Does there exist a constant $\sqrt[4]{2} < A < \sqrt2$ such that $\lfloor A^{2^n} \rfloor$ is a practical number for all $n \in \Bbb Z^+$? Does there exist a constant  $\sqrt[4]{2} < A < \sqrt2$ such that $\lfloor A^{2^n} \rfloor$ is a practical number for all $n \in \Bbb Z^+$?
I know we can exclude the range $[\sqrt[8]5,\sqrt[8]6)$, since these all have $5$ as their third term.
I've done a poor job so far of explaining exactly what I'm looking for, so let me try again. Let's call a practical number $q$ a Mills' practical number if there exists a real number $A$ and a positive integer $k$ such that $\lfloor A^{2^k} \rfloor = q$ and $\lfloor A^{2^n} \rfloor$ is practical $\forall n \in \Bbb Z^+$. Now call $A$ a primitive constant if there exists no smaller constant generating every Mills' practical number generated by $A$. Since every power of two is a practical number, it's clear that there will be infinitely many primitive constants for the practical numbers and every power of two will be a Mills' practical number, even without the proven analogue to Legendre's conjecture (see answers). Can we show that there do or do not exist primitive constants other than those that generate the powers of two? I'm not sure that the restrictions in the original question are consistent with what I'm actually looking for as described in this paragraph. If you have corrections or a more proper statement of the problem I would appreciate your contribution. 

From http://en.wikipedia.org/wiki/Practical_number:

Hausman & Shapiro (1984) showed that there always exists a practical number in the interval $[x^2, (x + 1)^2]$ for any positive real $x$, a result analogous to Legendre's conjecture for primes.

This implies the existence of a base-$2$ analogue to Mills' constant for practical numbers.
