All About 2018 -- A

Define $g(x)=f(x)-x$ .

Because $f(n)=f(n-1)-f(n-2)+n$

$f(n)-n=(f(n-1)-(n-1))-(f(n-2)-(n-2))+1,g(n)=g(n-1)-g(n-2)+1$ .So,

$g(1)=0,g(2)=-1,g(3)=0,g(4)=2,g(5)=3,g(6)=2$

$g(7)=0,g(8)=-1,g(9)=0,g(10)=2,g(11)=3,g(12)=2$

$\vdots$

$g(2017)=0,g(2018)=-1,...$

$f(2018)=g(2018)+2018=2017$

$\begin{aligned} f(n) & = f(n-1)-f(n-2)+n \\ f(n+1) & = f(n) - f(n-1) + n+1 = - f(n-2) + 2n+1 \\ f(n+2) & = - f(n-1)+2n+3 \\ f(n+3) & = - f(n) + 2n + 5 \\ f(n+4) & = - f(n+1)+2n + 7 = f(n-2) + 6 & \small \color{#3D99F6} \text{Replace }n-2 \text{ with }n \\ \implies f(n+6) & = f(n) + 6 \\ f(6m+a) & = 6m + f(a) & \small \color{#3D99F6} \text {where }a < 6, m \in \mathbb N \\ \implies f(N) & = 6 \left\lfloor \frac N6\right\rfloor + f(N \bmod 6) \\ f(2018) & = 6\times 336 + f(2) = 2016 + 1 = \boxed{2017} \end{aligned}$

If one lists out the values of $f(n)$ for the first twenty natural numbers, as I did, you may notice an interesting pattern. At the numbers $n = 1, 3, 7, 9, 13, 15, 19, ...$ , $f(n) = n$ . Notice how the spacing between these numbers alternates between 2 and 4 ( $3 - 1 = 2, 7 - 3 = 4, 9 - 7 = 2, 13 - 9 = 4, ...$ ). I conjectured that this pattern would continue.

Continuing this pattern we eventually find that $f(n) = n$ when $n = 2017$ and when $n = 2019$ . Thus, $f(2019) = f(2018) - f(2017) + 2019 \rightarrow 2019 = f(2018) - 2017 + 2019 \rightarrow \boxed{f(2018) = 2017}$