and we get:- $2f(2t)= f(2t-1) +2f(t)$ From previous eqation:- $2f(2t) = f(2t-1)+f(2t+1)$ Hence we see that terms of $f(n)$ are in $AP$ . Also we can see that $f(3) = 2 f(1)=4$ From the definition of AP $f(n) = f(1) +(n-1)d$ $f(3) = f(1) + 2d$ $4 = 2 + 2d$ $d=1$ Hence $f(n) = 2 + n-1$ $f(n) = n+1$

Plugging the simple cases: one can conjecture that f(n) = n + 1.

We use induction to prove this:

Base cases: n = 0, f(0) = 0 + 1 = 1 as plugged. n = 1, f(1) = 1 + 1 = 2 as given.

Assume f(n) = n + 1 for any n.

If n + 1 is even (implies that n is odd):

f(n + 1) = [f(n) + 2f([n + 1]/2)]/2

From f(n):

f(n) = 2 f([n-1]/2) as plugged from the given.

This implies:

f(n+1) = f([n - 1]/2) + f([n + 1]/2)

And if n + 1 is odd:

Plugging from the given gives f(n + 1) = 2 f(n/2).

Equating both f(n + 1) = f(n + 1) especially when n is part of integer set.

f([n - 1]/2) + f([n + 1]/2) = 2 f(n/2).

This is where we use Jensen's Inequality. Since f''(n) = 0, this is the equality case of Jensen. Hence, the induction is proved.

The first step is to realise that the odd part is a rather elaborate way of writing that $f(n)=2f(\frac{n-1}{2})$ . After calculating the image of the first few numbers, we start to suspect that this is a rather complicated way of describing the function $f(n)=n+1$ , which one can indeed show it is. As a result the given quotient equals $\frac{2012\cdot2013\cdot2014\cdot2015}{4\cdot5\cdot11\cdot13\cdot19\cdot53\cdot61\cdot503}$ , where all factors in the denominator are chosen nicely and cansel out good part of the numerator. All that's left in the end is $2\cdot3\cdot31 = \boxed{186}$ .

Java:

public class brilliant{
    public static void main(String args[]){
        double f[] = new double[2015];
        f[1] = 2;
        for(int i=2;i<2015;i++){
            if(i%2==0) f[i]=(f[i-1]+2*f[i/2])/2;
            else f[i] = 2*f[i/2];
        }
        System.out.println(f[2011]*f[2012]*f[2013]*f[2014]/f[3]/f[4]/f[10]/f[12]/f[18]/f[52]/f[60]/f[502]);
    }
}

Very simple question, just put up the values: f(0)=1 f(1)=2 f(2)=3

Hence, f(n)=n+1