How do we classify all possible extensions of the Fibonacci recursion to the complex plane?

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?

• Wow -1 really fast, is there something obvious I'm missing here? – user78249 Jul 18 '17 at 9:02