Find a number $A$ so that
- (1) $\lfloor A^{3^n} \rfloor$ is always odd for $n\geq 1$;($\lfloor x \rfloor$ is the largest integer not greater than $x.$)
- (2) $A>1$ and $A^{3^n}$ is never an integer for $n\geq 1$;(This is the place I have edited.)
- (3) There is a closed-form formula for $A.$(Series and integral can also be seen as a closed form.)
Mills' constant satisfies (1) and (2) but not (3) (so far).
We can construct a number satisfies (1) and (2) more easily than Mills' constant:
Let $a_1=1,a_{n+1}=a_n^3+2~(n\geq1).$ Denote $$A=\lim_{n\to \infty}a_n^{\frac{1}{3^n}}.$$ Then we can prove that $\lfloor A^{3^n} \rfloor=a_n,$ hence $A>1$ and $a_n$ is odd. The proof is very like to this. But I cannot find a closed-form formula for $A$ again.
Thanks in advance !
For any integer $m \ge 2$, let $A = m + \sqrt{m^2-1}$ and define a sequence $(T_n)_{n\in\mathbb{N}}$ by:
$$T_n = A^n + A^{-n}$$
It is easy to see $T_0 = 2$, $T_1 = 2m$ and $T_{n+2} = 2mT_{n+1} - T_{n}$ for any $n \ge 0$. From this, we can conclude $T_n$ is an even integer for any $n \ge 0$.
It is also clear $A > 1$ and hence for all $n \ge 1$, we have:
$$\begin{align}0 < A^{-(3^n)} < 1 \implies & T_{3^n} - 1 < A^{3^n} = T_{3^n} - A^{-(3^n)} < T_{3^n}\\ \implies & \lfloor A^{3^n} \rfloor = T_{3^n} - 1 \text{ is an odd integer.} \end{align}$$