12/30/2015, 09:49 AM
I believe I may have found a closed form for the power series of the third tetrate function as well.
I'm not sure if these are known, but I just used the elementary properties of binomials and Stirling numbers to derive these:
\(
\begin{equation}
{}^{3}x =
\sum_{k=0}^{\infty}
\log(x)^k
\sum_{j=0}^{k}
\sum_{i=0}^{k - j - 1}
\frac{(k - j - i)^j j^i}{(k - j - i)!j!i!}
\end{equation}
\)
\(
\begin{equation}
{}^{3}x =
\sum_{k=0}^{\infty}
(x - 1)^k
\sum_{j=0}^{k}
\sum_{J=0}^{j}
\sum_{i=0}^{k}
\sum_{I=0}^{i}
{\left[{i \atop I}\right]}
{\left[{j \atop J}\right]}
{\left({J \atop {k - j - i}}\right)}
\frac{J^I}{j!i!}
\end{equation}
\)
The first one (logarithmic power series) reminds me of something in one of Galidakis' papers about tetration, but I don't remember which paper. The second one is derived from the fact that the generating function of the signed Stirling numbers the first kind is \( (1 + x)^z \).
I'm not sure if these are known, but I just used the elementary properties of binomials and Stirling numbers to derive these:
\(
\begin{equation}
{}^{3}x =
\sum_{k=0}^{\infty}
\log(x)^k
\sum_{j=0}^{k}
\sum_{i=0}^{k - j - 1}
\frac{(k - j - i)^j j^i}{(k - j - i)!j!i!}
\end{equation}
\)
\(
\begin{equation}
{}^{3}x =
\sum_{k=0}^{\infty}
(x - 1)^k
\sum_{j=0}^{k}
\sum_{J=0}^{j}
\sum_{i=0}^{k}
\sum_{I=0}^{i}
{\left[{i \atop I}\right]}
{\left[{j \atop J}\right]}
{\left({J \atop {k - j - i}}\right)}
\frac{J^I}{j!i!}
\end{equation}
\)
The first one (logarithmic power series) reminds me of something in one of Galidakis' papers about tetration, but I don't remember which paper. The second one is derived from the fact that the generating function of the signed Stirling numbers the first kind is \( (1 + x)^z \).