As I was thinking about the Collatz Conjecture, it occurred to me that it would be helpful to confirm that $(2^{i_k} - 3^k)u - 3^{k-1} - \sum\limits_{s=1}^{k-2}3^{k-1-s}2^{i_s}$ is not a power of $2$.
I thought that this would be straight forward to prove but I am unable to complete the argument. Is there a straight forward way to show this or is this more difficult than it appears?
Let:
- $i_1, i_2, \dots, i_k$ be $k >3$ integers where:
$$i_k > i_{k-1} > i_{k-2} > \dots > i_2 > i_1 > 0$$
- $u$ be the minimum integer where:
$$(2^{i_k} - 3^k)u > 3^{k-1} + \sum\limits_{s=1}^{k-2}3^{k-1-s}2^{i_s}$$
Intuitively, I would assume that $(2^{i_k} - 3^k)u - 3^{k-1} - \sum\limits_{s=1}^{k-2}3^{k-1-s}2^{i_s}$ cannot be a power of $2$.
It would seem to me that a proof by contradiction should work where the first 4 steps might be:
(1) Assume it is a power of $2$
(2) $u$ must be odd since the sum must be even so that there exists $v$ with $u=2v+1$
(3) $(2^{i_k} - 3^k)(2v+1) - 3^{k-1} - \sum\limits_{s=1}^{k-2}3^{k-1-s}2^{i_s} = 2v(2^{i_k} - 3^k) + 2^{i_k} - 3^{k-1}4 - \sum\limits_{s=1}^{k-2}3^{k-1-s}2^{i_s}$
(4) Dividing by $2$ we have:
$$v(2^{i_k} - 3^k) + 2^{i_k-1} - 3^{k-1}2 - \sum\limits_{s=1}^{k-2}3^{k-1-s}2^{i_s-1}$$
At this point, the approach gets more complicated. If $i_1 = 1$, then $v$ is even. if $i_1 > 1$, then $v$ is odd.
I can see going through this process of breaking down $v$ into either $2w$ or $2w+1$ but I suspect that there is more standard and simpler way to prove that this sum is not a power of $2$.