fibonacci like

[tommy1729]

f(x) = f(x - 1) + f(x + i)

i = imaginary unit?

yes , i^2 = -1.

then its not a recursion, i.e. you dont have a unique solution for natural number arguments.
I dont think the usual methods apply.

i didnt claim unique solutions.

the usual method relates such equations with polynomials ;

f(x+a) is associated with x^a , but since x - x^-1 - x^i isnt a polynomial , its trivially not solvable by the usual methods , at least not the usual methods " alone ".

if it was solvable by the usual method , i would have done so and would not have posted it here.

why isnt it a recursion ???

im looking for solutions that are analytic almost everywhere.

tommy1729