(08/05/2022, 04:00 AM)Daniel Wrote: ....
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 \(n=-1\:\textrm{then}\:\{0,1,-1,0,2,-3,-2 \ldots\}\)
Not to sound wrong, how do we convert this into \(F(t)\) such that:
\[
F(t+2) = F(t+1) + F(t)\\
\]
Such that \(F(t)\) is real analytic (analytic for \(t\in \mathbb{R}\) and maps to \(\mathbb{R}\)). As much as I see where you're going, I'm confused about how you map it to a fibonacci solution. It seems close to what you are saying though.