Number of zeros of an imprecise function

Define $f_k(x) := (x + 1)(x + 2) \dots (x + k)f(x + k)$ for $k > 0$ . Then the given equation can be written $f_2(x) + f_1(x) + f(x) = 0$ , or equivalently $f_2(x) = -f_1(x) - f(x)$ . Replacing $x$ with $x + k$ and then multiplying throughout by $(x + 1) \dots (x + k)$ , we have that for $k > 2$ , $f_k(x) = -f_{k - 1}(x) - f_{k - 2}(x)$ .

Therefore:

$g(x) = f_6(x) - 2f_3(x) + f(x) + 1/2018$

$\quad\quad = -f_5(x) - f_4(x) - 2f_3(x) + f(x) + 1/2018$

$\quad\quad = f_4(x) + f_3(x) - f_4(x) - 2f_3(x) + f(x) + 1/2018$

$\quad\quad = -f_3(x) + f(x) + 1/2018$

$\quad\quad = f_2(x) + f_1(x) + f(x) + 1/2018$

$\quad\quad = 1/2018$

for all real $x$ , so $g(x)$ has no real zeros.

Let us define the shift operator and the identity operator in the following way: $Tf(x)=f(x+1)$ and $If(x)=f(x)$ , respectively. Then the given functional equation can be written in the form $((x+2)(x+1)T^2+(x+1)T+I)f(x)=0.$ If you apply the operator $(x+1)T-I$ to both sides of the previous equation, we get the equation $((x+1)(x+2)(x+3)T^3-I)f(x)=0.$ Now applying the operator $(x+1)(x+2)(x+3)T^3-I)$ to both sides of the last equation, we get the equation $((x+1)(x+2)(x+3)(x+4)(x+5)(x+6)T^6- 2(x+1)(x+2)(x+3)T^3+I)f(x)=0.$ Therefore, the expression $(x+1)(x+2)(x+3)(x+4)(x+5)(x+6)f(x+6)- 2(x+1)(x+2)(x+3)f(x+3)+f(x)$ must be equal to zero for all possible values of $x$ . Then, the function $g(x)$ has not real zeros. Thus the answer to the problem is $\boxed{0}.$