Is there a non-constant function $f$ such that $f'(x) = f(x - 1)$?

Question

In discrete calculus, where the difference operator $\Delta f = f(x + 1) - f(x)$ takes the place of $\frac{d}{dx}$, Fibonacci sequences are given by the functions satisfying:

$$ \Delta f(x) = f(x - 1) $$

Is there a non-constant function such that $\frac{d}{dx}f(x) = f(x - 1)$? If it exists, it would be the "continuous analogue of the Fibonacci sequence" in this sense, which seems cool.

Answer 1

Choose constant $C$ that satisfies $C=e^{-C}$. Note that $C=W(1)\approx 0.567$. We take

$$f(x)=e^{Cx}$$ $$f'(x)=Ce^{Cx}$$ $$f(x-1)=e^{C(x-1)}=e^{Cx}e^{-C}=Ce^{Cx}$$