Reconstructing analytic tetration with a complex height from a thinner set of points This is a follow-up to my previous question An explicit series representation for the analytic tetration with complex height.
Recall the definition $(11)$ from there:
$$t(z) = \sum_{n=0}^\infty \sum_{k=0}^n (-1)^{n-k} \, q^{\binom {n-k} 2} {z \brack n}_q {n \brack k}_q ({^k a}).\tag1$$
Let us introduce additional parameters $p,s\in\mathbb Z^+$ into this formula (changes are shown in red):
$$t(\color{red}p,\color{red}s,z) = \sum_{n=0}^\infty \sum_{k=0}^n (-1)^{n-k} \, q^{\,\color{red}p \, \binom {n-k} 2} {(z-\color{red}s)/\color{red}p \brack n}_{q^{\color{red}p}} {n \brack k}_{q^{\color{red}p}} ({^{\color{red}p\,k+\color{red}s} a}).\tag2$$
Basically, it means that we build our function from a thinner and shifted sample of tetration values.

Conjecture 1. Values of integer parameters $p,s$ do not matter, i.e. $$\forall p,s\in\mathbb Z^+, \, t(p,s,z) = t(z).$$

Let us generalize the definition of $t(p,s,z)$ to real-valued parameters $p,s\in\mathbb R^+$ (note that we switch to continuous tetration $t(z)$ in the last factor):
$$t_{\mathbb R}(p,s,z) = \sum_{n=0}^\infty \sum_{k=0}^n (-1)^{n-k} \, q^{\,p \, \binom {n-k} 2} {(z-s)/p \brack n}_{q^p} {n \brack k}_{q^p} \color{red}t(p\,k+s).\tag3$$

Conjecture 2. Values of real parameters $p,s$ do not matter, i.e. $$\forall p,s\in\mathbb R^+, \, t_{\mathbb R}(p,s,z) = t(z).$$

Can we prove these conjectures? If yes, it means that we do not need all values of the discrete tetration (represented by the factor $(^k a)$ in $(1)$) to build a formula for the analytic tetration, but a very thin (although still infinite) subset of them already contains enough information to exactly reconstruct its values at all points in between (including skipped integer points). Can we make a step further and find a way to construct an explicit series for the analytic tetration that relies only on a finite set of values of the discrete tetration?
 A: Well, this problem can be handled exactly as the last one was handled. Let me give you a rough reasoning as to why. I won't elaborate in detail as quite literally my last answer handles this case with little generalization.
Any bounded function $\phi(z)$ in the right half plane $\Re(z) > -\delta$ for some $\delta > 0$ can be written as
$$\phi(z) = \sum_{n=0}^\infty \dbinom{z}{n}\sum_{m=0}^n (-1)^{n-m}\dbinom{n}{m} \phi(m)$$
and $\phi(pz+s)$ for $p,s \in \mathbb{R}^+$ is equally so bounded, so that
$$\phi(pz+s) = \sum_{n=0}^\infty\dbinom{z}{n}\sum_{m=0}^n (-1)^{n-m}\dbinom{n}{m} \phi(pm+s)$$
The sampling data can be spread out, and shrunk with no effect when $\phi$ is bounded. (This result can be stated much more generally, but I'll stick with this version.) The essential point I'm making is that the sampling data doesn't matter, as long as its a linear function and it's evenly spaced.
Now what you are essentially doing is the same thing you did before, except you are working with $(^{zp+s}a)$ in stead of $(^z a)$. The sampling data will not matter because the function you are interpolating is bounded. Therefore watch carefully my reasoning
$$\phi(z) = \sum_{n=0}^\infty \sum_{k=0}^n (-1)^{n-k} \, q^{p \binom {n-k} 2} {z \brack n}_{q^{p}} {n \brack k}_{q^{p}} ({^{p\,k+s} a})$$
This function has imaginary period $2\pi i /\log(q)p$ (just like $^{zp+s}a$). This function tends to a constant as its real argument grows, therefore it is bounded.
Now, assuming that $\phi(n) = (^{pn+s}a)$ then $\phi(z) = (^{pz+s}a)$. I didn't see you state that it agrees on the naturals, but I'm assuming that is why you asked this. This result follows for the exact same reason as it did in your last question. The functions agree on the naturals $\phi\Big{|}_{\mathbb{N}} = (^{pz+s}a)\Big{|}_{\mathbb{N}}$ and both functions are bounded on the right half plane, therefore $\phi(z) = (^{pz+s}a)$.
