This is another Collatz-related problem about trying to represent a number in a certain form. As is usually the case with the Collatz conjecture, this is probably not useful. My question is :
Can you construct $m,k, n \in \Bbb{Z^+}$ such that: $$\frac{2^n k -\sum_{i=0}^{n-1} 2^{n-1-i} \times 3^i}{3^n}= m<k$$
As I have said, this has some link to the Collatz conjecture as you may be able to tell. Any help with this would be much appreciated.
The sum on the left hand side can be simplified given that $$a^n-b^n=(a-b)\sum_{i=0}^{n-1}a^ib^{n-1-i}.$$ So the left hand side becomes $$\frac{2^nk-\sum_{i=0}^{n-1}2^{n-1-i}\times3^i}{3^n} =\frac{2^nk-(3^n-2^n)}{3^n} =2^n\frac{k+1}{3^n}-1.$$ This is an integer if and only if $k+1$ is a multiple of $3^n$, say $k=c3^n-1$. In this case we have $$m=c2^n-1<c3^n-1=k.$$