Play with functions 2

$f(x-1)$ + $f(x+1)$ = $\sqrt{3}$ $f(x)$

Putting $x=x+1$

$f(x)$ + $f(x+2)$ = $\sqrt{3}$ $f(x+1)$

Putting $x=x+2$ in eq $\boxed{1}$

$f(x+1)$ + $f(x+3)$ = $\sqrt{3}$ $f(x+2)$

$f(x-1)$ + 2 $f(x+1)$ + $f(x+3)$ = $\sqrt{3}$ [ $\sqrt{3}$ $f(x+1)$ ]

$f(x-1)$ + $f(x+3)$ = $f(x+1)$

Again putting $x=x+2$

$f(x+1)$ + $f(x+5)$ = $f(x+3)$

$f(x-1)$ + $f(x+5)$ = 0

$f(x-1)$ = - $f(x+5)$

$f(x)$ = - $f(x+6)$

$f(x)$ = $f(x+12)$

$\sum_{i=0}^{19}$ $f(5+12i)$ =20 $f(5)$ =200

So the sum of digits is $2$

Let us define the sequence $a_n=f(n).$ Then the given condition for the function $f(x)$ can be translated into the following recursive equation: $a_{n+1}=\sqrt{3}a_n-a_{n-1}$ Therefore, ${a_n}$ is a sequence defined by a linear recurrence. The characteristic polynomial of the recurrence is $r^2-\sqrt{3}r+1$ that has the complex zeros $\frac{\sqrt{3}}{2}\pm \frac{1}{2}i=\cos{\frac{\pi}{6}}+i\sin{\frac{\pi}{6}}.$ Then for such a recurrence, we can obtain that $a_n=C_1\cos {\frac{n\pi}{6}} +C_2 \sin {\frac{n\pi}{6}},$ where $C_1$ and $C_2$ are constants. You can see Linear Recurrence Relations. From the formula obtained for $a_n,$ we get that $a_{n+12}=a_{n}$ for any positive integer value of $n$ . Therefore, $\sum_{l=0}^{19} f(5+12l)= \sum_{l=0}^{19} a_{5+12l}= \sum_{l=0}^{19} a_{5}=20(10)=200.$ Then, we can conclude that the answer for the question is $2+0+0=\boxed{2}.$