So, I have been working on a problem for a while. I am trying to see what sort of maximum and minimum values it can attain. I have for a while conjectured that it attains a maximum value of $\displaystyle \left(\frac{4}{3}\right)^{n+1}$. I want to see if my proof follows. If anyone is willing to take a look, I would be most appreciative.

Define $\displaystyle A_n = \left\{2^{a_n}(2^{a_{n-1}}(\cdots(2^{a_1}(4^{a_0}-1)-3)-\cdots)-3^{n-1})-3^n\in 3^{n+1}\mathbb{Z}>0\mid a_i \in \mathbb{N}\right\}$ (Note: the $\displaystyle a_i$'s are all positive integers, but for not all positive integers produce elements of $\displaystyle 3^{n+1}\mathbb{Z}$, so it doesn't span all positive integers).

Claim: $\displaystyle \displaymode \frac{\prod_{i=0}^n 2^{a_i}}{2^{a_n}(\cdots)-3^n}\le \left(\frac{4}{3}\right)^{n+1}$ where the denominator is an element of $\displaystyle A_n$.

Proof by induction on $\displaystyle n$.

For $\displaystyle n=0$, this is simply $\displaystyle \frac{2^{a_0}}{2^{a_0}-1}$. Since $\displaystyle 2^{a_0}-1\in 3\mathbb{Z}$, it must be that $\displaystyle a_0\equiv 0 (\text{mod } 2)$, so $\displaystyle \frac{4^{a_0}}{4^{a_0}-1}\le \frac{4^1}{4^1-1} = \frac{4}{3}$. Assume the claim is true for all sets up to $\displaystyle A_{n-1}$. Then, if

$\displaystyle \displaymode \frac{\prod_{i=0}^n 2^{a_i}}{2^{a_n}(\cdots)-3^n}\le k$, then

$\displaystyle \displaymode \log_2{\left(\frac{\prod_{i=0}^n 2^{a_i}}{2^{a_n}(\cdots)-3^n}\right)\le \log_2(k)$

$\displaystyle \displaymode \log_2{\prod_{i=0}^n 2^{a_i}}-\log_2{(2^{a_n}(\cdots)-3^n)}\le \log_2(k)$

$\displaystyle \displaymode \log_2{\prod_{i=0}^n 2^{a_i}}-\log_2{2^{a_n}(\cdots)}+\log_2{2^{a_n}(\cdots)}-\log_2{(2^{a_n}(\cdots)-3^n)}\le \log_2(k)$

$\displaystyle \displaymode \log_2{\frac{\prod_{i=0}^{n-1} 2^{a_i}}{(2^{a_{n-1}}(\cdots)-3^{n-1})}}+\log_2{\frac{2^{a_n}(\cdots)}{2^{a_n}(\cdots )-3^n}\le \log_2(k)$

We can apply the induction hypothesis to the first term, and for the second term, we know that the bottom is an element of $\displaystyle A_n$ with a minimum value of $\displaystyle 3^{n+1}$ (since every element of $\displaystyle A_n$ is an element of $\displaystyle 3^{n+1}\mathbb{Z}$. And the top is then $\displaystyle 3^{n+1}+3^n$. So, evaluating this, we get:

$\displaystyle \displaymode \log_2{\frac{\prod_{i=0}^{n-1} 2^{a_i}}{2^{a_{n-1}}(\cdots)-3^{n-1}}}+\log_2{\frac{2^{a_n}(\cdots)}{2^{a_n}(\cdots)-3^n}\le \log_2{\left(\frac{4}{3}\right)^n}+\log_2{\frac{3^ n(3+1)}{3^{n+1}}}=\log_2{\left(\frac{4}{3}\right)^ {n+1}$. Eliminating the logs from both sides (2 to the power of each side) yields the inequality desired, and hence by induction, proves the claim.

Is this proof correct? Am I overlooking something?