I noticed that the equation in my OP gets kind of wonky if you try to start with an odd-numbered value for n, so I thought it would be interesting if I could somehow equate the values. What I eventually ended up doing was starting with F(5) and F(7). I used the equation to put them in terms of F(-1/2) and F(1/2), just to see what would happen. I took the (2F(1/2) - F(1/2 - 3)) and rearranged it until I had F(n) = (7F(n/2 - 3) + 4F(n/2 - 4)) F(n/2). For this equation, I was planning to sub n = 7. I also got F(n) = (4F(n/2 - 2) + 3F(n/2 - 3)) F(n/2). For this one, I subbed in n = 5. Doing that actually makes the two F numbers equal to each other: F(1/2) and F(-1/2) respectively, but for simplicity, I just started calling them x and y. Now our equations look like F(7) = (7x + 4y)*F(7/2) and F(5) = (4x + 3y)F(5/2). I equated the remaining F into terms of x and y, resulting in F(7) = (7x + 4y)(3x + 2y) and F(5) = (4x + 3y)(2x + y). For any two n where n1 = n2 + 2, this should still work. Anyway, I may be taking the wrong approach here, but instead of leaving them factored, I multiplied them together, getting F(7) = 21x^2 + 26xy + 8y^2 and F(5) = 8x^2 + 10xy + 3y^2. Now that I’m here, I thought it would be good to sub in the actual values for the two F numbers, giving me 13 and 5 each. It’s interesting that I happened to choose F(5) and F(7) here, because the two middle values 26 and 10 are each 2 the values of the F numbers. Anyway, I wanted to reduce the quadratics into an equation for x in terms of y, so I divided the integer side of each equation by itself, giving me (21/13)x^2 + 2xy + (8/13)y^2 = 1 and (8/5)x^2 + 2xy + (3/5)y^2 = 1. Now that they are equal, I can negate the middle value 2xy and just have (21/13)x^2 + (8/13)y^2 = (8/5)x^2 + (3/5)y^2. I then cross multiplied to end up with 105x^2 + 40y^2 = 104x^2 + 39y^2. I subtracted the right side from the left to get x^2 + y^2 = 0. It’s here that I realized this is going to get into imaginary numbers. Anyway, I applied the quadratic formula and ended up with x = yi. Replacing x and y with their original values gives me F(1/2) = F(-1/2) * i. I’m not sure if anyone’s tried to calculate fractional Fibonacci numbers, but I thought it would be fun to give it a shot anyway. I did it again for F(3/2) and got F(3/2) = F(1/2) ± F(1/2) * i, but writing out the steps would be pretty long. I think the fractional Fibonacci numbers should follow the same rules that the integer numbers do, but I’m not sure. I was just messing around with the numbers, but I don’t know enough to understand what I’ve done or if it follows an existing rule. I’d like some help putting this into context, so any input would be welcome 
There is an explicit formula for Fibonacci numbers called Binet’s Fibonacci Number Formula.
One of the terms involves taking a power of a negative number, so if you plug fractions into it you get complex numbers. It’s not obvious (to me) that it has to work for fractions but it certainly seems to. For instance, using Wolfram Alpha we can compute:
F(0.5) = 0.568864 - 0.351578i
Also:
F(-0.5) = 0.351578 + 0.568864i
F(1.5) = 0.920442 + 0.217287i
Aside from an extra (probably a typo) minus sign, your derivation that:
F(3/2) = F(1/2) + F(1/2) * i
does seem to hold:
F(3/2) = (0.568864 - 0.351578i) + (0.568864 - 0.351578i)i =
(0.568864 - 0.351578i) + (0.351578 + 0.568864i) =
0.920442 + 0.217286i
F(1.5) = F(-0.5) + F(0.5) also holds here.
For what it’s worth, the proof I gave earlier extends to the fractional context: as noted, we can do arithmetic with pairs of numbers component-wise, and have a basis for this consisting of 1 = (1, 1) and m = (phi, psi), where phi is the Golden Ratio and psi is its conjugate.
We can take logarithms to produce log(m) = (log(phi), log(psi)), and using this, determine m[sup]n[/sup] = exp(n * log(m)) = (exp(n * log(phi)), exp(n * log(psi))) for arbitrary n. [For what it’s worth, in the complex context, logarithms are not generally unique; Wolfram Alpha follows the convention of choosing them with imaginary component as close to 0 as possible].
Once we’ve determined m[sup]n[/sup], this will amount to some linear combination of 1 and m, and we can extract these coefficients and consider them F(n - 1) and F(n) [this is what Binet’s formula does]. The recurrence relating F(n - 1) and F(n) to F(n + 1) follows just from the fact that m[sup]2[/sup] = m + 1, and the relationship between F(n - 1) and F(n) to F(2n) follows just from the expansion of (F(n - 1) + F(n) * m)[sup]2[/sup], as before.
This leaves an ambiguity for negative numbers: more to the point, Wolfram Alpha considers their logarithm to have minimal positive imaginary component, rather than negative.
It’s worth noting here that i and -i are completely interchangeable: If you take any true mathematical statement or valid derivation and swap i and -i everywhere either of them appears, you’ll get another true statement or valid derivation. In particular, if you start with a problem that’s purely real in its statement, and i ends up showing up in the solution, there’s another equally valid solution with -i instead.
That’s right. And in the same way, m and 1 - m are also completely interchangeable… if m is extraordinary, the only operations one is working with are rational functions with ordinary coefficients, and the only relation one is working with is equality. If you introduce other things, like a linear ordering or our part-extracter H, then they cease to be interchangeable, of course. But as far as purely field-theoretic statements go, 1.618… and -0.618… are completely interchangeable.
Similarly, i and -i cease to be interchangeable if you introduce the part-extracter Im or such things. Indistinguishability is relative to one’s toolkit.
Anyway, F(0.5) might just as well be taken to be -0.568864 - 0.351578i or -0.568864 + 0.351578i or 0.568864 + 0.351578i. This is related to the fact that numbers of the form a + b * m where a and b are complex (and m isn’t) typically have four square roots (the exceptions being along the lines where b/a is the Golden Ratio or its conjugate, where there are two square roots, and at zero, which has one square root). Carrying on making such choices for F(0.25), F(0.125), etc., we can construct multiple different exponential functions which pass through m at 1, giving us multiple different extensions of the Fibonacci sequence to fractional inputs and complex outputs, all satisfying the defining recurrence, the properties aforementioned in this thread, and all the other properties of that same sort.
Not that the particular extension calculated by Wolfram Alpha as above isn’t a perfectly reasonable convention to settle on. But no one’s actually forcing us to choose just one and ignore the rest!
I would argue that the Im function actually has an i hidden in its definition, and hence “swapping i and -i everywhere they occur” would also require swapping Im with a -Im function.
Fair enough. One can certainly treat the complex numbers within a background framework such that there is no ability to make any reference to any particular square root of -1 (rather, one could only ever introduce apparent such references via the quantifier “For any choice i of square root of -1, …”), and it is useful to keep this symmetry in mind. In the same way, for our “Fibonaccex” numbers, we should keep in mind that there’s no distinguishing between what is m and what is 1 - m, and correspondingly, our choice of m-part extracter H could just as well be replaced with its conjugate extracter -H.