Well after seeing your MO post I finally understood what you were getting at. Let me show you here how your function is in fact the standard Schroder iteration of exponentiation. I am talking specifically about this function
\( F(z) = \sum_{n=0}^\infty \sum_{m=0}^n (-1)^{n-m}\binom{z}{n}_q\binom{n}{m}_qq^{\binom{n-m}{2}} (^m a) \)
where \( 1 < a < e^{1/e} \) and \( q = \frac{d}{dx}|_{x=\omega} a^x \) where \( \omega =-W(\log(a))/\log(a) \)
Recall that
\( \exp_a^{\circ z}(x) = \Psi^{-1}(q^z\Psi(x)) \)
for \( 1 < a < e^{1/e} \), for \( x \) in a neighborhood of \( \omega =-W(\log(a))/\log(a) \). This can be extended to \( x = 1 \) so that we have a solution \( ^z a \) such that it is periodic with imaginary period and tends to \( \omega \) as the real argument of z grows.
Functions like this are subject to Ramanujan's master theorem, something I wrote on the MO post. This essentially means we can take a contour integral
\( f(x) =\frac{1}{2\pi i}\int_{\sigma - i \infty}^{\sigma + i \infty} \Gamma(z)(^{1-z}a)x^{-z}\,dz \)
which converges and due to mellin's inversion theorem satisfies
\( \int_0^\infty f(x)x^{z-1}\,dx =\Gamma(z)(^{1-z}a) \)
Now, using contour integration, the function f is equivalent to
\( f(x) =\sum_{n=0}^\infty\,Res_{z=-n}(\Gamma(z)(^{1-z}a)) \)
which is quite beautifully
\( f(x) =\sum_{n=0}^\infty (^{n+1} a) \frac{(-x)^n}{n!} \)
So therefore
\( \int_0^\infty f(x)x^{-z}\,dx =\Gamma(1-z)(^za) \)
Even better, this integral can be analytically continued by breaking \( \int_0^\infty = \int_0^1 + \int_1^\infty \) so that
\( (1)\,\,\,\,\,\Gamma(1-z)(^z a) =\sum_{n=0}^\infty (^{n+1} a)\frac{(-1)^n}{n!(n+1-z)} + \int_1^\infty f(x)x^{-z}\,dx \)
Now this solution is the sole solution to the following system of equations
\( F(n) = (^n a) \) for all \( n\in\mathbb{N} \)
\( F(z) = O(e^{\rho|\Re(z)| + \tau|\Im(z)|}) \) when \( \Re(z) > 0 \) and \( \rho,\tau \in \mathbb{R}^+ \) with the additional restriction that \( \tau < \pi/2 \)
This follows because if we had a second function it would be subject to all of the above steps and in the final expression (1) defining it it would necessarily equal the original function.
Your solution \( F \) is first of all periodic with imaginary period. This follows because \( \binom{z}{n}_q \) is periodic in \( z \), and your expression is an infinite sum of these terms. So it is bounded as \( \Im(z) \) grows.
Secondly your solution tends to \( -W(\log(a))/\log(a) \) as the real part of z grows, because \( \binom{z}{n}_q \to C(q) \) as \( \Re(z) \to \infty \), leaving only a constant in your series. Therefore your solution is bounded when \( \Re(z) > 0 \), and secondly interpolates the natural iterates, \( F(n) = (^na) \). Therefore your function is indeed the standard tetration function that I laid out above.
Quite frankly I am amazed by this equation you've put in front of me. In case you didn't know, I think this should in fact work on a much larger scale and produces a new way of iterating functions.
If \( \phi(\xi):G \to G \) and \( \phi(\xi_0) = \xi_0 \) and \( 0 < q =\phi'(\xi_0) < 1 \), then your expression works to find \( \phi^{\circ z}(\xi):I \to I \) for \( \Re(z) > 0 \) where \( \phi^{\circ z_0}(\phi^{\circ z_1}(\xi)) = \phi^{\circ z_0 + z_1}(\xi) \) , and \( I \) is the immediate basin about \( \xi_0 \) (largest connected domain about \( \xi_0 \) where \( \lim_{n\to\infty}\phi^{\circ n}(\xi) = \xi_0 \)).
IF
\( F(z,\xi) = \sum_{n=0}^\infty \sum_{m=0}^n (-1)^{n-m}\binom{z}{n}_q \binom{n}{m}_q q^{\binom{n}{2}} \phi^{\circ m}(\xi) \)
and
\( F(n,\xi) = \phi^{\circ n}(\xi) \)
then
\( F(z,\xi) = \phi^{\circ z}(\xi) \)
This follows exactly how your tetration function is shown to be tetration! The big question, is how do you show \( F(n,\xi) = \phi^{\circ n}(\xi) \)?
