By solving the recurrence, we get \[F_n^{(a)} = F_{n-1}a + F_{n-2}.\]Therefore, $b$ is fibonatic iff there exist $n>3$ and $a > 0$ so that $b = F_{n-1}a + F_{n-2}$. Thus, the set of the fibonatic numbers is \[\bigcup_{n=4}^\infty \{F_{n-1} a + F_{n-2} : a > 0\}.\]Define \[d(N) = \frac{\#(b \le N : 30b \text{ is fibonatic})}{N},\]the portion of numbers from $\{30, 60, \dots, 30N\}$ that are fibonatic, and \[d_n(N) = \frac{\#(b \le N : 30b \in \{F_{n-1} a+ F_{n-2} : a > 0\})}{N},\]the portion of numbers from $\{30, 60, \dots, 30N\}$ that are in $\{F_{n-1} a+ F_{n-2} : a > 0\}$. Note that, by union bound, \[d(N) \le \sum_{n=4}^\infty d_n(N).\]Also, $d_4(N) = d_5(N) = d_6(N) = d_7(N) = 0$, and $d_n(N) \leq \frac{1}{F_{n-1}}$, as $N \to \infty$. Finally, as $N \to \infty$, \[d(N) \le \sum_{n=8}^\infty d_n(N) \leq \sum_{n=8}^\infty \frac{1}{F_{n-1}},\]which we can show to be smaller than $.99$, by bounding $F_{n-1}$ with powers of $\phi$. Thus, the density of fibonatic numbers in $\{30, 60, \dots\}$ is at most $.99$, thus the density of non-fibonatic numbers in $\{30, 60, \dots\}$ is at least $.01$, which means there are infinitely many non-fibonatic numbers.

By induction, $F^{(a)}_n = F_{n-1}a + F_{n-2}, \, \forall n \geq 3$. Consequently, the sequences $F^{(1)}_n, F^{(2)}_n, F^{(3)}_n, \dots$ form APs with difference $F_{n-1}$ for each $n\geq 3$. We begin by assuming that there are finitely many even non-fibonatic integers (since clearly every odd integer is fibonatic), then we will look at even numbers in those APs and show that their density is too low for this to be true. Suppose there are $m_0$ non-fibonatic integers. Let $m$ be an even integer to be chosen later, but for now we restrict it to being greater than the greatest non-fibonatic number. The aforementioned APs must contain every even integer lower than or equal $m$ except non-fibonatic numbers. As such, they contain $S \geq \frac{m}{2} - m_0$ distinct even numbers $\leq m$. But let's actually count them and see this can't be the case for large enough $m$: The general term of each AP is of the form $F_{n-2} + F_{n-1}a$. Notice at least one of $F_{n-1}$ and $F_{n-2}$ is odd, so the AP is either always odd or alternates between even/odd. Not only that, we need $a \leq \frac{m - F_{n-2}}{F_{n-1}} < \frac{m}{F_{n-1}}$. After some point, this can't be an integer, say, after $n_0$. By the three previous facts, for each $n \leq n_0$, we have at most $\frac12 \cdot \frac{m}{F_{n-1}}$ distinct even integers in the AP with first term $F_n$ (we can simply divide by 2 because the inequality was strict to begin with, so no need to worry about left-over 1's). Therefore, we must have (recall we only care about $n > 3$): $$\frac{m}{2}\left(\frac{1}{F_4} + \frac{1}{F_5} + \cdots + \frac{1}{F_{n_0}}\right) \geq \frac{m}{2} - m_0$$Next we show that $\frac{1}{F_4} + \frac{1}{F_5} + \cdots + \frac{1}{F_{n_0}} < 1$, meaning the RHS will eventually surpass the LHS for large enough $m$, which is absurd. To do this, recall that $F_n > \varphi^{n-2}$ (easy to prove with induction) for $n\geq 3$. So, $$\frac{1}{F_4} + \frac{1}{F_5} + \cdots + \frac{1}{F_{n_0}} < \frac{1}{\varphi^2} + \frac{1}{\varphi^3} + \cdots = \frac{1}{\varphi^2}\left(1 + \frac{1}{\varphi} + \frac{1}{\varphi^2} \cdots\right) = \frac{1}{\varphi^2}\cdot\frac{\varphi}{\varphi - 1} = \frac{1}{\varphi^2 - \varphi} = 1$$ And we are done. $\square$ PS: Might not have needed the assumption that $m$ is even. It felt natural at first because I noticed the problem was the density of even numbers, but it might be unecessary, so removing that might lead into an entirely analogous proof. Too lazy to actually check that, tho.