Christmas Streak 26/88: Disgraceful Size

Firstly before we start anything, $\dfrac{a_{2017}}{1000}-\left\lfloor\dfrac{a_{2017}}{1000}\right\rfloor$ expresses the fractional part of $\dfrac{a_{2017}}{1000}.$

And cause $a_{2017}$ is a natural number, we see that what we're trying to get is the remainder when $a_{2017}$ is divided by $1000.$

Since $a_1$ is already divisible by $8,$ we can all know that $a_{2017}\equiv 0 \pmod{8}.$

Then all we need to know is the remainder of $a_{2017}$ when it's divided by $125.$

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Let $2^{b_n}\equiv a_n \pmod{125},$ and then we see that $a_1=2^5$ and $a_2\equiv 2^{10}\pmod{125}.$

Therefore we can set $b_1=5,~b_2=10,~b_{n+2}=13b_{n+1}+37b_{n}.$

Since $2^{100}=2^{\phi(125)}\equiv 1 \pmod{125},$ we can see that all we need is the remainder when $b_{2017}$ is divided by $100.$

Let's study about $b_{2017}$ with mod 4 and mod 25 then.

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$b_{n+2}=b_{n+1}+b_{n} \pmod{4},$ and since

$(b_1,~b_2,~b_3,~b_4,~b_5,~b_6,~b_7,~b_8,~b_9,~\cdots)=(1,~2,~3,~1,~0,~1,~1,~2,~3,~\cdots),$ we can see that $b_{n+6}\equiv b_{n}\pmod{4}.$

Therefore $b_{2017}\equiv b_{1}=1 \pmod{4}.$

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We all know that $b_{n}$ is always divisible by $5,$ since $b_1\equiv b_2\equiv 0\pmod{5}.$

So we'll set $b_n=5c_n,$ and all we need is the remainder when $c_{2017}$ is divided by $5.$

$c_{n+2}\equiv 3c_{n+1}+2c_n\pmod{5}.$

$(c_1,~c_2,~c_3,~c_4,~\cdots,~c_{23},~c_{24},~c_{25},~\cdots) \\ =(1,~2,~3,~3,~0,~1,~3,~1,~4,~4,~0,~3,~4,~3,~2,~2,~0,~4,~2,~4,~1,~1,~0,~2,~1,~2,~3,~\cdots) \pmod{5}.$

Therefore $c_{n+24}\equiv c_n\pmod{5},$ and $c_{2017}\equiv c_1=1 \pmod{5}.$

And so $b_{2017}\equiv 5\pmod{25}.$

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From above we see that $b_{2017}\equiv 5\pmod{100},$ by the Chinese Remainder Theorem .

Then $a_{2017}\equiv 2^{b_{2017}}\equiv 32 \pmod{125}.$

Therefore, using the Chinese Remainder Theorem again, we get $a_{2017}\equiv \boxed{32} \pmod{1000}.$