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$$F'(x) \sim \lambda^x$$

because $a_1 = 1$. Since another solution is just $F(z+\theta(z))$ for $\theta$ periodic, we can most likely form a contradiction by using our knowledge of the uniqueness of the exponential function under these conditions.

$$F'(x) \sim C\lambda^x$$

Since another solution is just $F(z+\theta(z))$ for $\theta$ periodic, we can most likely form a contradiction by using our knowledge of the uniqueness of the exponential function under these conditions.

Not sure if this is an answer (I may be misreading you) but a good suggestion of how to show (1) and (2) (and perhaps an actual solution), and a good start for (3). Consider the Newton series formula for tetration. This series converges for all $\Re(z) > -2$, and is very similar to your formula

Consider the rather beautiful fact that your solution seems to be bounded in the right half plane. I have a hunch on to how to show this, but I'm not certain yet, I'll look closer at q-analogs and how to modify the proof of the Newton series expansion (I would love it if someone could find the q-analog Newton Series I'm talking about). Essentially since $(^za)$ is bounded in the right half plane, this forces its finite differences to have nice decay. I imagine the same thing occurs with $q$ finite differences--except I'm not certain. This would essentially be like expanding the standard solution in a Taylor series except we're expanding it with a q-Newton series.

Edit: If the OP is referring to the standard gaussian binomials, I can show its bounded. (BELOW)

Nonetheless, let us assume $t$ is bounded for the sake of discussion. The first obvious fact is $t(n) = (^n a)$. Next consider the wondrous power of a little known identity theorem for functions bounded in the right half plane. Namely, if $F(z)$ and $G(z)$ are bounded in the right half plane and $F\Big{|}_{\mathbb{N}} = G\Big{|}_{\mathbb{N}}$ then $F = G$. This follows because of Ramanujan's master theorem.

All in all, I mostly just had conjectures to add and weigh in my two cents, but I think all the conjectures you put are true and that your solution is in fact the 'standard' solution, which is: the periodic one, the bounded one, the Schroder one, and hopefully, the completely monotone one.

REALLY BIG EDIT:

Looking at your expansion I noticed your solution is trivially bounded if we are actually using Gaussian Binomial coefficients (as I'm reading them) and not how you write them (there seems to be an inconsistency), and therefore (1) and (2) are necessarily solved by the above.

Observe if

$[z]_q = \frac{1-q^z}{1-q}$, $[z]_q$ is periodic.

$${z \brack n}_q = \frac{[z]_q!}{[n]_q![z-n]_q!} = \frac{[z]_q[z-1]_q...[z-n]_q}{[1]_q[2]_q...[n]_q}$$

And each of these functions have a period, namely $2\pi i/\log(q)$..

Therefore your final expansion has an imaginary period, it tends to $-W(\log(a))/\log(a)$ as $\Re(z) \to \infty$;thus, it is bounded in the right half plane. By all of the above, your function is the standard tetration function (If it converges).

I have no further comment on the completely monotone part though.

Not sure if this is an answer but a good suggestion of how to show (1) and (2) (and perhaps an actual solution), and a good start for (3). Consider the Newton series formula for tetration. This series converges for all $\Re(z) > -2$, and is very similar to your formula

Consider the rather beautiful fact that your solution is bounded in the right half plane.

Observe if

$[z]_q = \frac{1-q^z}{1-q}$, $[z]_q$ is periodic in $z$.

$${z \brack n}_q = \frac{[z]_q!}{[n]_q![z-n]_q!} = \frac{[z]_q[z-1]_q...[z-n+1]_q}{[1]_q[2]_q...[n]_q}$$

And each of these functions have a period, namely $2\pi i/\log(q)$..

Therefore your final expansion has an imaginary period, it tends to $-W(\log(a))/\log(a)$ as $\Re(z) \to \infty$; thus, it is bounded in the right half plane.

From this it follows your function is the standard tetration function (If it converges).

So $t$ is bounded. The first obvious fact is $t(n) = (^n a)$. Next consider the wondrous power of a little known identity theorem for functions bounded in the right half plane. Namely, if $F(z)$ and $G(z)$ are bounded in the right half plane and $F\Big{|}_{\mathbb{N}} = G\Big{|}_{\mathbb{N}}$ then $F = G$. This follows because of Ramanujan's master theorem.

All in all, I mostly just had to weigh in my two cents, but I think all the conjectures you put are true and that your solution is in fact the 'standard' solution, which is: the periodic one, the bounded one, the Schroder one, and hopefully, the completely monotone one. Hopefully the rough proof I gave of (1) and (2) is good enough for you.

Now, since this solution is bounded in the half plane $\Re(z) > 0$ it does have a uniqueness criterion--quite a good one at that too. It is the only solution to the functional equation to do such. In fact, if a solution $F$ exists where $F(z) = O(e^{\rho|\Re(z)| + \tau|\Im(z)|})$ for $\tau < \pi/2$ then $F$ is still the above solution.

What you have presented is a q-analog of the Newton series I just put above. Sadly I cannot find a rigorous paper detailing these types of series; I don't remember where I have seen them, but I have seen them. This suggests to me, that really you are just interpolating $(^na)$ in the same manner the Newton series does, except you are using q-analogs.

Consider the rather beautiful fact that your solution seems to be bounded in the right half plane. I have a hunch on to how to show this, but I'm not certain yet, I'll look closer at q-analogs and how to modify the proof of the Newton series expansion (I would love it if someone could find the q-analog Newton Series I'm talking about). Essentially since $(^za)$ is bounded in the right half plane, this forces its finite differences to have nice decay. I imagine the same thing occurs with $q$ finite differences--except I'm not certain. This would essentially be like expanding the standard solution in a Taylor series except we're expanding it with a q-Newton series.

(The op claims this expression diverges, an alternate expression is

$$\frac{1}{\Gamma(1-z)}\Big{(}\sum_{n=0}^\infty (^{n+1}a)\frac{(-1)^n}{n!(n+1-z)} + \int_1^\infty \big{(}\sum_{n=0}^\infty (^{n+1}a) \frac{(-x)^n}{n!}\big{)}x^{-z}\,dx\Big{)}$$

of which the rest of this discussion still applies.)

Now, since this solution is bounded in the half plane $\Re(z) > 0$ it does have a uniqueness criterion--quite a good one at that too. It is the only solution to the functional equation to do such. In fact, if a solution $F$ exists where $F(z) = O(e^{\rho|\Re(z)| + \tau|\Im(z)|})$ for $\tau < \pi/2$ then $F$ is still the above solution.

What you have presented is a q-analog of the Newton series I just put above. Sadly I cannot find a rigorous paper detailing these types of series; I don't remember where I have seen them, but I have seen them. This suggests to me, that really you are just interpolating $(^na)$ in the same manner the Newton series does, except you are using q-analogs. It is necessary though that $q$ be fixed to the value you put (explanation below).

Consider the rather beautiful fact that your solution seems to be bounded in the right half plane. I have a hunch on to how to show this, but I'm not certain yet, I'll look closer at q-analogs and how to modify the proof of the Newton series expansion (I would love it if someone could find the q-analog Newton Series I'm talking about). Essentially since $(^za)$ is bounded in the right half plane, this forces its finite differences to have nice decay. I imagine the same thing occurs with $q$ finite differences--except I'm not certain. This would essentially be like expanding the standard solution in a Taylor series except we're expanding it with a q-Newton series.

Edit: If the OP is referring to the standard gaussian binomials, I can show its bounded. (BELOW)