Sequencing

For the linear recurrence relation $S_n = S_{n-1}-S_{n-2}$ , the characteristic equation is:

$\begin{aligned} r^2 & = r-1 \\ \implies r^2 - r + 1 & = 0 \\ \implies r & = \frac {1 \pm \sqrt 3i}2 = e^{\pm \frac \pi 3 i} \end{aligned}$

$\begin{aligned} \implies S_n & = c_1e^{\frac \pi 3ni} + c_2 e^{-\frac \pi 3ni} \\ S_0 & = c_1 + c_2 = 0 \\ \implies c_2 & =-c_1 \\ S_1 & = c_1 \left(e^{\frac \pi 3i} - e^{-\frac \pi 3i}\right) = 1 \\ c_1 \sqrt 3 i & = 1 \\ \implies c_1 & = - \frac i{\sqrt 3} \\ \implies S_n & = \frac i{\sqrt 3} \left( e^{-\frac \pi 3ni} - e^{\frac \pi 3ni} \right) \\ & = \frac 2{\sqrt 3} \sin \frac {n\pi} 3 \\ \implies S_{2017} & = \frac 2{\sqrt 3} \sin \frac {2017\pi} 3 \\ & = \frac 2{\sqrt 3} \sin \frac \pi 3 \\ & = \boxed{1} \end{aligned}$

Calculating some terms we see that S=[0,1,1,0,-1,-1,0,1,...] This repeats with period 6 so the answer is the same as the 2017 mod 6 th term. 2017=1 mod 6 so the answer is one.

Let us solve this linear recursion by the following method. It has the characteristic equation $r^2 - r + 1 = 0$ , or $r = \frac{1 \pm \sqrt{3}i}{2}$ . This leads to the general form:

$S_n = A(\frac{1 + \sqrt{3}i}{2})^{n} + B(\frac{1 - \sqrt{3}i}{2})^{n} = Ae^{i \frac{n\pi}{3}} + Be^{-i \frac{n\pi}{3}}$

Implementing the initial conditions $S_0 = 0, S_1 = 1$ now solves the real constants $A$ and $B$ :

$0 = A + B$

$1 = A(\frac{1 + \sqrt{3}i}{2}) + B(\frac{1 - \sqrt{3}i}{2})$

which yields $A = \frac{1}{\sqrt{3}i}$ and $B = -\frac{1}{\sqrt{3}i}$ . We ultimately end up with:

$S_n = \frac{1}{\sqrt{3}i} \cdot [e^{i \frac{n\pi}{3}} - e^{-i \frac{n\pi}{3}}] = \frac{2}{2 \sqrt{3}i} \cdot [e^{i \frac{n\pi}{3}} - e^{-i \frac{n\pi}{3}}] = \frac{2}{\sqrt{3}} \cdot [\frac{e^{i \frac{n\pi}{3}} - e^{-i \frac{n\pi}{3}}}{2i}] = \frac{2}{\sqrt{3}} sin(\frac{n\pi}{3})$

Computing $S_{2017}$ gives $\frac{2}{\sqrt{3}} sin(\frac{2017\pi}{3}) = \frac{2}{\sqrt{3}} \cdot \frac{\sqrt{3}}{2} = \boxed{1}.$

S2017 = S2016-S2015

(S2015-S2014)-S2015

-S2014

-(S2013-S2012)

-(S2012-S2011-S2012)

S2011

Therefore,S2017=S2017-6x,Where x is a

integer and 6x<2017

Let x be 336, Therefore S2017=S2017-2016

S2017=S1=1