What a great formula!
\( F(z) = \sum_{n=0}^\infty \sum_{m=0}^n (-1)^{n-m}\binom{z}{n}_q\binom{n}{m}_qq^{\binom{n-m}{2}} (^m a) \)
where \( 1 < a < e^{1/e} \) and \( q = \frac{d}{dx}|_{x=\omega} a^x \) where \( \omega =-W(\log(a))/\log(a) \)
Recall that
\( \exp_a^{\circ z}(x) = \Psi^{-1}(q^z\Psi(x)) \)
for \( 1 < a < e^{1/e} \), for \( x \) in a neighborhood of \( \omega =-W(\log(a))/\log(a) \). This can be extended to \( x = 1 \) so that we have a solution \( ^z a \) such that it is periodic with imaginary period and tends to \( \omega \) as the real argument of z grows.
Functions like this are subject to Ramanujan's master theorem, something I wrote on the MO post. This essentially means we can take a contour integral
\( f(x) =\frac{1}{2\pi i}\int_{\sigma - i \infty}^{\sigma + i \infty} \Gamma(z)(^{1-z}a)x^{-z}\,dz \)
which converges and due to mellin's inversion theorem satisfies
\( \int_0^\infty f(x)x^{z-1}\,dx =\Gamma(z)(^{1-z}a) \)
Now, using contour integration, the function f is equivalent to
\( f(x) =\sum_{n=0}^\infty\,Res_{z=-n}(\Gamma(z)(^{1-z}a)) \)
which is quite beautifully
\( f(x) =\sum_{n=0}^\infty (^{n+1} a) \frac{(-x)^n}{n!} \)
So therefore
\( \int_0^\infty f(x)x^{-z}\,dx =\Gamma(1-z)(^za) \)
Even better, this integral can be analytically continued by breaking \( \int_0^\infty = \int_0^1 + \int_1^\infty \) so that
\( (1)\,\,\,\,\,\Gamma(1-z)(^z a) =\sum_{n=0}^\infty (^{n+1} a)\frac{(-1)^n}{n!(n+1-z)} + \int_1^\infty f(x)x^{-z}\,dx \)
Now this solution is the sole solution to the following system of equations
\( F(n) = (^n a) \) for all \( n\in\mathbb{N} \)
\( F(z) = O(e^{\rho|\Re(z)| + \tau|\Im(z)|}) \) when \( \Re(z) > 0 \) and \( \rho,\tau \in \mathbb{R}^+ \) with the additional restriction that \( \tau < \pi/2 \)
This follows because if we had a second function it would be subject to all of the above steps and in the final expression (1) defining it it would necessarily equal the original function.
Your solution \( F \) is first of all periodic with imaginary period. This follows because \( \binom{z}{n}_q \) is periodic in \( z \), and your expression is an infinite sum of these terms. So it is bounded as \( \Im(z) \) grows.
Secondly your solution tends to \( -W(\log(a))/\log(a) \) as the real part of z grows, because \( \binom{z}{n}_q \to C(q) \) as \( \Re(z) \to \infty \), leaving only a constant in your series. Therefore your solution is bounded when \( \Re(z) > 0 \), and secondly interpolates the natural iterates, \( F(n) = (^na) \). Therefore your function is indeed the standard tetration function that I laid out above.
Quite frankly I am amazed by this equation you've put in front of me. In case you didn't know, I think this should in fact work on a much larger scale and produces a new way of iterating functions.
If \( \phi(\xi):G \to G \) and \( \phi(\xi_0) = \xi_0 \) and \( 0 < q =\phi'(\xi_0) < 1 \), then your expression works to find \( \phi^{\circ z}(\xi):I \to I \) for \( \Re(z) > 0 \) where \( \phi^{\circ z_0}(\phi^{\circ z_1}(\xi)) = \phi^{\circ z_0 + z_1}(\xi) \) , and \( I \) is the immediate basin about \( \xi_0 \) (largest connected domain about \( \xi_0 \) where \( \lim_{n\to\infty}\phi^{\circ n}(\xi) = \xi_0 \)).
IF
\( F(z,\xi) = \sum_{n=0}^\infty \sum_{m=0}^n (-1)^{n-m}\binom{z}{n}_q \binom{n}{m}_q q^{\binom{n}{2}} \phi^{\circ m}(\xi) \)
and
\( F(n,\xi) = \phi^{\circ n}(\xi) \)
then
\( F(z,\xi) = \phi^{\circ z}(\xi) \)
This follows exactly how your tetration function is shown to be tetration! The big question, is how do you show \( F(n,\xi) = \phi^{\circ n}(\xi) \)?
What a great formula!