Hmm. We have

If we assume a solution of the form

exists, like in the Fibonacci numbers, we can get the equation

Dividing both sides by

gives

or

.

There is a caveat, however:

is multivalued. It is more useful, then, to recast this equation in terms of

,

and then the solutions for the functional equation are given by

for any

-value satisfying the above exponential equation. The functional equation is linear, so any linear combination of such solutions will be another solution, and since there are infinitely many such

-values, we can even consider infinite sums

with arbitrary

, provided this sum converges. Since there are infinitely many constants

, one could say the equation is like it has "infinitely many initial conditions".

This graph shows the function

on the complex plane. You can see the roots, the values of

used to construct solutions of the functional equation. The scale runs between

on each axis (x = real, y = imag). One set of roots seems to lie along the line

, while the other seems to lie along a slight curve (curving not visible here) that is asymptotic to the imaginary axis

.

For example, we could take two terms with the roots given by

and

and coefficients

. This function is plotted below at the same scale. Numerical calculation can be done to verify it really does solve

. I do not believe there is a closed form solution for these

-values in terms of any conventional special functions, but I could be wrong (and if I am, I'd like to know what the closed solution is.).

Note that these may not be the only possible solutions -- remember that the very simple case

has all 1-periodic functions as solutions. Not sure what the appropriate analogy is here. But the above could be thought of as a sort of "canonical" solution like how Binet's formula solves the Fibonacci numbers.