Take the straight forward Fibonacci equation


$$F\_0 = F\_1 = 1$$
$$F\_{n-2} + F\_{n-1} = F\_n$$


Let's consider a holomorphic function $F: \mathbb{C} \to \mathbb{C}$ such that


$$F(z)\Big{|}\_{\mathbb{N}} = F\_n$$
$$F(z-2) + F(z-1) = F(z)$$


Let's call such $F$, those that satisfy the Fibonacci equation in the complex plane. It is very easy to produce such functions. Taking the Binet identity


$$F\_n = \frac{\phi^{n} - \psi^n}{\phi - \psi}$$


where


$$\phi = \frac{1 + \sqrt{5}}{2}$$
$$\psi = \frac{1 - \sqrt{5}}{2}$$


It follows for any $k,j \in \mathbb{Z}$, using the standard branch of $\phi^z$ and $\psi^z$ that the functions


$$F\_{jk}(z) = \frac{e^{2\pi i j z}\phi^z -e^{2 \pi i k z}\psi^z}{\phi - \psi}$$ 


are a solution of the Fibonacci equation in the complex plane. These can't be all solutions though. Namely if $F$ and $G$ are solutions where we've simply chosen different $k$ and $j$ for each one, then


$$\frac{F}{2} + \frac{G}{2}$$ 


is equally a solution, which corresponds to no function from our list. This additionally implies that the infinite sum


$$\mathcal{F} = \sum\_{j=-\infty}^\infty \sum\_{k=-\infty}^\infty a\_{jk}F\_{jk}(z)$$


is a solution to the Fibonacci equation in the complex plane if it converges everywhere and


$$\sum\_{j=-\infty}^\infty \sum\_{k=-\infty}^\infty a\_{jk} = 1$$



> 
> Are these all of the solutions?
> 
> 
> 



How do we classify all possible extensions of the Fibonacci recursion to the complex plane?
Yes, these are all entire solutions. This follows from a general theorem which says that linear combinations of exponential solutions of such equations (linear homogeneous with constant coefficients) are dense
in the set of all solutions, see, for example Gelfond, Calculus of finite differences, Chap V, sect. 7. 


