Fibonacci as iteration of fractional linear function

[Daniel]

Is there a way to perform, let's say, "a crescent iteration" on the fibonacci sequence, so that we somehow map it to the reals?

We can always multiply by a 1-periodic function \(\theta(z)\), would it be possible to make \(\theta(z)F(z)\) real valued?

I've always wondered that, but I could never think of a solution; now seems as good a time to ask as any.

See Fibonacci almost to the bottom of the page for a real iteration of the Fibonacci series.

Could you elaborate, Daniel? Sorry, I'm not too sure what's going on here.

I understand you are writing:

\[
f(z) = \sum_{n=1}^\infty f_n \frac{z^n}{n!}\\
\]

Where now we are taking a parabolic iteration:

\[
f^{\circ t}(z)\\
\]

about \(0\), but how does this produce a fractional fibonacci that is real valued?

Not doubting you, just curious.

Sorry, I was on a combinatorial kick at the time. Due to the close connection between integer sequences and generating functions I displayed the iterated Fibonacci series in terms of an integer sequence. Let \(f(x)\) be the generating function for the Fibonacci series. The series associated with \(f^n(x)\) is then
\(\{0,1,n,n+n^2, -\frac{n}{2}+\frac{5n^2}{2}+n^3, -\frac{n}{3}+\frac{13n^3}{3}+n^4,\ldots\}\)

Fibonacci series
Let \(n=1\:\textrm{then}\:\{0,1,1,2,3,5,8 \ldots\}\)

OEIS A007440  Reversion of g.f. for Fibonacci numbers
Let \(n=-1\:\textrm{then}\:\{0,1,-1,0,2,-3,-2 \ldots\}\)