Hmm. We have
 = f(z - 1) + f(z + i))
If we assume a solution of the form
 = r^z)
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
)
,
u} - e^u = -1)
and then the solutions for the functional equation are given by
 = e^{uz})
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
 = \sum_{n=0}^{\infty} C_n e^{u_n z})
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
u} - e^u + 1)
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
.
[
attachment=717]
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
 = f(z-1) + f(z+i))
. 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.).
[
attachment=718]
Note that these may not be the only possible solutions -- remember that the very simple case
 = f(z-1))
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.