Could be tetration if this integral converges

[mike3]

I see what you're trying to do by making \( F(n) = (^n e) \) but that won't work. If we try it this way we will get the following formula:

\( \frac{1}{(^{-s} e)} = \frac{1}{F(-s)\Gamma(s)} \int_0^\infty e^{-w}w^{s-1}\,dw = \frac{1}{F(-s)} \)

So that all we are doing is recovering \( F \) which doesn't satisfy the recursion. I'm not sure how we can create an \( F \) that satisfies the right conditions but it is necessary that it does not interpolate tetration. I realize I forgot to say we need the extra condition:

\( \int_0^\infty |\sum_{n=0}^\infty F(n)\frac{(-w)^n}{n!(^{n+1} e)}| w^{\sigma - 1} < \infty \) for \( 0 < \sigma < 1 \)

This ensures recursion. It also expresses that we need a more complicated choice for \( F(n) \). I really do believe our best hope is the second condition on the exponential boundedness of \( |\frac{F(-z)}{(^{-z} e)}| \)

However, \( F(n) = {^{n+1}} e \) would make this condition hold. So if that doesn't work, then there must be additional conditions beyond that.

Also, if it's not supposed to interpolate tetration, then what about just interpolating a sequence which grows as fast or faster than tetration but is not tetration? E.g. use the interpolation method I mentioned but with a different sequence, e.g. perhaps not tetration itself but something close to it, so that the function in the absolute value still behaves like an exponential or otherwise decays to 0. What about \( F(n) \) as I gave but with a small perturbation term like \( F(n) + 2^{-n} \), or something? Then it no longer interpolates tetration but the terms in the series will asymptotically approach those of the exponential function, so it should have similar asymptotic behavior, I'd think, in particular the decay to 0.

Also, with regards to the "bounded" approach, the given \( F(n) \) that I mentioned seemed like it might be a good candidate for that too -- in particular, it demonstrates strong growth as \( \Re(z) \rightarrow +\infty \), regardless of the value of \( \Im(z) \), so it might be able to cancel out tetration's (or reciprocal tetration's) growth. However, it has zeros in the right half-plane, which would mean dividing by it would produce nasty poles, i.e. the quotient would fail to be holomorphic in the right half-plane and be only meromorphic. So the question then becomes as to whether there's a function with such growth with no zeros in the right half-plane.

Does the final tetration result not depend on the choice of \( F \)?

YES it does not depend in every situation I've come across. By this I mean:

if \( \beta(w) = \sum_{n=0} \phi(n) \frac{w^n}{n!} \) and \( \alpha(w) = \sum_{n=0} \psi(n) \frac{w^n}{n!} \) and \( \gamma(w) = \sum_{n=0}^\infty \phi(n) \psi(n)\frac{w^n}{n!} \) \( |\phi(z)|,|\psi(z)|, |\phi(z)\psi(z)| < C e^{\alpha |\Im(z)| + \rho |\Re(z)|} \)
for \( 0 \le \alpha < \pi/2 \) and \( \rho \ge 0 \)

Then

\( \frac{d^z}{dw^z}|_{w=0} \beta(w) = \phi(z) \)
\( \frac{d^z}{dw^z}|_{w=0} \alpha(w) = \psi(z) \)
\( \frac{d^z}{dw^z}|_{w=0} \gamma(w) = \phi(z)\psi(z) \)

So there are some strong results on uniqueness and it preserves a fair amount of data.

Hmm.