Christmas Streak 09/88: Subtract, Subtract, Subtract

Let $t_1=2017,~t_2=a.$

Then we see that

$\begin{aligned} & t_3=t_1-t_2 \\ & t_4=-t_1+2t_2 \\ &t_5=2t_1-3t_2 \\ & t_6=-3t_1+5t_2 \\ & t_7=5t_1-8t_2 \end{aligned}$

We deduce that $t_n=(-1)^{n+1}(t_1F_{n-2}-t_2F_{n-1}),$ where $F_n$ is a Fibonacci sequence .

Then solving this, we get:

$\cases{\begin{aligned} &\dfrac{t_1}{t_2}>\dfrac{F_{n-1}}{F_{n-2}} & (n\text{ is odd.}) \\\\ &\dfrac{t_1}{t_2}<\dfrac{F_{n-1}}{F_{n-2}} & (n\text{ is even.})\end{aligned}}$

We know that

$\begin{aligned}&\frac{F_{2}}{F_{1}}=1<\frac{F_{4}}{F_{3}}=\frac{3}{2}<\frac{F_6}{F_4}=\frac{8}{5}<~\cdots \\\\ &<\frac{1+\sqrt{5}}{2} \\\\ &<~\cdots~<\frac{F_7}{F_6}=\frac{13}{8}<\frac{F_5}{F_4}=\frac{5}{3}<\frac{F_3}{F_2}=\frac{2}{1}. \end{aligned}$

So, if and only if $\dfrac{t_1}{t_2}=\dfrac{1+\sqrt{5}}{2},$ the length of the sequence is infinity.

Therefore, $\dfrac{2017}{k}=\dfrac{1+\sqrt{5}}{2}.$ Solving this yields $k=\dfrac{2017\cdot2}{1+\sqrt{5}}=1246.57455530\cdots.$

$\therefore~\lfloor1000k\rfloor=\boxed{1246574}.$

The recursion $a_{i+2} = a_i - a_{i+1}$ has general solution $a_i = C_1p_1^i + C_2p_2^i$ where $p_{1,2}$ are solutions of $p^2 = 1 - p$ . We find $p_1 = -\tfrac12 + \tfrac12\sqrt 5$ and $p_2 = -\tfrac12 - \tfrac12\sqrt 5$ .

Since $0 < p_1 < 1$ , the first term remains positive (provided $C_1 > 0$ ) and tends to zero. But $p_2 < 1$ , so that the second term alternates between positive and negative and diverges; it will dominate the sequence. Thus the sum $a_1 = C_1p_1^1 + C_2p_2^2$ will eventually include negative values, unless we choose $C_2 = 0$ . Therefore the sequence will remain positive indefinitely only if we have $a_i = 2017\cdot (-\tfrac12 +\tfrac12\sqrt 5)^{i-1};$ in that case $a_2 = 2017\cdot (-\tfrac12 + \tfrac12\sqrt 5) = 1246.574\dots$