1,1,2,3,5,8,... = Additive Fibonacci: a(n-1) * phi is asymptotic to a(n)
2,2,4,8,32,256,... = Multiplicative Fibonacci a(n)=a(n-1)*a(n-2): a(n-1) ^ phi is asymptotic to a(n)
2,2,4,16,65536,1.158*10^77... = Exponential Fibonacci a(n)=a(n-1)^a(n-2): a(n-1) ? phi is asymptotic to a(n)
Since exponential isn't commutative, I've chosen the option that grows more slowly to get more values. What operation can replace the "?" to produce a true "asymptotic to" equation?
On
With regard to the multiplicative Fibonacci, I find that there is no limit (i.e., independent of $n$) for $\lim_{n\to \infty}\frac{a_n}{a_{n-1}}$. In fact,
$$ \lim_{n\to \infty}\frac{a_n}{a_{n-1}}=a_1^{\varphi^{n-2}/\sqrt{5}} $$
Here is the derivation. First note, that we cannot use $a_0=0$, but we can take $a_0=1$. Then we must have $a_1>1$, lest we have a trivial solution. So with that in mind, let $f_n=\ln(a_n)$, so that we recover the classic Fibonacci recurrence, to wit
$$ f_n=f_{n-1}+f_{n-2};\quad f_0=\ln(1)=0, f_1=\ln(a_1) $$
The general solution to which is
$$ f_n=f_1\frac{\varphi^n-\psi^n}{\varphi-\psi} $$
where $\varphi,\psi=(1\pm\sqrt{5})/2$.
Then, we can construct the solution for $a_n$ as follows
$$ \begin{align} a_n&=e^{f_n}\\ &=e^{f_1\frac{\varphi^n-\psi^n}{\varphi-\psi}}\\ &=e^{\ln(a_1)\frac{\varphi^n-\psi^n}{\varphi-\psi}}\\ &=e^{\ln(a_1)^{\frac{\varphi^n-\psi^n}{\varphi-\psi}}}\\ &=a_1^{\frac{\varphi^n-\psi^n}{\varphi-\psi}}\\ \end{align} $$
We can now examine the limiting ratio. Noting that $\varphi^n$ and $\psi^n$ are increasing and decreasing, respectively, with $n$, we get
$$ \begin{align} \lim_{n\to \infty}\frac{a_n}{a_{n-1}}&=a_1^{\frac{\varphi^n-\varphi^{n-1}}{\sqrt(5}}\\ &=a_1^{\frac{\varphi^{n-1}(\varphi-1)}{\sqrt(5}}\\ &=a_1^{\frac{\varphi^{n-1}/\varphi}{\sqrt(5}}\\ &=a_1^{\varphi^{n-2}/\sqrt{5}}\\ \end{align} $$
I have verified this result (to the extent possible with such a rapidly growing sequence) for random real and complex values of $a_1$.
EDIT: I've completed a more complete solution by relaxing the condition $a_0=1$. In that case, the general solution for $f_n$ is as follows,
$$ f_n=\bigg(f_1-\frac{f_0}{2}\bigg)\frac{\varphi^n-\psi^n}{\varphi-\psi}+\frac{f_0}{2}\frac{\varphi^n+\psi^n}{\varphi+\psi} $$
The limiting ratio can then be shown to be
$$ \lim_{n\to \infty}\frac{a_n}{a_{n-1}}=\bigg(\frac{a_1}{a_0^\psi}\bigg)^{\varphi^{n-2}/\sqrt{5}}\\ $$
As previously, the results were tested with positive and negative complex values for both $a_0$ and $a_1$.
For the first few steps of the "exponential Fibonacci" sequence defined in this way, it is reasonable to write $a(n)$ in the form $2^{2^{b(n)}}$ for some $b(n)$. Then we have $$ 2^{2^{b(n)}} = (2^{2^{b(n-1)}})^{2^{2^{b(n-2)}}} = 2^{2^{b(n-1)} \cdot 2^{2^{b(n-2)}}} = 2^{2^{b(n-1) + 2^{b(n-2)}}} $$ and so we can directly compute $b(n) = b(n-1) + 2^{b(n-2)}$. (This is sequence A000643 in the OEIS.) This lets us find the first few terms of $a(n)$: they are $$ 2^{2^0}, 2^{2^0}, 2^{2^1}, 2^{2^2}, 2^{2^4}, 2^{2^8}, 2^{2^{24}}, 2^{2^{280}}, \dots $$ After this point, though, chaos ensues. We got reasonably tame numbers like $280$ (which was $24 + 2^8$) because at this point, $b(n-1)$ and $2^{b(n-2)}$ are reasonably similar in magnitude. This stops being true from here on out. The next term of the sequence will be $$2^{2^{280 + 2^{24}}}$$ and the sum $280 + 2^{24}$ is basically the same as $2^{24}$ by itself. (It is approximately $2^{24.00002}$.) The next term is $2^{2^{280 + 2^{24} + 2^{280}}}$ and, again, $2^{280}$ dominates over $280$ and $2^{24}$, even more strongly.
We can take the approximation $b(n) \approx 2^{b(n-2)}$ from this point onward, and we'll get the following easy-to-describe estimates for the rest of the exponential Fibonacci sequence: $$ 2^{2^{2^{24}}}, 2^{2^{2^{280}}}, 2^{2^{2^{2^{24}}}}, 2^{2^{2^{2^{280}}}}, 2^{2^{2^{2^{2^{24}}}}}, 2^{2^{2^{2^{2^{280}}}}}, \dots $$ In terms of relative error, this is going to be a very bad approximation. But it's a very good approximation in the following sense: the exact value of the sequence can be written in the same form, but with $24$ or $280$ replaced by a value like $24 + \epsilon$ or $280+\epsilon$, with the error terms positive but very very small. (By far the largest contribution to the error term happens at the very beginning; so the $24$-like numbers, for instance, will never exceed $24.00003$.)
So the approximation we can give here (which I hesitate to call an asymptotic approximation because the relative error no longer goes to $0$) is $a(n) \approx 2^{a(n-2)}$. There is no involvement of the golden ratio here, and there is no clean approximate relationship between $a(n)$ and $a(n-1)$.
(More precisely, if we wanted to have a relationship like $a(n) \approx f(a(n-1))$, the function $f$ would have to satisfy $f(f(x)) \approx 2^x$: it would be an approximate half-exponential function. There is no "reasonable" definition of such a thing: nothing we can do with familiar algebraic operations.)
I should add: it might seem silly that we've taken a recurrence like $a(n) = a(n-1)^{a(n-2)}$, and approximated it by $a(n) \approx 2^{a(n-2)}$. Surely, these things couldn't be more different - $a(n-1)$ is ridiculously bigger than $2$!
The explanation is that at the scale that the terms of the sequence reach, the difference in the base of the exponent is negligible. If you compare two numbers like $$2^{2^{2^{2^{2^{100}}}}} \quad \text{and} \quad 2^{2^{2^{2^{2^{101}}}}},$$ the difference between them because of a small change in the highest exponent will be much larger than the effect of changing the very lowest $2$ even to $1\,000\,000$.