# 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.

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Nice question, I guess this is a variation of Mills' constant? Is the analogue of the Legendre conjecture proven for practical numbers? – Dan Brumleve Aug 23 '13 at 5:16
@Dan Brumleve Yes, it is. – Jaycob Coleman Aug 23 '13 at 5:19
If the answer to my second question is also yes, that there is always a practical number between consecutive squares, then it should follow. – Dan Brumleve Aug 23 '13 at 5:21

## 1 Answer

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.

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Every power of two is a practical number, which means that $A=2^{\frac{2^k}8}$ is such a constant for positive integers $k$. Does the Legendre's conjecture analogue guarantee the existence of additional constants, or could these be the only ones? – Jaycob Coleman Aug 23 '13 at 5:27
It seems very unlikely that these are exceptions because the set of double-powers is so much sparser than the set of squares. I need to do more research to know for sure. – Dan Brumleve Aug 23 '13 at 5:36
@JaycobColeman The fact that there are any non-power-of-two practical numbers means that there must be other ones. For instance, start with the range $18\lt A \lt 19$; we know there is a practical number $n_1$ with $18^2\lt n_1\lt 19^2$, and then another $n_2$ with $n_1^2\lt n_2\lt (n_1+1)^2$, etc. The Mills-style argument goes through with any starting value. – Steven Stadnicki Aug 23 '13 at 5:37
@StevenStadnicki Are you saying that there should be infinitely many constants (not including their double-powers), or further that every interval $[n^2,(n+1)^2]$ contains at least one constant? Or have I misunderstood altogether? Also, I left out the constants resulting from other powers of two in the previous comment, which was dumb. Clearly $A$ can be any power of two. – Jaycob Coleman Aug 24 '13 at 8:13