I'm trying to figure out why $x_k\not\in\mathbb{N}$ apart from $x=1$ if $$x_k=\frac{\sum_{i=1}^k 2^{s_{i-1}}3^{k-i}}{2^{s_k}-3^k}\mbox{ where }s_n=\sum_{i=1}^ne_i\mbox{ and }e_n\in\mathbb{N}$$
Now, the above looks quite complicated, but just think of the set $s$ as being any set of ascending natural numbers, and $k$ is the length of set $s$. So, $x_k$ is actually a function with input $s$. $$x_k=f(s)=\frac{\sum_{i=1}^k 2^{s_{i-1}}3^{k-i}}{2^{s_k}-3^k}$$
The question is: why is 1 the only natural number $f(s)$ can be?
What I noticed is that the numerator is just a polynomial of degree $k-1$ and the denominator is a polynomial of degree $k$. Maybe that can be part of the argument? I tried something, and here it goes: Since the numerator and denominator are polynomials, we can factor them according to their degree, $$x_k=\frac{a(3-r_1)(3-r_2)...(3-r_{k-1})}{b(3-p_1)(3-p_2)...(3-p_{k-1})(3-p_k)}$$ Now, $a$ and $b$ are just the leading coefficients of the two polynomials, so they are just $1$ and $-1$ respectively. We can see that the denominator has always 1 more factor than the numerator, and in order for the answer to be natural we need the denominator to be a factor of the numerator. But this implies the numerator having all the factors of the denominator, and how can it if it has less? The only explanation I can see is that some of the denominator's factors are actually just $1$ or $-1$ (correct me here if I'm wrong).
Looking at the denominator, it actually has most of its roots non-real. Depending on the parity (even or odd) of $k$, the denominator has a maximum of two real roots: $\pm\sqrt[k]{2^{s_k}}$. $$x_k=\frac{(3-r_1)(3-r_2)...(3-r_{k-1})}{-(3-p_1)(3-p_2)...(3-p_{k-2})(3-\sqrt[k]{2^{s_k}})(3+\sqrt[k]{2^{s_k}})}\mbox{ if }k\equiv0\mod{2}$$
This implies that we can set $$3-\sqrt[k]{2^{s_k}}=\pm1$$ $$\sqrt[k]{2^{s_k}}=2\mbox{ or }4$$ $$\frac{s_k}{k}=1\mbox{ or }2$$ $$s_k=k\mbox{ or }2k$$
But $s_k=k$ implies $\sum_{i=1}^ke_i=k$ which forces all elements of the set $e$ to be 1, therefore all $s_n=n$ and pluggin this in, we get $$x_k=\frac{\sum_{i=1}^k2^{i-1}3^{k-i}}{2^k-3^k}$$ $$x_k=\frac{\sum_{i=1}^k2^{i-1}3^{k-i}}{(2-3)\sum_{i=1}^k2^{i-1}3^{k-i}}$$ $$x_k=-1$$
Also, the $s_k=2k$ condition gives $$x_k=\frac{\sum_{i=1}^k2^{s_{i-1}}3^{k-i}}{4^k-3^k}$$ $$x_k=\frac{\sum_{i=1}^k2^{s_{i-1}}3^{k-i}}{(4-3)\sum_{i=1}^k4^{i-1}3^{k-i}}$$ $$x_k=\frac{\sum_{i=1}^k2^{s_{i-1}}3^{k-i}}{\sum_{i=1}^k4^{i-1}3^{k-i}}$$
Now we're looking at the quotient of two polynomials of equal degree where one is a factor of the other, and they have the same leading coefficients, the only possible answer is 1 (right?). So that's where the 1 comes from.
So, the only thing missing here is how the non-real factors can work and provide ways for $x_k$ to be an integer, but then we still need to show that $x_k$ cannot be another natural number other than 1. Is the following valid? :
$$x_k=-\frac{(3-r_1)(3-r_2)...(3-r_{k-1})}{(3-p_1)(3-p_2)...(3-p_{k-1})(3-p_k)}$$ The numerator has all of its roots non-real, so the denominator must have all of those $k-1$ many non-real roots, and they must all cancel, leaving $$x_k=-\frac{1}{3-p_k}$$
We know that this implies that $k$ is odd, and that $p_k=\sqrt[k]{2^{s_k}}$. $$x_k=-\frac{1}{3-\sqrt[k]{2^{s_k}}}$$
But this implies what we already know. I know there's definitely a problem with my arguments, they are definitely missing something. I know this because if I plug in $s=\{1,3\}$ I get out $x_k=-5$, and if I plug in $s=\{1,2,3,5,6,7,11\}$ I get out $x_k=-17$. So there must be other ways for the non-real roots to cancel themselves, allowing for $x_k\in\mathbb{Z}$. But I can't see how, and if I can't see how I can't use this to show $x_k\not\in\mathbb{N}$ if $x\neq1$.