How well do you know $\sqrt{2017}?$

Define $f_0(x) = \begin{cases} 0 & \text{if}\ \lfloor x \rfloor \equiv 0\ \text{mod}\ 3 \\ 1 & \text{if}\ \lfloor x \rfloor \equiv 1\ \text{mod}\ 3 \\ 2 & \text{if}\ \lfloor x \rfloor \equiv 2\ \text{mod}\ 3 \end{cases}$ Then $f(x) = \alpha(f_0(x) + \beta)$ , and the given sum is $\alpha \left(\sum_{n=1}^\infty \frac{f_0(3^n\sqrt{2017})}{3^n} + \beta \sum_{n=1}^\infty \frac 1{3^n} \right).$ Since $f_0(3^{-n} N)$ is precisely the $n$ th digit in the ternary representation of $N$ , the bracketed term on the left is equal to the fractional part of $\sqrt{2017}$ , which is equal to $\sqrt{2017}-44$ .

Next, note that $\sum_{n=1}^\infty 1/3^n = 1/2$ .

Combining these, the sum simplifies as $\alpha\left(\left[\sqrt{2017} - 44\right] + \tfrac12\beta\right) = 0;$ $\beta = -2(\sqrt{2017} - 44) = 88 - 2\sqrt{2017},$ so we post $a + b + c = 88 + 2 + 2017 = \boxed{2107}$ .

Consider the ternary expansion of $\sqrt{2017}$ .

$\sqrt{2017} = 44 + \sum_{k=1}^{\infty} \frac{a_{k}}{3^{k}}$

Where $a_{k} \in \{0,1,2 \}$ . If follows that

$\lfloor 3^{n}\sqrt{2017} \rfloor = \lfloor 44 \times 3^{n} + 3^{n-1}a_{1} + 3^{n-2}a_{2} + \ldots + 3a_{n-1}+a_{n}+3^{-1}a_{n+1} + \ldots \rfloor = 44 \times 3^{n} + 3^{n-1}a_{1} + 3^{n-2}a_{2} + \ldots + a_{n}$

(Since $\sum_{i=n+1}^{\infty} \frac{a_{i}}{3^{i}}<1$ if $a_{j}<2$ for any $j>n+1$ . This is the case since otherwise $3^{n}\sqrt{2017}$ is an integer, which is impossible)

Since all terms except $a_{n}$ are multiplied by a power of $3$ , the value of $f \left( 3^{n}\sqrt{2017} \right)$ is determined exclusively by the value of $a_{n}$ . It is easy to then see that $f \left( 3^{n}\sqrt{2017} \right) = a_{n}+\beta$

We then get that our sum is

$0 = \sum_{n=1}^{\infty} \frac{f \left( 3^{n}\sqrt{2017} \right) }{3^{n}} = \sum_{n=1}^{\infty} \frac{a_{n} + \beta}{3^{n}} = \frac{\beta}{2} + \sum_{n=1}^{\infty} \frac{a_{n}}{3^{n}}$

Our last infinite sum is easily evaluated by how we defined $a_{n}$

$0 = \left( \sqrt{2017}-44 \right) + \frac{\beta}{2} \Rightarrow \beta = 88-2\sqrt{2017}$

And so we get that $a+b+c = \boxed{2107}$